PHP IntlCalendar getFirstDayOfWeek () Function

| |
Syntax:
  • Object Oriented Style
    int IntlCalendar::getFirstDayOfWeek (void )
  • Procedural style
    int intlcal_get_first_day_of_week (IntlCalendar $cal)
Parameters:This function uses a single parameter $cal,which contains the IntlCalendar resource.Return value:this the function returns one of the constants of the IntlCalendar field, for example IntlCalendar::DOW_SUNDAY, IntlCalendar::DOW_MONDAY,…, IntlCalendar::DOW_SATURDAY on success or FALSE on error.The program below illustrates the IntlCalendar::getFirstDayOfWeek ( ) in PHP:Program :
// Mouth update DateTime zone ini_set ( ’date .timezone’ , ’Asia / Calcutta’ ); ini_set ( ’date.timezone’ , ’UTC’ );
// Set the DateTime object $calendar1 = IntlCalendar::fromDateTime ( ’2019-09-22’ );
// Use the getFirstDayOfWeek() function to get
// first day of the week var_dump ( $calendar1 -> getFirstDayOfWeek());
// Get the week of the year var_dump ( $calendar1 -> get (IntlCalendar::FIELD_WEEK_OF_YEAR));
// Create a calendar instance $calendar2 = IntlCalendar::createInstance (NULL, ’en_US’ );
// Use the getFirstDayOfWeek() function to get
// first day of the week var_dump ( $calendar2 -> getFirstDayOfWeek());
// Set date in calendar $calendar2 -> set (2020, 05, 20);
// Check if the given month is a leap month or not var_dump ( $calendar2 -> get (IntlCalendar::FIELD_IS_LEAP_MONTH));
?>
Exit:
int (1 ) int (39) int (1) int (0)
Link: https://www.php.net/manual/en/intlcalendar.getfirstdayofweek.php

PHP IntlCalendar getFirstDayOfWeek () Function: StackOverflow Questions

How do I merge two dictionaries in a single expression (taking union of dictionaries)?

Question by Carl Meyer

I have two Python dictionaries, and I want to write a single expression that returns these two dictionaries, merged (i.e. taking the union). The update() method would be what I need, if it returned its result instead of modifying a dictionary in-place.

>>> x = {"a": 1, "b": 2}
>>> y = {"b": 10, "c": 11}
>>> z = x.update(y)
>>> print(z)
None
>>> x
{"a": 1, "b": 10, "c": 11}

How can I get that final merged dictionary in z, not x?

(To be extra-clear, the last-one-wins conflict-handling of dict.update() is what I"m looking for as well.)

Answer #1:

How can I merge two Python dictionaries in a single expression?

For dictionaries x and y, z becomes a shallowly-merged dictionary with values from y replacing those from x.

  • In Python 3.9.0 or greater (released 17 October 2020): PEP-584, discussed here, was implemented and provides the simplest method:

    z = x | y          # NOTE: 3.9+ ONLY
    
  • In Python 3.5 or greater:

    z = {**x, **y}
    
  • In Python 2, (or 3.4 or lower) write a function:

    def merge_two_dicts(x, y):
        z = x.copy()   # start with keys and values of x
        z.update(y)    # modifies z with keys and values of y
        return z
    

    and now:

    z = merge_two_dicts(x, y)
    

Explanation

Say you have two dictionaries and you want to merge them into a new dictionary without altering the original dictionaries:

x = {"a": 1, "b": 2}
y = {"b": 3, "c": 4}

The desired result is to get a new dictionary (z) with the values merged, and the second dictionary"s values overwriting those from the first.

>>> z
{"a": 1, "b": 3, "c": 4}

A new syntax for this, proposed in PEP 448 and available as of Python 3.5, is

z = {**x, **y}

And it is indeed a single expression.

Note that we can merge in with literal notation as well:

z = {**x, "foo": 1, "bar": 2, **y}

and now:

>>> z
{"a": 1, "b": 3, "foo": 1, "bar": 2, "c": 4}

It is now showing as implemented in the release schedule for 3.5, PEP 478, and it has now made its way into the What"s New in Python 3.5 document.

However, since many organizations are still on Python 2, you may wish to do this in a backward-compatible way. The classically Pythonic way, available in Python 2 and Python 3.0-3.4, is to do this as a two-step process:

z = x.copy()
z.update(y) # which returns None since it mutates z

In both approaches, y will come second and its values will replace x"s values, thus b will point to 3 in our final result.

