例如: myStr = 'z'
是与myList = ['a','b','c']
配对,使得输出如下:的Python:配对的每一个元素的单一元素列表中的
['z','a']
['z','b']
['z','c']
一个班轮将是巨大的!
我试着这样做:
print zip([myStr, x] for x in myList)
但产量并不像我想,在这里:
[(['z', 'a'],), (['z', 'b'],), (['z', 'c'],)]
例如: myStr = 'z'
是与myList = ['a','b','c']
配对,使得输出如下:的Python:配对的每一个元素的单一元素列表中的
['z','a']
['z','b']
['z','c']
一个班轮将是巨大的!
我试着这样做:
print zip([myStr, x] for x in myList)
但产量并不像我想,在这里:
[(['z', 'a'],), (['z', 'b'],), (['z', 'c'],)]
试试这个:
myList = ['a','b','c']
myStr = 'z'
res = [[myStr, x] for x in myList]
Itertools- izip-longest
>>>from itertools import izip_longest
>>>[list(i) for i in list(izip_longest(['z'],['a','b','c'],fillvalue ='z'))]
>>>[['z', 'a'], ['z', 'b'], ['z', 'c']]
from itertools import izip_longest
list(izip_longest([], ['a','b','c'], fillvalue='z'))
随着zip:
>>> zip(myStr*3,myList)
[('z', 'a'), ('z', 'b'), ('z', 'c')]
也许'[[myStr中,x]中对于x在myList中]'? – JCOC611
删除'zip'调用。 – hjpotter92