timeit

Python timeit doesn't works for list.remove operation

由 半世苍凉 提交于 2019-12-20 02:34:44
问题 I was trying to check the performance of remove operation in python list through the timeit module, but it throws a ValueError. In [4]: a = [1, 2, 3] In [5]: timeit a.remove(2) --------------------------------------------------------------------------- ValueError Traceback (most recent call last) <ipython-input-5-7b32a87ebb7a> in <module>() ----> 1 get_ipython().magic('timeit a.remove(2)') /home/amit/anaconda3/lib/python3.4/site-packages/IPython/core/interactiveshell.py in magic(self, arg_s)

timeit versus timing decorator

由 …衆ロ難τιáo~ 提交于 2019-12-17 04:22:40
问题 I'm trying to time some code. First I used a timing decorator: #!/usr/bin/env python import time from itertools import izip from random import shuffle def timing_val(func): def wrapper(*arg, **kw): '''source: http://www.daniweb.com/code/snippet368.html''' t1 = time.time() res = func(*arg, **kw) t2 = time.time() return (t2 - t1), res, func.__name__ return wrapper @timing_val def time_izip(alist, n): i = iter(alist) return [x for x in izip(*[i] * n)] @timing_val def time_indexing(alist, n):

How to use timeit module

由 为君一笑 提交于 2019-12-16 19:54:46
问题 I understand the concept of what timeit does but I am not sure how to implement it in my code. How can I compare two functions, say insertion_sort and tim_sort , with timeit ? 回答1: The way timeit works is to run setup code once and then make repeated calls to a series of statements. So, if you want to test sorting, some care is required so that one pass at an in-place sort doesn't affect the next pass with already sorted data (that, of course, would make the Timsort really shine because it

How to use timeit module

由 为君一笑 提交于 2019-12-16 19:54:18
问题 I understand the concept of what timeit does but I am not sure how to implement it in my code. How can I compare two functions, say insertion_sort and tim_sort , with timeit ? 回答1: The way timeit works is to run setup code once and then make repeated calls to a series of statements. So, if you want to test sorting, some care is required so that one pass at an in-place sort doesn't affect the next pass with already sorted data (that, of course, would make the Timsort really shine because it

timeit function with random list

由 心不动则不痛 提交于 2019-12-13 18:34:06
问题 Trying to time how long it takes to sort a random list: import random import timeit randoms = random.sample(xrange(100), 10) print randoms timeit.timeit('sorted(r)',setup = 'r = random.sample(xrange(100), 10)') Error: Traceback (most recent call last): File "C:/Users/Arthur/Desktop/Dropbox/uni stuff/cs/python/theory hmwk/random.py", line 6, in <module> timeit.timeit('sorted(r)',setup = 'r = random.sample(xrange(100), 10)') File "C:\Python27\lib\timeit.py", line 230, in timeit return Timer

Python3 - Using timeit to measure script execution time externally

由 ￡可爱￡侵袭症＋ 提交于 2019-12-11 16:52:52
问题 Let's say I have a python script script.py and I want to measure the execution time of it just as if I put the whole script in quotation marks and used: from timeit import Timer t = Timer(stmt='THE WHOLE SCRIPT') print(t.timeit()) I could do this by editing script.py , but I'd like not to. What I want is a second script measure.py , which can be executed like ./measure.py script.py , which will basically do the above with any script as an argument. How could I approach writing such a script

Issues with using timeit in ipython

由 自古美人都是妖i 提交于 2019-12-11 05:27:53
问题 I was quickly trying to time 2 functions in ipython, m1() and m2() doing the same task with 2 different implementation. In [23]: %timeit for x in range(100): m1(a) 10000 loops, best of 3: 57.6 us per loop In [24]: %timeit for x in range(100): m2(a) 10000 loops, best of 3: 108 us per loop Result: the first implementation is almost 2x faster. So far, so good. Out of curiousity, I changed the range of the for loop above, and now I am at a loss making sense of the output. In [25]: %timeit for x

ipython %timeit “local variable 'a' referenced before assignment”

由 穿精又带淫゛_ 提交于 2019-12-10 17:53:28
问题 I am trying to the run the following code but I get a local variable 'a' referenced before assignment. a = [x for x in range(10)] b = [x for x in range(10)] %timeit a+=b The statement works without the %timeit magic. Is there something I am missing? Thank you. 回答1: What do you expect it to do? Outside of the timeit it does: In [188]: a += b In [189]: a Out[189]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9] I tried initializing x , and got a near infinite loop, eventually

IPython %timeit magic - changing output from “mean & std” to “best of 3”

由 限于喜欢 提交于 2019-12-06 06:35:19
问题 I am using an IPython 6.2.1 integration with Eclipse/PyDev on Ubuntu. Python version is 3.5.2. I often see people timing a script like >>>%timeit for _ in range(1000): True 10000 loops, best of 3: 37.8 µs per loop When I perform the same operation, my output is >>>%timeit for _ in range(1000): True 20.8 µs ± 353 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each) Imho, "best of 3" is the better measure, so I would like to change my output. I read both, the IPython and the Python timeit

IPython %timeit magic - changing output from “mean & std” to “best of 3”

由 送分小仙女□ 提交于 2019-12-04 14:18:15
I am using an IPython 6.2.1 integration with Eclipse/PyDev on Ubuntu. Python version is 3.5.2. I often see people timing a script like >>>%timeit for _ in range(1000): True 10000 loops, best of 3: 37.8 µs per loop When I perform the same operation, my output is >>>%timeit for _ in range(1000): True 20.8 µs ± 353 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each) Imho, "best of 3" is the better measure, so I would like to change my output. I read both, the IPython and the Python timeit documentation. They both don't even mention, that the output could differ from "best of 3". Is this a
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