Not yet on Python 3.5, but want a single expression

If you are not yet on Python 3.5 or need to write backward-compatible code, and you want this in a single expression, the most performant while the correct approach is to put it in a function:

def merge_two_dicts(x, y):
    """Given two dictionaries, merge them into a new dict as a shallow copy."""
    z = x.copy()
    z.update(y)
    return z

and then you have a single expression:

z = merge_two_dicts(x, y)

You can also make a function to merge an arbitrary number of dictionaries, from zero to a very large number:

def merge_dicts(*dict_args):
    """
    Given any number of dictionaries, shallow copy and merge into a new dict,
    precedence goes to key-value pairs in latter dictionaries.
    """
    result = {}
    for dictionary in dict_args:
        result.update(dictionary)
    return result

This function will work in Python 2 and 3 for all dictionaries. e.g. given dictionaries a to g:

z = merge_dicts(a, b, c, d, e, f, g) 

and key-value pairs in g will take precedence over dictionaries a to f, and so on.

Critiques of Other Answers

Don"t use what you see in the formerly accepted answer:

z = dict(x.items() + y.items())

In Python 2, you create two lists in memory for each dict, create a third list in memory with length equal to the length of the first two put together, and then discard all three lists to create the dict. In Python 3, this will fail because you"re adding two dict_items objects together, not two lists -

>>> c = dict(a.items() + b.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: "dict_items" and "dict_items"

and you would have to explicitly create them as lists, e.g. z = dict(list(x.items()) + list(y.items())). This is a waste of resources and computation power.

Similarly, taking the union of items() in Python 3 (viewitems() in Python 2.7) will also fail when values are unhashable objects (like lists, for example). Even if your values are hashable, since sets are semantically unordered, the behavior is undefined in regards to precedence. So don"t do this:

>>> c = dict(a.items() | b.items())

This example demonstrates what happens when values are unhashable:

>>> x = {"a": []}
>>> y = {"b": []}
>>> dict(x.items() | y.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: "list"

Here"s an example where y should have precedence, but instead the value from x is retained due to the arbitrary order of sets:

>>> x = {"a": 2}
>>> y = {"a": 1}
>>> dict(x.items() | y.items())
{"a": 2}

Another hack you should not use:

z = dict(x, **y)

This uses the dict constructor and is very fast and memory-efficient (even slightly more so than our two-step process) but unless you know precisely what is happening here (that is, the second dict is being passed as keyword arguments to the dict constructor), it"s difficult to read, it"s not the intended usage, and so it is not Pythonic.

Here"s an example of the usage being remediated in django.

Dictionaries are intended to take hashable keys (e.g. frozensets or tuples), but this method fails in Python 3 when keys are not strings.

>>> c = dict(a, **b)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: keyword arguments must be strings

From the mailing list, Guido van Rossum, the creator of the language, wrote:

I am fine with declaring dict({}, **{1:3}) illegal, since after all it is abuse of the ** mechanism.

and

Apparently dict(x, **y) is going around as "cool hack" for "call x.update(y) and return x". Personally, I find it more despicable than cool.

It is my understanding (as well as the understanding of the creator of the language) that the intended usage for dict(**y) is for creating dictionaries for readability purposes, e.g.:

dict(a=1, b=10, c=11)

instead of

{"a": 1, "b": 10, "c": 11}

Response to comments

Despite what Guido says, dict(x, **y) is in line with the dict specification, which btw. works for both Python 2 and 3. The fact that this only works for string keys is a direct consequence of how keyword parameters work and not a short-coming of dict. Nor is using the ** operator in this place an abuse of the mechanism, in fact, ** was designed precisely to pass dictionaries as keywords.

Again, it doesn"t work for 3 when keys are not strings. The implicit calling contract is that namespaces take ordinary dictionaries, while users must only pass keyword arguments that are strings. All other callables enforced it. dict broke this consistency in Python 2:

>>> foo(**{("a", "b"): None})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() keywords must be strings
>>> dict(**{("a", "b"): None})
{("a", "b"): None}

This inconsistency was bad given other implementations of Python (PyPy, Jython, IronPython). Thus it was fixed in Python 3, as this usage could be a breaking change.

I submit to you that it is malicious incompetence to intentionally write code that only works in one version of a language or that only works given certain arbitrary constraints.

More comments:

dict(x.items() + y.items()) is still the most readable solution for Python 2. Readability counts.

My response: merge_two_dicts(x, y) actually seems much clearer to me, if we"re actually concerned about readability. And it is not forward compatible, as Python 2 is increasingly deprecated.

