2015-12-02 155 views
2

例如: myStr = 'z'是與myList = ['a','b','c']配對,使得輸出如下:的Python:配對的每一個元素的單一元素列表中的

['z','a'] 
['z','b'] 
['z','c'] 

一個班輪將是巨大的!

我試着這樣做:

print zip([myStr, x] for x in myList) 

但產量並不像我想,在這裏:

[(['z', 'a'],), (['z', 'b'],), (['z', 'c'],)] 
+0

也許'[[myStr中,x]中對於x在myList中]'? – JCOC611

+0

刪除'zip'調用。 – hjpotter92

回答

4

試試這個:

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

隨着zip

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