python 解析zip函数和itertools.zip_longest函数
一、zip:用来使列表一一对应,该函数返回一个以元组为元素的列表。返回的列表长度为最短列表的长度
1、当只有一个参数时:
yc = ['a', 'b', 'c', 'd']
dd = list(zip(yc))
print(dd)
输出结果:
[('a',), ('b',), ('c',), ('d',)]
2、当有两个参数时:
yc = ['a', 'b', 'c', 'd']
num = [1, 2, 3]
dd = list(zip(yc, num))
print(dd)
输出结果:
[('a', 1), ('b', 2), ('c', 3)]
3、当有多个参数时:
yc = ['a', 'b', 'c', 'd']
num = [1, 2, 3]
name = ['yangchao', 'xiaman']
dd = list(zip(yc, num, name))
print(dd)
输出结果:
[('a', 1, 'yangchao'), ('b', 2, 'xiaman')]
二、zip_longest具体可以用来对列表的一一对应,如果列表的长度不一致,则其会选择最长的那个列表,并将没有的填充为None(这个可以自己传参时设置)
1、当只有一个参数时:
import itertools
yc = ['a', 'b', 'c', 'd']
num = [1, 2, 3]
name = ['yangchao', 'xiaman']
dic =[ll for ll in itertools.zip_longest(yc)]
print(dic)
输出结果:
[('a',), ('b',), ('c',), ('d',)]
2、当有两个参数时:
import itertools
yc = ['a', 'b', 'c', 'd']
num = [1, 2, 3]
dic1 =[ll for ll in itertools.zip_longest(yc, num)]
print(dic1)
dic2 ={ll:nn for ll,nn in itertools.zip_longest(yc, num)}##这里可以对指定元组中的每一个元素进行操作
print(dic2)
输出结果:
[('a', 1), ('b', 2), ('c', 3), ('d', None)]
{'a': 1, 'b': 2, 'c': 3, 'd': None}
3、当有多个参数时:
import itertools
yc = ['a', 'b', 'c', 'd']
num = [1, 2, 3]
name = ['yangchao', 'xiaman']
dic =[ll for ll in itertools.zip_longest(yc, num, name)]
print(dic)
输出结果:
[('a', 1, 'yangchao'), ('b', 2, 'xiaman'), ('c', 3, None), ('d', None, None)]
三、最后 附上itertools.zip_longest源码:
def itertools.zip_longest(*args, fillvalue=None):
# itertools.zip_longest('ABCD', 'xy', fillvalue='-') --> Ax By C- D-
iterators = [iter(it) for it in args]
num_active = len(iterators)
if not num_active:
return
while True:
values = []
for i, it in enumerate(iterators):
try:
value = next(it)
except StopIteration:
num_active -= 1
if not num_active:
return
iterators[i] = repeat(fillvalue)
value = fillvalue
values.append(value)
yield tuple(values)
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