{**x, **y} does not seem to handle nested dictionaries. the contents of nested keys are simply overwritten, not merged [...] I ended up being burnt by these answers that do not merge recursively and I was surprised no one mentioned it. In my interpretation of the word "merging" these answers describe "updating one dict with another", and not merging.

Yes. I must refer you back to the question, which is asking for a shallow merge of two dictionaries, with the first"s values being overwritten by the second"s - in a single expression.

Assuming two dictionaries of dictionaries, one might recursively merge them in a single function, but you should be careful not to modify the dictionaries from either source, and the surest way to avoid that is to make a copy when assigning values. As keys must be hashable and are usually therefore immutable, it is pointless to copy them:

from copy import deepcopy

def dict_of_dicts_merge(x, y):
    z = {}
    overlapping_keys = x.keys() & y.keys()
    for key in overlapping_keys:
        z[key] = dict_of_dicts_merge(x[key], y[key])
    for key in x.keys() - overlapping_keys:
        z[key] = deepcopy(x[key])
    for key in y.keys() - overlapping_keys:
        z[key] = deepcopy(y[key])
    return z

Usage:

>>> x = {"a":{1:{}}, "b": {2:{}}}
>>> y = {"b":{10:{}}, "c": {11:{}}}
>>> dict_of_dicts_merge(x, y)
{"b": {2: {}, 10: {}}, "a": {1: {}}, "c": {11: {}}}

Coming up with contingencies for other value types is far beyond the scope of this question, so I will point you at my answer to the canonical question on a "Dictionaries of dictionaries merge".

Less Performant But Correct Ad-hocs

These approaches are less performant, but they will provide correct behavior. They will be much less performant than copy and update or the new unpacking because they iterate through each key-value pair at a higher level of abstraction, but they do respect the order of precedence (latter dictionaries have precedence)

You can also chain the dictionaries manually inside a dict comprehension:

{k: v for d in dicts for k, v in d.items()} # iteritems in Python 2.7

or in Python 2.6 (and perhaps as early as 2.4 when generator expressions were introduced):

dict((k, v) for d in dicts for k, v in d.items()) # iteritems in Python 2

itertools.chain will chain the iterators over the key-value pairs in the correct order:

from itertools import chain
z = dict(chain(x.items(), y.items())) # iteritems in Python 2

Performance Analysis

I"m only going to do the performance analysis of the usages known to behave correctly. (Self-contained so you can copy and paste yourself.)

from timeit import repeat
from itertools import chain

x = dict.fromkeys("abcdefg")
y = dict.fromkeys("efghijk")

def merge_two_dicts(x, y):
    z = x.copy()
    z.update(y)
    return z

min(repeat(lambda: {**x, **y}))
min(repeat(lambda: merge_two_dicts(x, y)))
min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
min(repeat(lambda: dict(chain(x.items(), y.items()))))
min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))

In Python 3.8.1, NixOS:

>>> min(repeat(lambda: {**x, **y}))
1.0804965235292912
>>> min(repeat(lambda: merge_two_dicts(x, y)))
1.636518670246005
>>> min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
3.1779992282390594
>>> min(repeat(lambda: dict(chain(x.items(), y.items()))))
2.740647904574871
>>> min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))
4.266070580109954
$ uname -a
Linux nixos 4.19.113 #1-NixOS SMP Wed Mar 25 07:06:15 UTC 2020 x86_64 GNU/Linux

Resources on Dictionaries

Answer #2:

In your case, what you can do is:

z = dict(list(x.items()) + list(y.items()))

This will, as you want it, put the final dict in z, and make the value for key b be properly overridden by the second (y) dict"s value:

>>> x = {"a":1, "b": 2}
>>> y = {"b":10, "c": 11}
>>> z = dict(list(x.items()) + list(y.items()))
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python 2, you can even remove the list() calls. To create z:

>>> z = dict(x.items() + y.items())
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python version 3.9.0a4 or greater, then you can directly use:

x = {"a":1, "b": 2}
y = {"b":10, "c": 11}
z = x | y
print(z)
{"a": 1, "c": 11, "b": 10}

Answer #3:

An alternative:

z = x.copy()
z.update(y)

Answer #4:

Another, more concise, option:

z = dict(x, **y)

Note: this has become a popular answer, but it is important to point out that if y has any non-string keys, the fact that this works at all is an abuse of a CPython implementation detail, and it does not work in Python 3, or in PyPy, IronPython, or Jython. Also, Guido is not a fan. So I can"t recommend this technique for forward-compatible or cross-implementation portable code, which really means it should be avoided entirely.

