# Генерация целых чисел

def gen_prime(x):
    multiples = []
    results = []
    for i in range(2, x+1):
        if i not in multiples:
            results.append(i)
            for j in range(i*i, x+1, i):
                multiples.append(j)

    return results


import timeit

# Засекаем время

start_time = timeit.default_timer()
gen_prime(3000)
print(timeit.default_timer() - start_time)

# start_time = timeit.default_timer()
# gen_prime(1001)
# print(timeit.default_timer() - start_time)

# Генерация целых чисел

def gen_prime(x):
    multiples = []
    results = []
    for i in range(2, x+1):
        if i not in multiples:
            results.append(i)
            for j in range(i*i, x+1, i):
                multiples.append(j)

    return results


import timeit

# Засекаем время

start_time = timeit.default_timer()
gen_prime(3000)
print(timeit.default_timer() - start_time)

# start_time = timeit.default_timer()
# gen_prime(1001)
# print(timeit.default_timer() - start_time)

让我们在以下每个语句中设置相同的字典并测试执行时间。

setup参数基本上是设置字典

编号是运行代码1000000次。不是设置，而是stmt

当你运行这个时，你会发现index比get快得多。您可以多次运行它来查看。

这段代码基本上是试图从字典中获取c的值。

import timeit

print('Getting value of C by index:', timeit.timeit(stmt="mydict['c']", setup="mydict={'a':5, 'b':6, 'c':7}", number=1000000))
print('Getting value of C by get:', timeit.timeit(stmt="mydict.get('c')", setup="mydict={'a':5, 'b':6, 'c':7}", number=1000000))

这是我的结果，你的结果会有所不同。

按索引:0.20900007452246427

get: 0.54841166886888

我发现使用timeit最简单的方法是从命令行:

鉴于test.py:

def InsertionSort(): ...
def TimSort(): ...

像这样运行timeit:

% python -mtimeit -s'import test' 'test.InsertionSort()'
% python -mtimeit -s'import test' 'test.TimSort()'

如果你想快速比较两个代码/函数块，你可以这样做:

import timeit

start_time = timeit.default_timer()
func1()
print(timeit.default_timer() - start_time)

start_time = timeit.default_timer()
func2()
print(timeit.default_timer() - start_time)

我告诉你一个秘密:使用timeit的最好方法是在命令行上。

在命令行上，timeit执行适当的统计分析:它告诉您最短的运行花费了多长时间。这很好，因为所有的时间误差都是正的。所以最短的时间误差最小。没有办法得到负错误，因为计算机的计算速度永远不会超过它的计算速度!

那么，命令行界面:

%~> python -m timeit "1 + 2"
10000000 loops, best of 3: 0.0468 usec per loop

这很简单，是吧?

你可以设置:

%~> python -m timeit -s "x = range(10000)" "sum(x)"
1000 loops, best of 3: 543 usec per loop

这也很有用!

如果你想要多行，你可以使用shell的自动延续或者使用单独的参数:

%~> python -m timeit -s "x = range(10000)" -s "y = range(100)" "sum(x)" "min(y)"
1000 loops, best of 3: 554 usec per loop

这给出了一个

x = range(1000)
y = range(100)

和时间

sum(x)
min(y)

---

如果你想要更长的脚本，你可能会倾向于在Python脚本中计时。我建议避免这样做，因为在命令行上进行分析和计时会更好。相反，我倾向于编写shell脚本:

 SETUP="

 ... # lots of stuff

 "

 echo Minmod arr1
 python -m timeit -s "$SETUP" "Minmod(arr1)"

 echo pure_minmod arr1
 python -m timeit -s "$SETUP" "pure_minmod(arr1)"

 echo better_minmod arr1
 python -m timeit -s "$SETUP" "better_minmod(arr1)"

 ... etc

由于多次初始化，这可能需要更长的时间，但通常这不是一个大问题。

---

但是如果你想在你的模块中使用timeit呢?

那么，最简单的方法就是:

def function(...):
    ...

timeit.Timer(function).timeit(number=NUMBER)

这给了您运行该次数的累计时间(不是最小时间!)。

为了得到一个好的分析，使用.repeat并取最小值:

min(timeit.Timer(function).repeat(repeat=REPEATS, number=NUMBER))

您通常应该将其与functools结合使用。用Partial代替lambda:…降低开销。因此，你可以有这样的东西:

from functools import partial

def to_time(items):
    ...

test_items = [1, 2, 3] * 100
times = timeit.Timer(partial(to_time, test_items)).repeat(3, 1000)

# Divide by the number of repeats
time_taken = min(times) / 1000

---

你还可以:

timeit.timeit("...", setup="from __main__ import ...", number=NUMBER)

这将使您更接近命令行中的接口，但不那么酷。"from __main__ import…"允许你在timeit创建的人工环境中使用主模块中的代码。

值得注意的是，这是Timer(…).timeit(…)的方便包装，因此在计时方面不是特别好。我个人更喜欢使用Timer(…).repeat(…)，就像我上面展示的那样。

---

警告

有一些关于时间的警告无处不在。

Overhead is not accounted for. Say you want to time x += 1, to find out how long addition takes: >>> python -m timeit -s "x = 0" "x += 1" 10000000 loops, best of 3: 0.0476 usec per loop Well, it's not 0.0476 µs. You only know that it's less than that. All error is positive. So try and find pure overhead: >>> python -m timeit -s "x = 0" "" 100000000 loops, best of 3: 0.014 usec per loop That's a good 30% overhead just from timing! This can massively skew relative timings. But you only really cared about the adding timings; the look-up timings for x also need to be included in overhead: >>> python -m timeit -s "x = 0" "x" 100000000 loops, best of 3: 0.0166 usec per loop The difference isn't much larger, but it's there. Mutating methods are dangerous. >>> python -m timeit -s "x = [0]*100000" "while x: x.pop()" 10000000 loops, best of 3: 0.0436 usec per loop But that's completely wrong! x is the empty list after the first iteration. You'll need to reinitialize: >>> python -m timeit "x = [0]*100000" "while x: x.pop()" 100 loops, best of 3: 9.79 msec per loop But then you have lots of overhead. Account for that separately. >>> python -m timeit "x = [0]*100000" 1000 loops, best of 3: 261 usec per loop Note that subtracting the overhead is reasonable here only because the overhead is a small-ish fraction of the time. For your example, it's worth noting that both Insertion Sort and Tim Sort have completely unusual timing behaviours for already-sorted lists. This means you will require a random.shuffle between sorts if you want to avoid wrecking your timings.