ceil | Python Methods and Functions

Allows you to determine the result of the math.ceil () function.

Allows you to determine the result of the math.ceil () function.

object .__ ceil __ (self)

self - A reference to the instance.

If this method is defined in a custom class, then it will be called when referring to math .ceil () .

` `` from math import ceil `

class MyClass:

def __init__ (self, val):

self.val = val

def __ceil __ (self):

return self.val + 1

my_var = MyClass (6)

ceil (my_var) # 7

Can someone explain this (straight from the docs- emphasis mine):

math.ceil(x)Return the ceiling of xas a float, the smallestintegervalue greater than or equal to x.

math.floor(x)Return the floor of xas a float, the largestintegervalue less than or equal to x.

Why would `.ceil`

and `.floor`

return floats when they are by definition supposed to calculate integers?

**EDIT:**

Well this got some very good arguments as to why they *should* return floats, and I was just getting used to the idea, when @jcollado pointed out that they in fact *do* return ints in Python 3...

I found out about the `//`

operator in Python which in Python 3 does division with floor.

Is there an operator which divides with ceil instead? (I know about the `/`

operator which in Python 3 does floating point division.)

Python has no built-in encryption schemes, no. You also should take encrypted data storage serious; trivial encryption schemes that one developer understands to be insecure and a toy scheme may well be mistaken for a secure scheme by a less experienced developer. If you encrypt, encrypt properly.

You don‚Äôt need to do much work to implement a proper encryption scheme however. First of all, *don‚Äôt re-invent the cryptography wheel*, use a trusted cryptography library to handle this for you. For Python 3, that trusted library is `cryptography`

.

I also recommend that encryption and decryption applies to *bytes*; encode text messages to bytes first; `stringvalue.encode()`

encodes to UTF8, easily reverted again using `bytesvalue.decode()`

.

Last but not least, when encrypting and decrypting, we talk about *keys*, not passwords. A key should not be human memorable, it is something you store in a secret location but machine readable, whereas a password often can be human-readable and memorised. You *can* derive a key from a password, with a little care.

But for a web application or process running in a cluster without human attention to keep running it, you want to use a key. Passwords are for when only an end-user needs access to the specific information. Even then, you usually secure the application with a password, then exchange encrypted information using a key, perhaps one attached to the user account.

The `cryptography`

library includes the Fernet recipe, a best-practices recipe for using cryptography. Fernet is an open standard,
with ready implementations in a wide range of programming languages and it packages AES CBC encryption for you with version information, a timestamp and an HMAC signature to prevent message tampering.

Fernet makes it very easy to encrypt and decrypt messages *and* keep you secure. It is the ideal method for encrypting data with a secret.

I recommend you use `Fernet.generate_key()`

to generate a secure key. You can use a password too (next section), but a full 32-byte secret key (16 bytes to encrypt with, plus another 16 for the signature) is going to be more secure than most passwords you could think of.

The key that Fernet generates is a `bytes`

object with URL and file safe base64 characters, so printable:

```
from cryptography.fernet import Fernet
key = Fernet.generate_key() # store in a secure location
print("Key:", key.decode())
```

To encrypt or decrypt messages, create a `Fernet()`

instance with the given key, and call the `Fernet.encrypt()`

or `Fernet.decrypt()`

, both the plaintext message to encrypt and the encrypted token are `bytes`

objects.

`encrypt()`

and `decrypt()`

functions would look like:

```
from cryptography.fernet import Fernet
def encrypt(message: bytes, key: bytes) -> bytes:
return Fernet(key).encrypt(message)
def decrypt(token: bytes, key: bytes) -> bytes:
return Fernet(key).decrypt(token)
```

Demo:

```
>>> key = Fernet.generate_key()
>>> print(key.decode())
GZWKEhHGNopxRdOHS4H4IyKhLQ8lwnyU7vRLrM3sebY=
>>> message = "John Doe"
>>> encrypt(message.encode(), key)
"gAAAAABciT3pFbbSihD_HZBZ8kqfAj94UhknamBuirZWKivWOukgKQ03qE2mcuvpuwCSuZ-X_Xkud0uWQLZ5e-aOwLC0Ccnepg=="
>>> token = _
>>> decrypt(token, key).decode()
"John Doe"
```

You can use a password instead of a secret key, provided you use a strong key derivation method. You do then have to include the salt and the HMAC iteration count in the message, so the encrypted value is not Fernet-compatible anymore without first separating salt, count and Fernet token:

```
import secrets
from base64 import urlsafe_b64encode as b64e, urlsafe_b64decode as b64d
from cryptography.fernet import Fernet
from cryptography.hazmat.backends import default_backend
from cryptography.hazmat.primitives import hashes
from cryptography.hazmat.primitives.kdf.pbkdf2 import PBKDF2HMAC
backend = default_backend()
iterations = 100_000
def _derive_key(password: bytes, salt: bytes, iterations: int = iterations) -> bytes:
"""Derive a secret key from a given password and salt"""
kdf = PBKDF2HMAC(
algorithm=hashes.SHA256(), length=32, salt=salt,
iterations=iterations, backend=backend)
return b64e(kdf.derive(password))
def password_encrypt(message: bytes, password: str, iterations: int = iterations) -> bytes:
salt = secrets.token_bytes(16)
key = _derive_key(password.encode(), salt, iterations)
return b64e(
b"%b%b%b" % (
salt,
iterations.to_bytes(4, "big"),
b64d(Fernet(key).encrypt(message)),
)
)
def password_decrypt(token: bytes, password: str) -> bytes:
decoded = b64d(token)
salt, iter, token = decoded[:16], decoded[16:20], b64e(decoded[20:])
iterations = int.from_bytes(iter, "big")
key = _derive_key(password.encode(), salt, iterations)
return Fernet(key).decrypt(token)
```

Demo:

```
>>> message = "John Doe"
>>> password = "mypass"
>>> password_encrypt(message.encode(), password)
b"9Ljs-w8IRM3XT1NDBbSBuQABhqCAAAAAAFyJdhiCPXms2vQHO7o81xZJn5r8_PAtro8Qpw48kdKrq4vt-551BCUbcErb_GyYRz8SVsu8hxTXvvKOn9QdewRGDfwx"
>>> token = _
>>> password_decrypt(token, password).decode()
"John Doe"
```

Including the salt in the output makes it possible to use a random salt value, which in turn ensures the encrypted output is guaranteed to be fully random regardless of password reuse or message repetition. Including the iteration count ensures that you can adjust for CPU performance increases over time without losing the ability to decrypt older messages.

