如何在 Python 中按元素连接两个列表？
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How to concatenate element-wise two lists in Python?
提问by Sanchit
I have two lists and I want to concatenate them element-wise. One of the list is subjected to string-formatting before concatenation.
我有两个列表，我想按元素连接它们。列表之一在连接之前进行字符串格式化。
For example :
例如 ：
a = [0, 1, 5, 6, 10, 11]
b = ['asp1', 'asp1', 'asp1', 'asp1', 'asp2', 'asp2']
In this case, ais subjected to string-formatting. That is, new aor aashould be :
在这种情况下，a受制于字符串格式。也就是说，新的a或aa应该是：
aa = [00, 01, 05, 06, 10, 11]
Final output should be :
最终输出应该是：
c = ['asp100', 'asp101', 'asp105', 'asp106', 'asp210', 'asp211']
Can somebody please tell me how to do that?
有人可以告诉我怎么做吗？
采纳答案by orlp
回答by Edgar Klerks
Than can be done elegantly with map and zip:
可以使用 map 和 zip 优雅地完成：
map(lambda (x,y): x+y, zip(list1, list2))
Example:
例子：
In [1]: map(lambda (x,y): x+y, zip([1,2,3,4],[4,5,6,7]))
Out[1]: [5, 7, 9, 11]
回答by Vorsprung
not using zip. I dunno, I think this is the obvious way to do it. Maybe I just learnt C first :)
不使用 zip。我不知道，我认为这是显而易见的方法。也许我只是先学了 C :)
c=[]
for i in xrange(len(a)):
c.append("%s%02d" % (b[i],a[i]))
回答by RMcG
Using zip
使用 zip
[m+str(n) for m,n in zip(b,a)]
output
输出
['asp10', 'asp11', 'asp15', 'asp16', 'asp210', 'asp211']
回答by MONTYHS
b = ['asp1', 'asp1', 'asp1', 'asp1', 'asp2', 'asp2']
aa = [0, 1, 5, 6, 10, 11]
new_list =[]
if len(aa) != len(b):
print 'list length mismatch'
else:
for each in range(0,len(aa)):
new_list.append(b[each] + str(aa[each]))
print new_list
回答by vaab
Other solution (preferring printf formatingstyle over .format()usage), it's also smaller:
其他解决方案（更喜欢printf 格式样式而不是.format()使用），它也更小：
>>> ["%s%02d" % t for t in zip(b, a)]
['asp100', 'asp101', 'asp105', 'asp106', 'asp210', 'asp211']
回答by IndPythCoder
inputs:
输入：
a = [0, 1, 5, 6, 10, 11]
b = ['asp1', 'asp1', 'asp1', 'asp1', 'asp2', 'asp2']
concat_func = lambda x,y: x + "" + str(y)
list(map(concat_func,b,a)) # list the map function
output:
输出：
['asp10', 'asp11', 'asp15', 'asp16', 'asp210', 'asp211']
回答by Veneet Reddy
If you wanted to concatenate arbitrary number of lists, you could do this:
如果你想连接任意数量的列表，你可以这样做：
In [1]: lists = [["a", "b", "c"], ["m", "n", "o"], ["p", "q", "r"]] # Or more
In [2]: lists
Out[2]: [['a', 'b', 'c'], ['m', 'n', 'o'], ['p', 'q', 'r']]
In [4]: list(map("".join, zip(*lists)))
Out[4]: ['amp', 'bnq', 'cor']