Answer #5:

This probably won"t be a popular answer, but you almost certainly do not want to do this. If you want a copy that"s a merge, then use copy (or deepcopy, depending on what you want) and then update. The two lines of code are much more readable - more Pythonic - than the single line creation with .items() + .items(). Explicit is better than implicit.

In addition, when you use .items() (pre Python 3.0), you"re creating a new list that contains the items from the dict. If your dictionaries are large, then that is quite a lot of overhead (two large lists that will be thrown away as soon as the merged dict is created). update() can work more efficiently, because it can run through the second dict item-by-item.

In terms of time:

>>> timeit.Timer("dict(x, **y)", "x = dict(zip(range(1000), range(1000)))
y=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
15.52571702003479
>>> timeit.Timer("temp = x.copy()
temp.update(y)", "x = dict(zip(range(1000), range(1000)))
y=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
15.694622993469238
>>> timeit.Timer("dict(x.items() + y.items())", "x = dict(zip(range(1000), range(1000)))
y=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
41.484580039978027

IMO the tiny slowdown between the first two is worth it for the readability. In addition, keyword arguments for dictionary creation was only added in Python 2.3, whereas copy() and update() will work in older versions.

PHP IntlCalendar getFirstDayOfWeek () Function: StackOverflow Questions

Why is reading lines from stdin much slower in C++ than Python?

I wanted to compare reading lines of string input from stdin using Python and C++ and was shocked to see my C++ code run an order of magnitude slower than the equivalent Python code. Since my C++ is rusty and I"m not yet an expert Pythonista, please tell me if I"m doing something wrong or if I"m misunderstanding something.


(TLDR answer: include the statement: cin.sync_with_stdio(false) or just use fgets instead.

TLDR results: scroll all the way down to the bottom of my question and look at the table.)


C++ code:

#include <iostream>
#include <time.h>

using namespace std;

int main() {
    string input_line;
    long line_count = 0;
    time_t start = time(NULL);
    int sec;
    int lps;

    while (cin) {
        getline(cin, input_line);
        if (!cin.eof())
            line_count++;
    };

    sec = (int) time(NULL) - start;
    cerr << "Read " << line_count << " lines in " << sec << " seconds.";
    if (sec > 0) {
        lps = line_count / sec;
        cerr << " LPS: " << lps << endl;
    } else
        cerr << endl;
    return 0;
}

// Compiled with:
// g++ -O3 -o readline_test_cpp foo.cpp

Python Equivalent:

#!/usr/bin/env python
import time
import sys

count = 0
start = time.time()

for line in  sys.stdin:
    count += 1

delta_sec = int(time.time() - start_time)
if delta_sec >= 0:
    lines_per_sec = int(round(count/delta_sec))
    print("Read {0} lines in {1} seconds. LPS: {2}".format(count, delta_sec,
       lines_per_sec))

Here are my results:

$ cat test_lines | ./readline_test_cpp
Read 5570000 lines in 9 seconds. LPS: 618889

$ cat test_lines | ./readline_test.py
Read 5570000 lines in 1 seconds. LPS: 5570000

I should note that I tried this both under Mac OS X v10.6.8 (Snow Leopard) and Linux 2.6.32 (Red Hat Linux 6.2). The former is a MacBook Pro, and the latter is a very beefy server, not that this is too pertinent.

$ for i in {1..5}; do echo "Test run $i at `date`"; echo -n "CPP:"; cat test_lines | ./readline_test_cpp ; echo -n "Python:"; cat test_lines | ./readline_test.py ; done
Test run 1 at Mon Feb 20 21:29:28 EST 2012
CPP:   Read 5570001 lines in 9 seconds. LPS: 618889
Python:Read 5570000 lines in 1 seconds. LPS: 5570000
Test run 2 at Mon Feb 20 21:29:39 EST 2012
CPP:   Read 5570001 lines in 9 seconds. LPS: 618889
Python:Read 5570000 lines in 1 seconds. LPS: 5570000
Test run 3 at Mon Feb 20 21:29:50 EST 2012
CPP:   Read 5570001 lines in 9 seconds. LPS: 618889
Python:Read 5570000 lines in 1 seconds. LPS: 5570000
Test run 4 at Mon Feb 20 21:30:01 EST 2012
CPP:   Read 5570001 lines in 9 seconds. LPS: 618889
Python:Read 5570000 lines in 1 seconds. LPS: 5570000
Test run 5 at Mon Feb 20 21:30:11 EST 2012
CPP:   Read 5570001 lines in 10 seconds. LPS: 557000
Python:Read 5570000 lines in  1 seconds. LPS: 5570000