A password alone *can* be as safe as a Fernet 32-byte random key, provided you generate a properly random password from a similar size pool. 32 bytes gives you 256 ^ 32 number of keys, so if you use an alphabet of 74 characters (26 upper, 26 lower, 10 digits and 12 possible symbols), then your password should be at least `math.ceil(math.log(256 ** 32, 74))`

== 42 characters long. However, a well-selected larger number of HMAC iterations can mitigate the lack of entropy somewhat as this makes it much more expensive for an attacker to brute force their way in.

Just know that choosing a shorter but still reasonably secure password won‚Äôt cripple this scheme, it just reduces the number of possible values a brute-force attacker would have to search through; make sure to pick a strong enough password for your security requirements.

An alternative is *not to encrypt*. Don"t be tempted to just use a low-security cipher, or a home-spun implementation of, say Vignere. There is no security in these approaches, but may give an inexperienced developer that is given the task to maintain your code in future the illusion of security, which is worse than no security at all.

If all you need is obscurity, just base64 the data; for URL-safe requirements, the `base64.urlsafe_b64encode()`

function is fine. Don"t use a password here, just encode and you are done. At most, add some compression (like `zlib`

):

```
import zlib
from base64 import urlsafe_b64encode as b64e, urlsafe_b64decode as b64d
def obscure(data: bytes) -> bytes:
return b64e(zlib.compress(data, 9))
def unobscure(obscured: bytes) -> bytes:
return zlib.decompress(b64d(obscured))
```

This turns `b"Hello world!"`

into `b"eNrzSM3JyVcozy_KSVEEAB0JBF4="`

.

If all you need is a way to make sure that the data can be trusted to be *unaltered* after having been sent to an untrusted client and received back, then you want to sign the data, you can use the `hmac`

library for this with SHA1 (still considered secure for HMAC signing) or better:

```
import hmac
import hashlib
def sign(data: bytes, key: bytes, algorithm=hashlib.sha256) -> bytes:
assert len(key) >= algorithm().digest_size, (
"Key must be at least as long as the digest size of the "
"hashing algorithm"
)
return hmac.new(key, data, algorithm).digest()
def verify(signature: bytes, data: bytes, key: bytes, algorithm=hashlib.sha256) -> bytes:
expected = sign(data, key, algorithm)
return hmac.compare_digest(expected, signature)
```

Use this to sign data, then attach the signature with the data and send that to the client. When you receive the data back, split data and signature and verify. I"ve set the default algorithm to SHA256, so you"ll need a 32-byte key:

```
key = secrets.token_bytes(32)
```

You may want to look at the `itsdangerous`

library, which packages this all up with serialisation and de-serialisation in various formats.

Fernet builds on AEC-CBC with a HMAC signature to ensure integrity of the encrypted data; a malicious attacker can"t feed your system nonsense data to keep your service busy running in circles with bad input, because the ciphertext is signed.

The Galois / Counter mode block cipher produces ciphertext and a *tag* to serve the same purpose, so can be used to serve the same purposes. The downside is that unlike Fernet there is no easy-to-use one-size-fits-all recipe to reuse on other platforms. AES-GCM also doesn"t use padding, so this encryption ciphertext matches the length of the input message (whereas Fernet / AES-CBC encrypts messages to blocks of fixed length, obscuring the message length somewhat).

AES256-GCM takes the usual 32 byte secret as a key:

```
key = secrets.token_bytes(32)
```

then use

```
import binascii, time
from base64 import urlsafe_b64encode as b64e, urlsafe_b64decode as b64d
from cryptography.hazmat.primitives.ciphers import Cipher, algorithms, modes
from cryptography.hazmat.backends import default_backend
from cryptography.exceptions import InvalidTag
backend = default_backend()
def aes_gcm_encrypt(message: bytes, key: bytes) -> bytes:
current_time = int(time.time()).to_bytes(8, "big")
algorithm = algorithms.AES(key)
iv = secrets.token_bytes(algorithm.block_size // 8)
cipher = Cipher(algorithm, modes.GCM(iv), backend=backend)
encryptor = cipher.encryptor()
encryptor.authenticate_additional_data(current_time)
ciphertext = encryptor.update(message) + encryptor.finalize()
return b64e(current_time + iv + ciphertext + encryptor.tag)
def aes_gcm_decrypt(token: bytes, key: bytes, ttl=None) -> bytes:
algorithm = algorithms.AES(key)
try:
data = b64d(token)
except (TypeError, binascii.Error):
raise InvalidToken
timestamp, iv, tag = data[:8], data[8:algorithm.block_size // 8 + 8], data[-16:]
if ttl is not None:
current_time = int(time.time())
time_encrypted, = int.from_bytes(data[:8], "big")
if time_encrypted + ttl < current_time or current_time + 60 < time_encrypted:
# too old or created well before our current time + 1 h to account for clock skew
raise InvalidToken
cipher = Cipher(algorithm, modes.GCM(iv, tag), backend=backend)
decryptor = cipher.decryptor()
decryptor.authenticate_additional_data(timestamp)
ciphertext = data[8 + len(iv):-16]
return decryptor.update(ciphertext) + decryptor.finalize()
```

I"ve included a timestamp to support the same time-to-live use-cases that Fernet supports.

This is the approach that All –Ü—ï V–∞–∏—ñ—Çy follows, albeit incorrectly. This is the `cryptography`

version, but note that I *include the IV in the ciphertext*, it should not be stored as a global (reusing an IV weakens the security of the key, and storing it as a module global means it"ll be re-generated the next Python invocation, rendering all ciphertext undecryptable):

```
import secrets
from base64 import urlsafe_b64encode as b64e, urlsafe_b64decode as b64d
from cryptography.hazmat.primitives.ciphers import Cipher, algorithms, modes
from cryptography.hazmat.backends import default_backend
backend = default_backend()
def aes_cfb_encrypt(message, key):
algorithm = algorithms.AES(key)
iv = secrets.token_bytes(algorithm.block_size // 8)
cipher = Cipher(algorithm, modes.CFB(iv), backend=backend)
encryptor = cipher.encryptor()
ciphertext = encryptor.update(message) + encryptor.finalize()
return b64e(iv + ciphertext)
def aes_cfb_decrypt(ciphertext, key):
iv_ciphertext = b64d(ciphertext)
algorithm = algorithms.AES(key)
size = algorithm.block_size // 8
iv, encrypted = iv_ciphertext[:size], iv_ciphertext[size:]
cipher = Cipher(algorithm, modes.CFB(iv), backend=backend)
decryptor = cipher.decryptor()
return decryptor.update(encrypted) + decryptor.finalize()
```

This lacks the added armoring of an HMAC signature and there is no timestamp; you‚Äôd have to add those yourself.

