python:遍历具有列表值的字典
问题描述
给定一个列表字典,例如
Given a dictionary of lists, such as
d = {'1':[11,12], '2':[21,21]}
哪个更pythonic或更可取:
Which is more pythonic or otherwise preferable:
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?
or is there something else to consider?
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.
EDIT, in case a list might be useful (e.g., standard dicts don't preserve order), this might be appropriate, although it's much slower.
d2 = d.items()
for k in d2:
for x in d2[1]:
# whatever with k, x
解决方案
这是一个速度测试,为什么不呢:
Here's a speed test, why not:
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:
cProfile example output:
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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