Tiny benchmark addendum and recap

For completeness, I thought I"d update the read speed for the same file on the same box with the original (synced) C++ code. Again, this is for a 100M line file on a fast disk. Here"s the comparison, with several solutions/approaches:

Implementation Lines per second
python (default) 3,571,428
cin (default/naive) 819,672
cin (no sync) 12,500,000
fgets 14,285,714
wc (not fair comparison) 54,644,808

Answer #1:

tl;dr: Because of different default settings in C++ requiring more system calls.

By default, cin is synchronized with stdio, which causes it to avoid any input buffering. If you add this to the top of your main, you should see much better performance:

std::ios_base::sync_with_stdio(false);

Normally, when an input stream is buffered, instead of reading one character at a time, the stream will be read in larger chunks. This reduces the number of system calls, which are typically relatively expensive. However, since the FILE* based stdio and iostreams often have separate implementations and therefore separate buffers, this could lead to a problem if both were used together. For example:

int myvalue1;
cin >> myvalue1;
int myvalue2;
scanf("%d",&myvalue2);

If more input was read by cin than it actually needed, then the second integer value wouldn"t be available for the scanf function, which has its own independent buffer. This would lead to unexpected results.

To avoid this, by default, streams are synchronized with stdio. One common way to achieve this is to have cin read each character one at a time as needed using stdio functions. Unfortunately, this introduces a lot of overhead. For small amounts of input, this isn"t a big problem, but when you are reading millions of lines, the performance penalty is significant.

Fortunately, the library designers decided that you should also be able to disable this feature to get improved performance if you knew what you were doing, so they provided the sync_with_stdio method.

Answer #2:

Just out of curiosity I"ve taken a look at what happens under the hood, and I"ve used dtruss/strace on each test.

C++

./a.out < in
Saw 6512403 lines in 8 seconds.  Crunch speed: 814050

syscalls sudo dtruss -c ./a.out < in

CALL                                        COUNT
__mac_syscall                                   1
<snip>
open                                            6
pread                                           8
mprotect                                       17
mmap                                           22
stat64                                         30
read_nocancel                               25958

Python

./a.py < in
Read 6512402 lines in 1 seconds. LPS: 6512402

syscalls sudo dtruss -c ./a.py < in

CALL                                        COUNT
__mac_syscall                                   1
<snip>
open                                            5
pread                                           8
mprotect                                       17
mmap                                           21
stat64                                         29

Answer #3:

I"m a few years behind here, but:

In "Edit 4/5/6" of the original post, you are using the construction:

$ /usr/bin/time cat big_file | program_to_benchmark

This is wrong in a couple of different ways:

  1. You"re actually timing the execution of cat, not your benchmark. The "user" and "sys" CPU usage displayed by time are those of cat, not your benchmarked program. Even worse, the "real" time is also not necessarily accurate. Depending on the implementation of cat and of pipelines in your local OS, it is possible that cat writes a final giant buffer and exits long before the reader process finishes its work.

  2. Use of cat is unnecessary and in fact counterproductive; you"re adding moving parts. If you were on a sufficiently old system (i.e. with a single CPU and -- in certain generations of computers -- I/O faster than CPU) -- the mere fact that cat was running could substantially color the results. You are also subject to whatever input and output buffering and other processing cat may do. (This would likely earn you a "Useless Use Of Cat" award if I were Randal Schwartz.

A better construction would be:

$ /usr/bin/time program_to_benchmark < big_file

In this statement it is the shell which opens big_file, passing it to your program (well, actually to time which then executes your program as a subprocess) as an already-open file descriptor. 100% of the file reading is strictly the responsibility of the program you"re trying to benchmark. This gets you a real reading of its performance without spurious complications.

I will mention two possible, but actually wrong, "fixes" which could also be considered (but I "number" them differently as these are not things which were wrong in the original post):

A. You could "fix" this by timing only your program:

$ cat big_file | /usr/bin/time program_to_benchmark

B. or by timing the entire pipeline:

$ /usr/bin/time sh -c "cat big_file | program_to_benchmark"

These are wrong for the same reasons as #2: they"re still using cat unnecessarily. I mention them for a few reasons:

  • they"re more "natural" for people who aren"t entirely comfortable with the I/O redirection facilities of the POSIX shell

  • there may be cases where cat is needed (e.g.: the file to be read requires some sort of privilege to access, and you do not want to grant that privilege to the program to be benchmarked: sudo cat /dev/sda | /usr/bin/time my_compression_test --no-output)

  • in practice, on modern machines, the added cat in the pipeline is probably of no real consequence.