The above also illustrates how easy it is to combine basic cryptography building blocks incorrectly; All –Ü—ï V–∞–∏—ñ—Çy‚Äòs incorrect handling of the IV value can lead to a data breach or all encrypted messages being unreadable because the IV is lost. Using Fernet instead protects you from such mistakes.

If you previously implemented AES ECB encryption and need to still support this in Python 3, you can do so still with `cryptography`

too. **The same caveats apply, ECB is not secure enough for real-life applications**. Re-implementing that answer for Python 3, adding automatic handling of padding:

```
from base64 import urlsafe_b64encode as b64e, urlsafe_b64decode as b64d
from cryptography.hazmat.primitives.ciphers import Cipher, algorithms, modes
from cryptography.hazmat.primitives import padding
from cryptography.hazmat.backends import default_backend
backend = default_backend()
def aes_ecb_encrypt(message, key):
cipher = Cipher(algorithms.AES(key), modes.ECB(), backend=backend)
encryptor = cipher.encryptor()
padder = padding.PKCS7(cipher.algorithm.block_size).padder()
padded = padder.update(msg_text.encode()) + padder.finalize()
return b64e(encryptor.update(padded) + encryptor.finalize())
def aes_ecb_decrypt(ciphertext, key):
cipher = Cipher(algorithms.AES(key), modes.ECB(), backend=backend)
decryptor = cipher.decryptor()
unpadder = padding.PKCS7(cipher.algorithm.block_size).unpadder()
padded = decryptor.update(b64d(ciphertext)) + decryptor.finalize()
return unpadder.update(padded) + unpadder.finalize()
```

Again, this lacks the HMAC signature, and you shouldn‚Äôt use ECB anyway. The above is there merely to illustrate that `cryptography`

can handle the common cryptographic building blocks, even the ones you shouldn‚Äôt actually use.

The TensorFlow Convolution example gives an overview about the difference between `SAME`

and `VALID`

:

For the

`SAME`

padding, the output height and width are computed as:`out_height = ceil(float(in_height) / float(strides[1])) out_width = ceil(float(in_width) / float(strides[2]))`

And

For the

`VALID`

padding, the output height and width are computed as:`out_height = ceil(float(in_height - filter_height + 1) / float(strides[1])) out_width = ceil(float(in_width - filter_width + 1) / float(strides[2]))`

Most answers suggested `round`

or `format`

. `round`

sometimes rounds up, and in my case I needed the *value* of my variable to be rounded down and not just displayed as such.

```
round(2.357, 2) # -> 2.36
```

I found the answer here: How do I round a floating point number up to a certain decimal place?

```
import math
v = 2.357
print(math.ceil(v*100)/100) # -> 2.36
print(math.floor(v*100)/100) # -> 2.35
```

or:

```
from math import floor, ceil
def roundDown(n, d=8):
d = int("1" + ("0" * d))
return floor(n * d) / d
def roundUp(n, d=8):
d = int("1" + ("0" * d))
return ceil(n * d) / d
```

The ceil (ceiling) function:

```
import math
print(int(math.ceil(4.2)))
```

The difference between `import module`

and `from module import foo`

is mainly subjective. Pick the one you like best and be consistent in your use of it. Here are some points to help you decide.

`import module`

**Pros:**- Less maintenance of your
`import`

statements. Don"t need to add any additional imports to start using another item from the module

- Less maintenance of your
**Cons:**- Typing
`module.foo`

in your code can be tedious and redundant (tedium can be minimized by using`import module as mo`

then typing`mo.foo`

)

- Typing

`from module import foo`

**Pros:**- Less typing to use
`foo`

- More control over which items of a module can be accessed

- Less typing to use
**Cons:**- To use a new item from the module you have to update your
`import`

statement - You lose context about
`foo`

. For example, it"s less clear what`ceil()`

does compared to`math.ceil()`

- To use a new item from the module you have to update your

Either method is acceptable, but **don"t** use `from module import *`

.

For any reasonable large set of code, if you `import *`

you will likely be cementing it into the module, unable to be removed. This is because it is difficult to determine what items used in the code are coming from "module", making it easy to get to the point where you think you don"t use the `import`

any more but it"s extremely difficult to be sure.

**Warning:** `timeit`

results may vary due to differences in hardware or
version of Python.

Below is a script which compares a number of implementations:

- ambi_sieve_plain,
- rwh_primes,
- rwh_primes1,
- rwh_primes2,
- sieveOfAtkin,
- sieveOfEratosthenes,
- sundaram3,
- sieve_wheel_30,
- ambi_sieve (requires numpy)
- primesfrom3to (requires numpy)
- primesfrom2to (requires numpy)

Many thanks to stephan for bringing sieve_wheel_30 to my attention. Credit goes to Robert William Hanks for primesfrom2to, primesfrom3to, rwh_primes, rwh_primes1, and rwh_primes2.

Of the plain Python methods tested, **with psyco**, for n=1000000,
**rwh_primes1** was the fastest tested.

```
+---------------------+-------+
| Method | ms |
+---------------------+-------+
| rwh_primes1 | 43.0 |
| sieveOfAtkin | 46.4 |
| rwh_primes | 57.4 |
| sieve_wheel_30 | 63.0 |
| rwh_primes2 | 67.8 |
| sieveOfEratosthenes | 147.0 |
| ambi_sieve_plain | 152.0 |
| sundaram3 | 194.0 |
+---------------------+-------+
```

Of the plain Python methods tested, **without psyco**, for n=1000000,
**rwh_primes2** was the fastest.

```
+---------------------+-------+
| Method | ms |
+---------------------+-------+
| rwh_primes2 | 68.1 |
| rwh_primes1 | 93.7 |
| rwh_primes | 94.6 |
| sieve_wheel_30 | 97.4 |
| sieveOfEratosthenes | 178.0 |
| ambi_sieve_plain | 286.0 |
| sieveOfAtkin | 314.0 |
| sundaram3 | 416.0 |
+---------------------+-------+
```

Of all the methods tested, *allowing numpy*, for n=1000000,
**primesfrom2to** was the fastest tested.

```
+---------------------+-------+
| Method | ms |
+---------------------+-------+
| primesfrom2to | 15.9 |
| primesfrom3to | 18.4 |
| ambi_sieve | 29.3 |
+---------------------+-------+
```

Timings were measured using the command:

```
python -mtimeit -s"import primes" "primes.{method}(1000000)"
```

with `{method}`

replaced by each of the method names.

