用Python將兩個列表轉(zhuǎn)換為字典想象一下你有：keys = ['name', 'age', 'food']values = ['Monty', 42, 'spam']制作以下詞典的最簡單方法是什么？a_dict = {'name' : 'Monty', 'age' : 42, 'food' : 'spam'}
4 回答
繁花如伊
TA貢獻(xiàn)2012條經(jīng)驗 獲得超12個贊
試試這個：
>>> import itertools
>>> keys = ('name', 'age', 'food')
>>> values = ('Monty', 42, 'spam')
>>> adict = dict(itertools.izip(keys,values))
>>> adict
{'food': 'spam', 'age': 42, 'name': 'Monty'}
在Python 2中，與zip.
滄海一幻覺
TA貢獻(xiàn)1824條經(jīng)驗 獲得超5個贊
想象一下你有： keys = ('name', 'age', 'food')values = ('Monty', 42, 'spam')制作以下詞典的最簡單方法是什么？
dict = {'name' : 'Monty', 'age' : 42, 'food' : 'spam'}
最具表現(xiàn)力的-Python2.7和3，DECT理解：
new_dict = {k: v for k, v in zip(keys, values)}zipizip
from itertools import izip as zip
new_dict = {k: v for k, v in zip(keys, values)}Python 2，適合<=2.6
izipitertoolszipizip
from itertools import izip new_dict = dict(izip(keys, values))
Python 3
zipitertools
new_dict = dict(zip(keys, values))
所有案件的結(jié)果：
>>> new_dict{'age': 42, 'name': 'Monty', 'food': 'spam'}說明：
dict
>>> help(dict)class dict(object)
| dict() -> new empty dictionary | dict(mapping) -> new dictionary initialized from a mapping object's
| (key, value) pairs
| dict(iterable) -> new dictionary initialized as if via:
| d = {}
| for k, v in iterable:
| d[k] = v
| dict(**kwargs) -> new dictionary initialized with the name=value pairs
| in the keyword argument list. For example: dict(one=1, two=2)>>> zip(keys, values)[('name', 'Monty'), ('age', 42), ('food', 'spam')]>>> list(zip(keys, values))[('name', 'Monty'), ('age', 42), ('food', 'spam')]zip
>>> zip(keys, values)<zip object at 0x7f0e2ad029c8>
zip
績效較差的替代辦法：
generator_expression = ((k, v) for k, v in zip(keys, values))dict(generator_expression)
dict((k, v) for k, v in zip(keys, values))
dict([(k, v) for k, v in zip(keys, values)])
業(yè)績審查：
>>> min(timeit.repeat(lambda: {k: v for k, v in zip(keys, values)}))
0.7836067057214677
>>> min(timeit.repeat(lambda: dict(zip(keys, values))))
1.0321204089559615
>>> min(timeit.repeat(lambda: {keys[i]: values[i] for i in range(len(keys))}))
1.0714934510178864
>>> min(timeit.repeat(lambda: dict([(k, v) for k, v in zip(keys, values)])))
1.6110592018812895
>>> min(timeit.repeat(lambda: dict((k, v) for k, v in zip(keys, values))))
1.7361853648908436
不負(fù)相思意
TA貢獻(xiàn)1777條經(jīng)驗 獲得超10個贊
>>> keys = ('name', 'age', 'food')
>>> values = ('Monty', 42, 'spam')
>>> dict(zip(keys, values))
{'food': 'spam', 'age': 42, 'name': 'Monty'}
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