使用键列表从Python字典中收集值

Question

Dict = {'w1': 56, 'w2': 19, 'w3': 77, 'w4': 45, 'w5': 31}
myWords = ['w1','w4','w5']

OutputList=[]
for items in myWords:
    tmps = Dict[items]
    OutputList.append(tmps)

我的问题是,我们可以在不使用for循环的情况下从具有特定列表("myWord")的字典中收集值(输出)吗？

Answer 1

这operator.itemgetter是为了什么:

>>> import operator
>>> Dict = {'w1': 56, 'w2': 19, 'w3': 77, 'w4': 45, 'w5': 31}
>>> myWords = ['w1','w4','w5']
>>> operator.itemgetter(*myWords)(Dict)
[56, 45, 31]

Answer 2

您可以使用列表理解:

OutputList = [Dict[x] for x in myWords]

Answer 3

以下是几种不同方法的基准:

from __future__ import print_function
import timeit
from operator import itemgetter

def f1(d, l):
    '''map'''
    return list(map(d.get, l))

def f2(d, l):
    '''itemgetter'''
    return itemgetter(*l)(d)

def f3(d, l):
    '''list comprehension'''
    return [d[k] for k in l]

def f4(d, l):
    '''WRONG, but map and filter'''
    return list(map(lambda k: d[k], filter(d.get, l)))

def f5(d, l):
    '''simple for loop'''
    rtr=[]
    for e in l:
        rtr.append(d[e])
    return rtr  

def f6(d, l):
    '''CORRECTED map, filter '''    
    return list(map(lambda k: d[k], filter(d.__contains__, l))) 

if __name__ == '__main__':
    s=10000000
    d={'W{}'.format(k):k for k in range(s)} 
    l=['W{}'.format(x) for x in range(0,s,4)]

    times=[]                
    for f in (f1,f2,f3,f4,f5,f6):
        times.append((f.__doc__, timeit.timeit('f(d,l)', setup="from __main__ import f, d, l", number=10)))

    for e in sorted(times, key=itemgetter(1)):
         print('{:30}{:10.3f} seconds'.format(*e))

对于Python 2.7,打印:

itemgetter                         4.109 seconds
list comprehension                 4.467 seconds
map                                5.450 seconds
simple for loop                    6.132 seconds
CORRECTED map, filter             11.283 seconds
WRONG, but map and filter         11.852 seconds

Python 3.4:

itemgetter                         5.196 seconds
list comprehension                 5.224 seconds
map                                5.923 seconds
simple for loop                    6.548 seconds
WRONG, but map and filter          9.080 seconds
CORRECTED map, filter              9.931 seconds

PyPy:

list comprehension                 4.450 seconds
map                                4.718 seconds
simple for loop                    5.962 seconds
itemgetter                         7.952 seconds
WRONG, but map and filter          8.962 seconds
CORRECTED map, filter              9.909 seconds

你可以看到,即使使用类似大小的字典(1,000,000个元素)来表示OP的状态,一个简单的'for'循环与更高级的方法竞争.列表理解非常有竞争力.

你还可以看到看起来很花哨的东西并不那么棒.

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