primes.py:

```
#!/usr/bin/env python
import psyco; psyco.full()
from math import sqrt, ceil
import numpy as np
def rwh_primes(n):
# https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Returns a list of primes < n """
sieve = [True] * n
for i in xrange(3,int(n**0.5)+1,2):
if sieve[i]:
sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
return [2] + [i for i in xrange(3,n,2) if sieve[i]]
def rwh_primes1(n):
# https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Returns a list of primes < n """
sieve = [True] * (n/2)
for i in xrange(3,int(n**0.5)+1,2):
if sieve[i/2]:
sieve[i*i/2::i] = [False] * ((n-i*i-1)/(2*i)+1)
return [2] + [2*i+1 for i in xrange(1,n/2) if sieve[i]]
def rwh_primes2(n):
# https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Input n>=6, Returns a list of primes, 2 <= p < n """
correction = (n%6>1)
n = {0:n,1:n-1,2:n+4,3:n+3,4:n+2,5:n+1}[n%6]
sieve = [True] * (n/3)
sieve[0] = False
for i in xrange(int(n**0.5)/3+1):
if sieve[i]:
k=3*i+1|1
sieve[ ((k*k)/3) ::2*k]=[False]*((n/6-(k*k)/6-1)/k+1)
sieve[(k*k+4*k-2*k*(i&1))/3::2*k]=[False]*((n/6-(k*k+4*k-2*k*(i&1))/6-1)/k+1)
return [2,3] + [3*i+1|1 for i in xrange(1,n/3-correction) if sieve[i]]
def sieve_wheel_30(N):
# http://zerovolt.com/?p=88
""" Returns a list of primes <= N using wheel criterion 2*3*5 = 30
Copyright 2009 by zerovolt.com
This code is free for non-commercial purposes, in which case you can just leave this comment as a credit for my work.
If you need this code for commercial purposes, please contact me by sending an email to: info [at] zerovolt [dot] com."""
__smallp = ( 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,
61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139,
149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227,
229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311,
313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401,
409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491,
499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599,
601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683,
691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797,
809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887,
907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997)
wheel = (2, 3, 5)
const = 30
if N < 2:
return []
if N <= const:
pos = 0
while __smallp[pos] <= N:
pos += 1
return list(__smallp[:pos])
# make the offsets list
offsets = (7, 11, 13, 17, 19, 23, 29, 1)
# prepare the list
p = [2, 3, 5]
dim = 2 + N // const
tk1 = [True] * dim
tk7 = [True] * dim
tk11 = [True] * dim
tk13 = [True] * dim
tk17 = [True] * dim
tk19 = [True] * dim
tk23 = [True] * dim
tk29 = [True] * dim
tk1[0] = False
# help dictionary d
# d[a , b] = c ==> if I want to find the smallest useful multiple of (30*pos)+a
# on tkc, then I need the index given by the product of [(30*pos)+a][(30*pos)+b]
# in general. If b < a, I need [(30*pos)+a][(30*(pos+1))+b]
d = {}
for x in offsets:
for y in offsets:
res = (x*y) % const
if res in offsets:
d[(x, res)] = y
# another help dictionary: gives tkx calling tmptk[x]
tmptk = {1:tk1, 7:tk7, 11:tk11, 13:tk13, 17:tk17, 19:tk19, 23:tk23, 29:tk29}
pos, prime, lastadded, stop = 0, 0, 0, int(ceil(sqrt(N)))
# inner functions definition
def del_mult(tk, start, step):
for k in xrange(start, len(tk), step):
tk[k] = False
# end of inner functions definition
cpos = const * pos
while prime < stop:
# 30k + 7
if tk7[pos]:
prime = cpos + 7
p.append(prime)
lastadded = 7
for off in offsets:
tmp = d[(7, off)]
start = (pos + prime) if off == 7 else (prime * (const * (pos + 1 if tmp < 7 else 0) + tmp) )//const
del_mult(tmptk[off], start, prime)
# 30k + 11
if tk11[pos]:
prime = cpos + 11
p.append(prime)
lastadded = 11
for off in offsets:
tmp = d[(11, off)]
start = (pos + prime) if off == 11 else (prime * (const * (pos + 1 if tmp < 11 else 0) + tmp) )//const
del_mult(tmptk[off], start, prime)
# 30k + 13
if tk13[pos]:
prime = cpos + 13
p.append(prime)
lastadded = 13
for off in offsets:
tmp = d[(13, off)]
start = (pos + prime) if off == 13 else (prime * (const * (pos + 1 if tmp < 13 else 0) + tmp) )//const
del_mult(tmptk[off], start, prime)
# 30k + 17
if tk17[pos]:
prime = cpos + 17
p.append(prime)
lastadded = 17
for off in offsets:
tmp = d[(17, off)]
start = (pos + prime) if off == 17 else (prime * (const * (pos + 1 if tmp < 17 else 0) + tmp) )//const
del_mult(tmptk[off], start, prime)
# 30k + 19
if tk19[pos]:
prime = cpos + 19
p.append(prime)
lastadded = 19
for off in offsets:
tmp = d[(19, off)]
start = (pos + prime) if off == 19 else (prime * (const * (pos + 1 if tmp < 19 else 0) + tmp) )//const
del_mult(tmptk[off], start, prime)
# 30k + 23
if tk23[pos]:
prime = cpos + 23
p.append(prime)
lastadded = 23
for off in offsets:
tmp = d[(23, off)]
start = (pos + prime) if off == 23 else (prime * (const * (pos + 1 if tmp < 23 else 0) + tmp) )//const
del_mult(tmptk[off], start, prime)
# 30k + 29
if tk29[pos]:
prime = cpos + 29
p.append(prime)
lastadded = 29
for off in offsets:
tmp = d[(29, off)]
start = (pos + prime) if off == 29 else (prime * (const * (pos + 1 if tmp < 29 else 0) + tmp) )//const
del_mult(tmptk[off], start, prime)
# now we go back to top tk1, so we need to increase pos by 1
pos += 1