But I say that last thing with some hesitation. If we examine the last result in "Edit 5" --

$ /usr/bin/time cat temp_big_file | wc -l
0.01user 1.34system 0:01.83elapsed 74%CPU ...

-- this claims that cat consumed 74% of the CPU during the test; and indeed 1.34/1.83 is approximately 74%. Perhaps a run of:

$ /usr/bin/time wc -l < temp_big_file

would have taken only the remaining .49 seconds! Probably not: cat here had to pay for the read() system calls (or equivalent) which transferred the file from "disk" (actually buffer cache), as well as the pipe writes to deliver them to wc. The correct test would still have had to do those read() calls; only the write-to-pipe and read-from-pipe calls would have been saved, and those should be pretty cheap.

Still, I predict you would be able to measure the difference between cat file | wc -l and wc -l < file and find a noticeable (2-digit percentage) difference. Each of the slower tests will have paid a similar penalty in absolute time; which would however amount to a smaller fraction of its larger total time.

In fact I did some quick tests with a 1.5 gigabyte file of garbage, on a Linux 3.13 (Ubuntu 14.04) system, obtaining these results (these are actually "best of 3" results; after priming the cache, of course):

$ time wc -l < /tmp/junk
real 0.280s user 0.156s sys 0.124s (total cpu 0.280s)
$ time cat /tmp/junk | wc -l
real 0.407s user 0.157s sys 0.618s (total cpu 0.775s)
$ time sh -c "cat /tmp/junk | wc -l"
real 0.411s user 0.118s sys 0.660s (total cpu 0.778s)

Notice that the two pipeline results claim to have taken more CPU time (user+sys) than real wall-clock time. This is because I"m using the shell (bash)"s built-in "time" command, which is cognizant of the pipeline; and I"m on a multi-core machine where separate processes in a pipeline can use separate cores, accumulating CPU time faster than realtime. Using /usr/bin/time I see smaller CPU time than realtime -- showing that it can only time the single pipeline element passed to it on its command line. Also, the shell"s output gives milliseconds while /usr/bin/time only gives hundredths of a second.

So at the efficiency level of wc -l, the cat makes a huge difference: 409 / 283 = 1.453 or 45.3% more realtime, and 775 / 280 = 2.768, or a whopping 177% more CPU used! On my random it-was-there-at-the-time test box.

I should add that there is at least one other significant difference between these styles of testing, and I can"t say whether it is a benefit or fault; you have to decide this yourself:

When you run cat big_file | /usr/bin/time my_program, your program is receiving input from a pipe, at precisely the pace sent by cat, and in chunks no larger than written by cat.

When you run /usr/bin/time my_program < big_file, your program receives an open file descriptor to the actual file. Your program -- or in many cases the I/O libraries of the language in which it was written -- may take different actions when presented with a file descriptor referencing a regular file. It may use mmap(2) to map the input file into its address space, instead of using explicit read(2) system calls. These differences could have a far larger effect on your benchmark results than the small cost of running the cat binary.

Of course it is an interesting benchmark result if the same program performs significantly differently between the two cases. It shows that, indeed, the program or its I/O libraries are doing something interesting, like using mmap(). So in practice it might be good to run the benchmarks both ways; perhaps discounting the cat result by some small factor to "forgive" the cost of running cat itself.

How do you read from stdin?

I"m trying to do some of the code golf challenges, but they all require the input to be taken from stdin. How do I get that in Python?

Answer #1:

You could use the fileinput module:

import fileinput

for line in fileinput.input():
    pass

fileinput will loop through all the lines in the input specified as file names given in command-line arguments, or the standard input if no arguments are provided.

Note: line will contain a trailing newline; to remove it use line.rstrip()

Answer #2:

There"s a few ways to do it.

  • sys.stdin is a file-like object on which you can call functions read or readlines if you want to read everything or you want to read everything and split it by newline automatically. (You need to import sys for this to work.)

  • If you want to prompt the user for input, you can use raw_input in Python 2.X, and just input in Python 3.

  • If you actually just want to read command-line options, you can access them via the sys.argv list.

You will probably find this Wikibook article on I/O in Python to be a useful reference as well.

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