2015-12-02 149 views
2

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'],)] 
+0

也许'[[myStr中,x]中对于x在myList中]'? – JCOC611

+0

删除'zip'调用。 – hjpotter92

回答

4

试试这个:

myList = ['a','b','c'] 
myStr = 'z' 
res = [[myStr, x] for x in myList] 
0

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']] 
0
from itertools import izip_longest 
list(izip_longest([], ['a','b','c'], fillvalue='z')) 
1

随着zip

>>> zip(myStr*3,myList) 
[('z', 'a'), ('z', 'b'), ('z', 'c')] 
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