cpos = const * pos
# 30k + 1
if tk1[pos]:
prime = cpos + 1
p.append(prime)
lastadded = 1
for off in offsets:
tmp = d[(1, off)]
start = (pos + prime) if off == 1 else (prime * (const * pos + tmp) )//const
del_mult(tmptk[off], start, prime)
# time to add remaining primes
# if lastadded == 1, remove last element and start adding them from tk1
# this way we don"t need an "if" within the last while
if lastadded == 1:
p.pop()
# now complete for every other possible prime
while pos < len(tk1):
cpos = const * pos
if tk1[pos]: p.append(cpos + 1)
if tk7[pos]: p.append(cpos + 7)
if tk11[pos]: p.append(cpos + 11)
if tk13[pos]: p.append(cpos + 13)
if tk17[pos]: p.append(cpos + 17)
if tk19[pos]: p.append(cpos + 19)
if tk23[pos]: p.append(cpos + 23)
if tk29[pos]: p.append(cpos + 29)
pos += 1
# remove exceeding if present
pos = len(p) - 1
while p[pos] > N:
pos -= 1
if pos < len(p) - 1:
del p[pos+1:]
# return p list
return p
def sieveOfEratosthenes(n):
"""sieveOfEratosthenes(n): return the list of the primes < n."""
# Code from: <[email protected]>, Nov 30 2006
# http://groups.google.com/group/comp.lang.python/msg/f1f10ced88c68c2d
if n <= 2:
return []
sieve = range(3, n, 2)
top = len(sieve)
for si in sieve:
if si:
bottom = (si*si - 3) // 2
if bottom >= top:
break
sieve[bottom::si] = [0] * -((bottom - top) // si)
return [2] + [el for el in sieve if el]
def sieveOfAtkin(end):
"""sieveOfAtkin(end): return a list of all the prime numbers <end
using the Sieve of Atkin."""
# Code by Steve Krenzel, <[email protected]>, improved
# Code: https://web.archive.org/web/20080324064651/http://krenzel.info/?p=83
# Info: http://en.wikipedia.org/wiki/Sieve_of_Atkin
assert end > 0
lng = ((end-1) // 2)
sieve = [False] * (lng + 1)
x_max, x2, xd = int(sqrt((end-1)/4.0)), 0, 4
for xd in xrange(4, 8*x_max + 2, 8):
x2 += xd
y_max = int(sqrt(end-x2))
n, n_diff = x2 + y_max*y_max, (y_max << 1) - 1
if not (n & 1):
n -= n_diff
n_diff -= 2
for d in xrange((n_diff - 1) << 1, -1, -8):
m = n % 12
if m == 1 or m == 5:
m = n >> 1
sieve[m] = not sieve[m]
n -= d
x_max, x2, xd = int(sqrt((end-1) / 3.0)), 0, 3
for xd in xrange(3, 6 * x_max + 2, 6):
x2 += xd
y_max = int(sqrt(end-x2))
n, n_diff = x2 + y_max*y_max, (y_max << 1) - 1
if not(n & 1):
n -= n_diff
n_diff -= 2
for d in xrange((n_diff - 1) << 1, -1, -8):
if n % 12 == 7:
m = n >> 1
sieve[m] = not sieve[m]
n -= d
x_max, y_min, x2, xd = int((2 + sqrt(4-8*(1-end)))/4), -1, 0, 3
for x in xrange(1, x_max + 1):
x2 += xd
xd += 6
if x2 >= end: y_min = (((int(ceil(sqrt(x2 - end))) - 1) << 1) - 2) << 1
n, n_diff = ((x*x + x) << 1) - 1, (((x-1) << 1) - 2) << 1
for d in xrange(n_diff, y_min, -8):
if n % 12 == 11:
m = n >> 1
sieve[m] = not sieve[m]
n += d
primes = [2, 3]
if end <= 3:
return primes[:max(0,end-2)]
for n in xrange(5 >> 1, (int(sqrt(end))+1) >> 1):
if sieve[n]:
primes.append((n << 1) + 1)
aux = (n << 1) + 1
aux *= aux
for k in xrange(aux, end, 2 * aux):
sieve[k >> 1] = False
s = int(sqrt(end)) + 1
if s % 2 == 0:
s += 1
primes.extend([i for i in xrange(s, end, 2) if sieve[i >> 1]])
return primes
def ambi_sieve_plain(n):
s = range(3, n, 2)
for m in xrange(3, int(n**0.5)+1, 2):
if s[(m-3)/2]:
for t in xrange((m*m-3)/2,(n>>1)-1,m):
s[t]=0
return [2]+[t for t in s if t>0]
def sundaram3(max_n):
# https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/2073279#2073279
numbers = range(3, max_n+1, 2)
half = (max_n)//2
initial = 4
for step in xrange(3, max_n+1, 2):
for i in xrange(initial, half, step):
numbers[i-1] = 0
initial += 2*(step+1)
if initial > half:
return [2] + filter(None, numbers)
################################################################################
# Using Numpy:
def ambi_sieve(n):
# http://tommih.blogspot.com/2009/04/fast-prime-number-generator.html
s = np.arange(3, n, 2)
for m in xrange(3, int(n ** 0.5)+1, 2):
if s[(m-3)/2]:
s[(m*m-3)/2::m]=0
return np.r_[2, s[s>0]]
def primesfrom3to(n):
# https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Returns a array of primes, p < n """
assert n>=2
sieve = np.ones(n/2, dtype=np.bool)
for i in xrange(3,int(n**0.5)+1,2):
if sieve[i/2]:
sieve[i*i/2::i] = False
return np.r_[2, 2*np.nonzero(sieve)[0][1::]+1]
def primesfrom2to(n):
# https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Input n>=6, Returns a array of primes, 2 <= p < n """
sieve = np.ones(n/3 + (n%6==2), dtype=np.bool)
sieve[0] = False
for i in xrange(int(n**0.5)/3+1):
if sieve[i]:
k=3*i+1|1
sieve[ ((k*k)/3) ::2*k] = False
sieve[(k*k+4*k-2*k*(i&1))/3::2*k] = False
return np.r_[2,3,((3*np.nonzero(sieve)[0]+1)|1)]
if __name__=="__main__":
import itertools
import sys
def test(f1,f2,num):
print("Testing {f1} and {f2} return same results".format(
f1=f1.func_name,
f2=f2.func_name))
if not all([a==b for a,b in itertools.izip_longest(f1(num),f2(num))]):
sys.exit("Error: %s(%s) != %s(%s)"%(f1.func_name,num,f2.func_name,num))
n=1000000
test(sieveOfAtkin,sieveOfEratosthenes,n)
test(sieveOfAtkin,ambi_sieve,n)
test(sieveOfAtkin,ambi_sieve_plain,n)
test(sieveOfAtkin,sundaram3,n)
test(sieveOfAtkin,sieve_wheel_30,n)
test(sieveOfAtkin,primesfrom3to,n)
test(sieveOfAtkin,primesfrom2to,n)
test(sieveOfAtkin,rwh_primes,n)
test(sieveOfAtkin,rwh_primes1,n)
test(sieveOfAtkin,rwh_primes2,n)
```

Running the script tests that all implementations give the same result.

You can just do upside-down floor division:

```
def ceildiv(a, b):
return -(-a // b)
```

This works because Python"s division operator does floor division (unlike in C, where integer division truncates the fractional part).

This also works with Python"s big integers, because there"s no (lossy) floating-point conversion.

Here"s a demonstration:

```
>>> from __future__ import division # a/b is float division
>>> from math import ceil
>>> b = 3
>>> for a in range(-7, 8):
... print(["%d/%d" % (a, b), int(ceil(a / b)), -(-a // b)])
...
["-7/3", -2, -2]
["-6/3", -2, -2]
["-5/3", -1, -1]
["-4/3", -1, -1]
["-3/3", -1, -1]
["-2/3", 0, 0]
["-1/3", 0, 0]
["0/3", 0, 0]
["1/3", 1, 1]
["2/3", 1, 1]
["3/3", 1, 1]
["4/3", 2, 2]
["5/3", 2, 2]
["6/3", 2, 2]
["7/3", 3, 3]
```

Matplotlib uses a dictionary from its colors.py module.

To print the names use:

```
# python2:
import matplotlib
for name, hex in matplotlib.colors.cnames.iteritems():
print(name, hex)
# python3:
import matplotlib
for name, hex in matplotlib.colors.cnames.items():
print(name, hex)
```

This is the complete dictionary:

```
cnames = {
"aliceblue": "#F0F8FF",
"antiquewhite": "#FAEBD7",
"aqua": "#00FFFF",
"aquamarine": "#7FFFD4",
"azure": "#F0FFFF",
"beige": "#F5F5DC",
"bisque": "#FFE4C4",
"black": "#000000",
"blanchedalmond": "#FFEBCD",
"blue": "#0000FF",
"blueviolet": "#8A2BE2",
"brown": "#A52A2A",
"burlywood": "#DEB887",
"cadetblue": "#5F9EA0",
"chartreuse": "#7FFF00",
"chocolate": "#D2691E",
"coral": "#FF7F50",
"cornflowerblue": "#6495ED",
"cornsilk": "#FFF8DC",
"crimson": "#DC143C",
"cyan": "#00FFFF",
"darkblue": "#00008B",
"darkcyan": "#008B8B",
"darkgoldenrod": "#B8860B",
"darkgray": "#A9A9A9",
"darkgreen": "#006400",
"darkkhaki": "#BDB76B",
"darkmagenta": "#8B008B",
"darkolivegreen": "#556B2F",
"darkorange": "#FF8C00",
"darkorchid": "#9932CC",
"darkred": "#8B0000",
"darksalmon": "#E9967A",
"darkseagreen": "#8FBC8F",
"darkslateblue": "#483D8B",
"darkslategray": "#2F4F4F",
"darkturquoise": "#00CED1",
"darkviolet": "#9400D3",
"deeppink": "#FF1493",
"deepskyblue": "#00BFFF",
"dimgray": "#696969",
"dodgerblue": "#1E90FF",
"firebrick": "#B22222",
"floralwhite": "#FFFAF0",
"forestgreen": "#228B22",
"fuchsia": "#FF00FF",
"gainsboro": "#DCDCDC",
"ghostwhite": "#F8F8FF",
"gold": "#FFD700",
"goldenrod": "#DAA520",
"gray": "#808080",
"green": "#008000",
"greenyellow": "#ADFF2F",
"honeydew": "#F0FFF0",
"hotpink": "#FF69B4",
"indianred": "#CD5C5C",
"indigo": "#4B0082",
"ivory": "#FFFFF0",
"khaki": "#F0E68C",
"lavender": "#E6E6FA",
"lavenderblush": "#FFF0F5",
"lawngreen": "#7CFC00",
"lemonchiffon": "#FFFACD",
"lightblue": "#ADD8E6",
"lightcoral": "#F08080",
"lightcyan": "#E0FFFF",
"lightgoldenrodyellow": "#FAFAD2",
"lightgreen": "#90EE90",
"lightgray": "#D3D3D3",
"lightpink": "#FFB6C1",
"lightsalmon": "#FFA07A",
"lightseagreen": "#20B2AA",
"lightskyblue": "#87CEFA",
"lightslategray": "#778899",
"lightsteelblue": "#B0C4DE",
"lightyellow": "#FFFFE0",
"lime": "#00FF00",
"limegreen": "#32CD32",
"linen": "#FAF0E6",
"magenta": "#FF00FF",
"maroon": "#800000",
"mediumaquamarine": "#66CDAA",
"mediumblue": "#0000CD",
"mediumorchid": "#BA55D3",
"mediumpurple": "#9370DB",
"mediumseagreen": "#3CB371",
"mediumslateblue": "#7B68EE",
"mediumspringgreen": "#00FA9A",
"mediumturquoise": "#48D1CC",
"mediumvioletred": "#C71585",
"midnightblue": "#191970",
"mintcream": "#F5FFFA",
"mistyrose": "#FFE4E1",
"moccasin": "#FFE4B5",
"navajowhite": "#FFDEAD",
"navy": "#000080",
"oldlace": "#FDF5E6",
"olive": "#808000",
"olivedrab": "#6B8E23",
"orange": "#FFA500",
"orangered": "#FF4500",
"orchid": "#DA70D6",
"palegoldenrod": "#EEE8AA",
"palegreen": "#98FB98",
"paleturquoise": "#AFEEEE",
"palevioletred": "#DB7093",
"papayawhip": "#FFEFD5",
"peachpuff": "#FFDAB9",
"peru": "#CD853F",
"pink": "#FFC0CB",
"plum": "#DDA0DD",
"powderblue": "#B0E0E6",
"purple": "#800080",
"red": "#FF0000",
"rosybrown": "#BC8F8F",
"royalblue": "#4169E1",
"saddlebrown": "#8B4513",
"salmon": "#FA8072",
"sandybrown": "#FAA460",
"seagreen": "#2E8B57",
"seashell": "#FFF5EE",
"sienna": "#A0522D",
"silver": "#C0C0C0",
"skyblue": "#87CEEB",
"slateblue": "#6A5ACD",
"slategray": "#708090",
"snow": "#FFFAFA",
"springgreen": "#00FF7F",
"steelblue": "#4682B4",
"tan": "#D2B48C",
"teal": "#008080",
"thistle": "#D8BFD8",
"tomato": "#FF6347",
"turquoise": "#40E0D0",
"violet": "#EE82EE",
"wheat": "#F5DEB3",
"white": "#FFFFFF",
"whitesmoke": "#F5F5F5",
"yellow": "#FFFF00",
"yellowgreen": "#9ACD32"}
```

