Question

给出列表字典，例如

d = {'1':[11,12], '2':[21,21]}

哪个更pythonic或更优选：

for k in d:
    for x in d[k]:
        # whatever with k, x

或

for k, dk in d.iteritems():
    for x in dk:
        # whatever with k, x

还是还有其他需要考虑的事情吗？

编辑，如果列表可能有用（例如，标准字母不保留顺序），这可能是合适的，尽管速度要慢得多。

d2 = d.items()
for k in d2:
        for x in d2[1]:
            # whatever with k, x

Answer 1

这是速度测试，为什么不呢：

import random
numEntries = 1000000
d = dict(zip(range(numEntries), [random.sample(range(0, 100), 2) for x in range(numEntries)]))

def m1(d):
    for k in d:
        for x in d[k]:
            pass

def m2(d):
    for k, dk in d.iteritems():
        for x in dk:
            pass

import cProfile

cProfile.run('m1(d)')

print

cProfile.run('m2(d)')

# Ran 3 trials:
# m1: 0.205, 0.194, 0.193: average 0.197 s
# m2: 0.176, 0.166, 0.173: average 0.172 s

# Method 1 takes 15% more time than method 2

cProfile示例输出：

         3 function calls in 0.194 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.194    0.194 <string>:1(<module>)
        1    0.194    0.194    0.194    0.194 stackoverflow.py:7(m1)
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}



         4 function calls in 0.179 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.179    0.179 <string>:1(<module>)
        1    0.179    0.179    0.179    0.179 stackoverflow.py:12(m2)
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
        1    0.000    0.000    0.000    0.000 {method 'iteritems' of 'dict' objects}

Answer 2

我考虑了几种方法：

import itertools

COLORED_THINGS = {'blue': ['sky', 'jeans', 'powerline insert mode'],
                  'yellow': ['sun', 'banana', 'phone book/monitor stand'],
                  'red': ['blood', 'tomato', 'test failure']}

def forloops():
    """ Nested for loops. """
    for color, things in COLORED_THINGS.items():
        for thing in things:
            pass

def iterator():
    """ Use itertools and list comprehension to construct iterator. """
    for color, thing in (
        itertools.chain.from_iterable(
            [itertools.product((k,), v) for k, v in COLORED_THINGS.items()])):
        pass

def iterator_gen():
    """ Use itertools and generator to construct iterator. """
    for color, thing in (
        itertools.chain.from_iterable(
            (itertools.product((k,), v) for k, v in COLORED_THINGS.items()))):
        pass

我使用ipython和memory_profiler来测试性能：

>>> %timeit forloops()
1000000 loops, best of 3: 1.31 µs per loop

>>> %timeit iterator()
100000 loops, best of 3: 3.58 µs per loop

>>> %timeit iterator_gen()
100000 loops, best of 3: 3.91 µs per loop

>>> %memit -r 1000 forloops()
peak memory: 35.79 MiB, increment: 0.02 MiB

>>> %memit -r 1000 iterator()
peak memory: 35.79 MiB, increment: 0.00 MiB

>>> %memit -r 1000 iterator_gen()
peak memory: 35.79 MiB, increment: 0.00 MiB

正如您所看到的，该方法对峰值内存使用没有可观察到的影响，但嵌套的for循环对于速度是无与伦比的（更不用说可读性）。

Answer 3

这是列表理解方法。嵌套...

r = [[i for i in d[x]] for x in d.keys()]
print r

[[11, 12], [21, 21]]

Answer 4

Brionius代码的结果：

         3 function calls in 0.173 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.173    0.173 <string>:1(<module>)
        1    0.173    0.173    0.173    0.173 speed.py:5(m1)
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Prof
iler' objects}


         4 function calls in 0.185 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.185    0.185 <string>:1(<module>)
        1    0.185    0.185    0.185    0.185 speed.py:10(m2)
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Prof
iler' objects}
        1    0.000    0.000    0.000    0.000 {method 'iteritems' of 'dict' obje
cts}

python：使用列表值迭代字典

4 个答案: