python:遍历具有列表值的字典

Question

给定一个列表字典，例如

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

哪个更pythonic或更可取:

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

或者还有什么需要考虑的？

编辑，如果列表可能有用(例如，标准字典不保留顺序)，这可能是合适的，尽管它要慢得多。

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

Answer 1

这是一个速度测试，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}