Basic approximations in Python

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Approximation means estimating the value of something that is not quite accurate, but almost correct. It plays a vital role in science and technology. Let’s start with the most common example. Have you ever used the exact pi value? Of course not. It is an infinite irrational number with a very long meaning. If we continue to write the exact value of Pi , perhaps even this article will not be enough for this:

 3.14159 26535 89793 23846 26433 83279 ... 

So here where approximation plays a role. We usually approximate Pi as 3.14 or in rational terms 22/7 . When you got to high school, you probably saw a wider application of approximations in mathematics, which use differentials to approximate the values ‚Äã‚Äãof quantities, such as (36.6) ^ 1/2 or (0.009) ^ 1/3. In computer science, we can use approximation to find the value or approximate the value of something using loops.

For example: approximation of the cube root of any number. Take a look at the process below:

# Python program to approximate
# cube root of 27

guess = 0.0

cube = 27

increment = 0.0001

epsilon = 0.1


# Find an approximate value

while abs (guess * * 3 - cube) & gt; = epsilon:

guess + = increment


# Check approximate value

if abs (guess * * 3 - cube) & gt; = epsilon:

print ( "Failed on the cube root of" , cube)

else :

print (guess, " is close to the cube root of " , cube)

Output of the above code:

 2.9963000000018987 is close to the cube root of 27 

As we can see, 2.99 is not the exact value of (27) ^ 1/3 but is very close to the exact value 3. This is what we call approximation. Here we have used a series of calculations to approximate the value. First, we declare a variable guess = 0.0 which we will keep incrementing in a loop until it approaches the cube root of 27. Another variable epsilon is chosen as little as possible to get more accurate meaning. The line while abs (guess ** 3 - cube) & gt; = epsilon: will take care of this. If it breaks out of the loop with a value greater than epsilon , it means that we have already crossed the approximate value and failed in the test. Otherwise, it will return the guess value.

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Basic approximations in Python find: Questions

Finding the index of an item in a list

5 answers

Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index (1) in Python?

3740

Answer #1

>>> ["foo", "bar", "baz"].index("bar")
1

Reference: Data Structures > More on Lists

Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can"t remember the last time I used it in anger. It"s been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about list.index follow. It is probably worth initially taking a look at the documentation for it:

list.index(x[, start[, end]])

Return zero-based index in the list of the first item whose value is equal to x. Raises a ValueError if there is no such item.

The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If your list is long, and you don"t know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give index a hint. For instance, in this snippet, l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:

>>> import timeit
>>> timeit.timeit("l.index(999_999)", setup="l = list(range(0, 1_000_000))", number=1000)
9.356267921015387
>>> timeit.timeit("l.index(999_999, 999_990, 1_000_000)", setup="l = list(range(0, 1_000_000))", number=1000)
0.0004404920036904514
 

Only returns the index of the first match to its argument

A call to index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2

Most places where I once would have used index, I now use a list comprehension or generator expression because they"re more generalizable. So if you"re considering reaching for index, take a look at these excellent Python features.

Throws if element not present in list

A call to index results in a ValueError if the item"s not present.

>>> [1, 1].index(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: 2 is not in list

If the item might not be present in the list, you should either

  1. Check for it first with item in my_list (clean, readable approach), or
  2. Wrap the index call in a try/except block which catches ValueError (probably faster, at least when the list to search is long, and the item is usually present.)

3740

Answer #2

One thing that is really helpful in learning Python is to use the interactive help function:

>>> help(["foo", "bar", "baz"])
Help on list object:

class list(object)
 ...

 |
 |  index(...)
 |      L.index(value, [start, [stop]]) -> integer -- return first index of value
 |

which will often lead you to the method you are looking for.

3740

Answer #3

The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use enumerate():

for i, j in enumerate(["foo", "bar", "baz"]):
    if j == "bar":
        print(i)

The index() function only returns the first occurrence, while enumerate() returns all occurrences.

As a list comprehension:

[i for i, j in enumerate(["foo", "bar", "baz"]) if j == "bar"]

Here"s also another small solution with itertools.count() (which is pretty much the same approach as enumerate):

from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ["foo", "bar", "baz"]) if j == "bar"]

This is more efficient for larger lists than using enumerate():

$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 174 usec per loop
$ python -m timeit "[i for i, j in enumerate(["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 196 usec per loop

log

Python"s equivalent of && (logical-and) in an if-statement

5 answers

delete By delete

Here"s my code:

def front_back(a, b):
  # +++your code here+++
  if len(a) % 2 == 0 && len(b) % 2 == 0:
    return a[:(len(a)/2)] + b[:(len(b)/2)] + a[(len(a)/2):] + b[(len(b)/2):] 
  else:
    #todo! Not yet done. :P
  return

I"m getting an error in the IF conditional.
What am I doing wrong?

934

Answer #1

You would want and instead of &&.

934

Answer #2

Python uses and and or conditionals.

i.e.

if foo == "abc" and bar == "bac" or zoo == "123":
  # do something

log

How do you get the logical xor of two variables in Python?

5 answers

Zach Hirsch By Zach Hirsch

How do you get the logical xor of two variables in Python?

For example, I have two variables that I expect to be strings. I want to test that only one of them contains a True value (is not None or the empty string):

str1 = raw_input("Enter string one:")
str2 = raw_input("Enter string two:")
if logical_xor(str1, str2):
    print "ok"
else:
    print "bad"

The ^ operator seems to be bitwise, and not defined on all objects:

>>> 1 ^ 1
0
>>> 2 ^ 1
3
>>> "abc" ^ ""
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for ^: "str" and "str"
794

Answer #1

If you"re already normalizing the inputs to booleans, then != is xor.

bool(a) != bool(b)

We hope this article has helped you to resolve the problem. Apart from Basic approximations in Python, check other find-related topics.

Want to excel in Python? See our review of the best Python online courses 2022. If you are interested in Data Science, check also how to learn programming in R.

By the way, this material is also available in other languages:



Olivia Richtgofen

Berlin | 2022-11-27

Simply put and clear. Thank you for sharing. Basic approximations in Python and other issues with String Variables was always my weak point 😁. I am just not quite sure it is the best method

Anna Galleotti

San Francisco | 2022-11-27

Simply put and clear. Thank you for sharing. Basic approximations in Python and other issues with String Variables was always my weak point 😁. Will use it in my bachelor thesis

Dmitry Lehnman

Warsaw | 2022-11-27

Maybe there are another answers? What Basic approximations in Python exactly means?. Will use it in my bachelor thesis

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