You could plot them like this:

```
import matplotlib.pyplot as plt
import matplotlib.patches as patches
import matplotlib.colors as colors
import math
fig = plt.figure()
ax = fig.add_subplot(111)
ratio = 1.0 / 3.0
count = math.ceil(math.sqrt(len(colors.cnames)))
x_count = count * ratio
y_count = count / ratio
x = 0
y = 0
w = 1 / x_count
h = 1 / y_count
for c in colors.cnames:
pos = (x / x_count, y / y_count)
ax.add_patch(patches.Rectangle(pos, w, h, color=c))
ax.annotate(c, xy=pos)
if y >= y_count-1:
x += 1
y = 0
else:
y += 1
plt.show()
```

I have an approach which I think is interesting and a bit different from the rest. The main difference in my approach, compared to some of the others, is in how the image segmentation step is performed--I used the DBSCAN clustering algorithm from Python"s scikit-learn; it"s optimized for finding somewhat amorphous shapes that may not necessarily have a single clear centroid.

At the top level, my approach is fairly simple and can be broken down into about 3 steps. First I apply a threshold (or actually, the logical "or" of two separate and distinct thresholds). As with many of the other answers, I assumed that the Christmas tree would be one of the brighter objects in the scene, so the first threshold is just a simple monochrome brightness test; any pixels with values above 220 on a 0-255 scale (where black is 0 and white is 255) are saved to a binary black-and-white image. The second threshold tries to look for red and yellow lights, which are particularly prominent in the trees in the upper left and lower right of the six images, and stand out well against the blue-green background which is prevalent in most of the photos. I convert the rgb image to hsv space, and require that the hue is either less than 0.2 on a 0.0-1.0 scale (corresponding roughly to the border between yellow and green) or greater than 0.95 (corresponding to the border between purple and red) and additionally I require bright, saturated colors: saturation and value must both be above 0.7. The results of the two threshold procedures are logically "or"-ed together, and the resulting matrix of black-and-white binary images is shown below:

You can clearly see that each image has one large cluster of pixels roughly corresponding to the location of each tree, plus a few of the images also have some other small clusters corresponding either to lights in the windows of some of the buildings, or to a background scene on the horizon. The next step is to get the computer to recognize that these are separate clusters, and label each pixel correctly with a cluster membership ID number.

For this task I chose DBSCAN. There is a pretty good visual comparison of how DBSCAN typically behaves, relative to other clustering algorithms, available here. As I said earlier, it does well with amorphous shapes. The output of DBSCAN, with each cluster plotted in a different color, is shown here:

There are a few things to be aware of when looking at this result. First is that DBSCAN requires the user to set a "proximity" parameter in order to regulate its behavior, which effectively controls how separated a pair of points must be in order for the algorithm to declare a new separate cluster rather than agglomerating a test point onto an already pre-existing cluster. I set this value to be 0.04 times the size along the diagonal of each image. Since the images vary in size from roughly VGA up to about HD 1080, this type of scale-relative definition is critical.

Another point worth noting is that the DBSCAN algorithm as it is implemented in scikit-learn has memory limits which are fairly challenging for some of the larger images in this sample. Therefore, for a few of the larger images, I actually had to "decimate" (i.e., retain only every 3rd or 4th pixel and drop the others) each cluster in order to stay within this limit. As a result of this culling process, the remaining individual sparse pixels are difficult to see on some of the larger images. Therefore, for display purposes only, the color-coded pixels in the above images have been effectively "dilated" just slightly so that they stand out better. It"s purely a cosmetic operation for the sake of the narrative; although there are comments mentioning this dilation in my code, rest assured that it has nothing to do with any calculations that actually matter.

Once the clusters are identified and labeled, the third and final step is easy: I simply take the largest cluster in each image (in this case, I chose to measure "size" in terms of the total number of member pixels, although one could have just as easily instead used some type of metric that gauges physical extent) and compute the convex hull for that cluster. The convex hull then becomes the tree border. The six convex hulls computed via this method are shown below in red:

The source code is written for Python 2.7.6 and it depends on numpy, scipy, matplotlib and scikit-learn. I"ve divided it into two parts. The first part is responsible for the actual image processing:

```
from PIL import Image
import numpy as np
import scipy as sp
import matplotlib.colors as colors
from sklearn.cluster import DBSCAN
from math import ceil, sqrt
"""
Inputs:
rgbimg: [M,N,3] numpy array containing (uint, 0-255) color image
hueleftthr: Scalar constant to select maximum allowed hue in the
yellow-green region
huerightthr: Scalar constant to select minimum allowed hue in the
blue-purple region
satthr: Scalar constant to select minimum allowed saturation
valthr: Scalar constant to select minimum allowed value
monothr: Scalar constant to select minimum allowed monochrome
brightness
maxpoints: Scalar constant maximum number of pixels to forward to
the DBSCAN clustering algorithm
proxthresh: Proximity threshold to use for DBSCAN, as a fraction of
the diagonal size of the image
Outputs:
borderseg: [K,2,2] Nested list containing K pairs of x- and y- pixel
values for drawing the tree border
X: [P,2] List of pixels that passed the threshold step
labels: [Q,2] List of cluster labels for points in Xslice (see
below)
Xslice: [Q,2] Reduced list of pixels to be passed to DBSCAN
"""
def findtree(rgbimg, hueleftthr=0.2, huerightthr=0.95, satthr=0.7,
valthr=0.7, monothr=220, maxpoints=5000, proxthresh=0.04):
# Convert rgb image to monochrome for
gryimg = np.asarray(Image.fromarray(rgbimg).convert("L"))
# Convert rgb image (uint, 0-255) to hsv (float, 0.0-1.0)
hsvimg = colors.rgb_to_hsv(rgbimg.astype(float)/255)
# Initialize binary thresholded image
binimg = np.zeros((rgbimg.shape[0], rgbimg.shape[1]))
# Find pixels with hue<0.2 or hue>0.95 (red or yellow) and saturation/value
# both greater than 0.7 (saturated and bright)--tends to coincide with
# ornamental lights on trees in some of the images
boolidx = np.logical_and(
np.logical_and(
np.logical_or((hsvimg[:,:,0] < hueleftthr),
(hsvimg[:,:,0] > huerightthr)),
(hsvimg[:,:,1] > satthr)),
(hsvimg[:,:,2] > valthr))
# Find pixels that meet hsv criterion
binimg[np.where(boolidx)] = 255
# Add pixels that meet grayscale brightness criterion
binimg[np.where(gryimg > monothr)] = 255
# Prepare thresholded points for DBSCAN clustering algorithm
X = np.transpose(np.where(binimg == 255))
Xslice = X
nsample = len(Xslice)
if nsample > maxpoints:
# Make sure number of points does not exceed DBSCAN maximum capacity
Xslice = X[range(0,nsample,int(ceil(float(nsample)/maxpoints)))]
# Translate DBSCAN proximity threshold to units of pixels and run DBSCAN
pixproxthr = proxthresh * sqrt(binimg.shape[0]**2 + binimg.shape[1]**2)
db = DBSCAN(eps=pixproxthr, min_samples=10).fit(Xslice)
labels = db.labels_.astype(int)
# Find the largest cluster (i.e., with most points) and obtain convex hull
unique_labels = set(labels)
maxclustpt = 0
for k in unique_labels:
class_members = [index[0] for index in np.argwhere(labels == k)]
if len(class_members) > maxclustpt:
points = Xslice[class_members]
hull = sp.spatial.ConvexHull(points)
maxclustpt = len(class_members)
borderseg = [[points[simplex,0], points[simplex,1]] for simplex
in hull.simplices]
return borderseg, X, labels, Xslice
```

