python:通过带有列表值的字典迭代?

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给定一个列表字典,如

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

这是更pythonic或其他更好:

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

要么

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

还是有其他要考虑的事情?

d2 = d.items()
for k in d2:
        for x in d2[1]:
            # whatever with k, x
提问于
用户回答回答于

我的结果来自Brionius代码:

         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}
用户回答回答于

这是一个速度测试:

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}

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