and the second part is a user-level script which calls the first file and generates all of the plots above:

```
#!/usr/bin/env python
from PIL import Image
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from findtree import findtree
# Image files to process
fname = ["nmzwj.png", "aVZhC.png", "2K9EF.png",
"YowlH.png", "2y4o5.png", "FWhSP.png"]
# Initialize figures
fgsz = (16,7)
figthresh = plt.figure(figsize=fgsz, facecolor="w")
figclust = plt.figure(figsize=fgsz, facecolor="w")
figcltwo = plt.figure(figsize=fgsz, facecolor="w")
figborder = plt.figure(figsize=fgsz, facecolor="w")
figthresh.canvas.set_window_title("Thresholded HSV and Monochrome Brightness")
figclust.canvas.set_window_title("DBSCAN Clusters (Raw Pixel Output)")
figcltwo.canvas.set_window_title("DBSCAN Clusters (Slightly Dilated for Display)")
figborder.canvas.set_window_title("Trees with Borders")
for ii, name in zip(range(len(fname)), fname):
# Open the file and convert to rgb image
rgbimg = np.asarray(Image.open(name))
# Get the tree borders as well as a bunch of other intermediate values
# that will be used to illustrate how the algorithm works
borderseg, X, labels, Xslice = findtree(rgbimg)
# Display thresholded images
axthresh = figthresh.add_subplot(2,3,ii+1)
axthresh.set_xticks([])
axthresh.set_yticks([])
binimg = np.zeros((rgbimg.shape[0], rgbimg.shape[1]))
for v, h in X:
binimg[v,h] = 255
axthresh.imshow(binimg, interpolation="nearest", cmap="Greys")
# Display color-coded clusters
axclust = figclust.add_subplot(2,3,ii+1) # Raw version
axclust.set_xticks([])
axclust.set_yticks([])
axcltwo = figcltwo.add_subplot(2,3,ii+1) # Dilated slightly for display only
axcltwo.set_xticks([])
axcltwo.set_yticks([])
axcltwo.imshow(binimg, interpolation="nearest", cmap="Greys")
clustimg = np.ones(rgbimg.shape)
unique_labels = set(labels)
# Generate a unique color for each cluster
plcol = cm.rainbow_r(np.linspace(0, 1, len(unique_labels)))
for lbl, pix in zip(labels, Xslice):
for col, unqlbl in zip(plcol, unique_labels):
if lbl == unqlbl:
# Cluster label of -1 indicates no cluster membership;
# override default color with black
if lbl == -1:
col = [0.0, 0.0, 0.0, 1.0]
# Raw version
for ij in range(3):
clustimg[pix[0],pix[1],ij] = col[ij]
# Dilated just for display
axcltwo.plot(pix[1], pix[0], "o", markerfacecolor=col,
markersize=1, markeredgecolor=col)
axclust.imshow(clustimg)
axcltwo.set_xlim(0, binimg.shape[1]-1)
axcltwo.set_ylim(binimg.shape[0], -1)
# Plot original images with read borders around the trees
axborder = figborder.add_subplot(2,3,ii+1)
axborder.set_axis_off()
axborder.imshow(rgbimg, interpolation="nearest")
for vseg, hseg in borderseg:
axborder.plot(hseg, vseg, "r-", lw=3)
axborder.set_xlim(0, binimg.shape[1]-1)
axborder.set_ylim(binimg.shape[0], -1)
plt.show()
```

The following are rough guidelines and educated guesses based on experience. You should `timeit`

or profile your concrete use case to get hard numbers, and those numbers may occasionally disagree with the below.

A list comprehension is usually a tiny bit faster than the precisely equivalent `for`

loop (that actually builds a list), most likely because it doesn"t have to look up the list and its `append`

method on every iteration. However, a list comprehension still does a bytecode-level loop:

```
>>> dis.dis(<the code object for `[x for x in range(10)]`>)
1 0 BUILD_LIST 0
3 LOAD_FAST 0 (.0)
>> 6 FOR_ITER 12 (to 21)
9 STORE_FAST 1 (x)
12 LOAD_FAST 1 (x)
15 LIST_APPEND 2
18 JUMP_ABSOLUTE 6
>> 21 RETURN_VALUE
```

Using a list comprehension in place of a loop that *doesn"t* build a list, nonsensically accumulating a list of meaningless values and then throwing the list away, is often *slower* because of the overhead of creating and extending the list. List comprehensions aren"t magic that is inherently faster than a good old loop.

As for functional list processing functions: While these are written in C and probably outperform equivalent functions written in Python, they are *not* necessarily the fastest option. Some speed up is expected **if** the function is written in C too. But most cases using a `lambda`

(or other Python function), the overhead of repeatedly setting up Python stack frames etc. eats up any savings. Simply doing the same work in-line, without function calls (e.g. a list comprehension instead of `map`

or `filter`

) is often slightly faster.

Suppose that in a game that I"m developing I need to draw complex and huge maps using for loops. This question would be definitely relevant, for if a list-comprehension, for example, is indeed faster, it would be a much better option in order to avoid lags (Despite the visual complexity of the code).

Chances are, if code like this isn"t already fast enough when written in good non-"optimized" Python, no amount of Python level micro optimization is going to make it fast enough and you should start thinking about dropping to C. While extensive micro optimizations can often speed up Python code considerably, there is a low (in absolute terms) limit to this. Moreover, even before you hit that ceiling, it becomes simply more cost efficient (15% speedup vs. 300% speed up with the same effort) to bite the bullet and write some C.

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