我有一个这样的清单:
a = ["a","b","c","d","e"]
如何在列表中的每个项目的末尾插入一个单词并将其作为列表本身?
预期产出:
a = [("a","TEST"),("b","TEST"),("c","TEST"),("d","TEST"),("e","TEST")]
我尝试了很多方法,但没有运气。
答案 0 :(得分:9)
创建一个包含项目和所需单词的元组:
word = "TEST"
a = ["a","b","c","d","e"]
new_a = [(i, word) for i in a]
输出:
[('a', 'TEST'), ('b', 'TEST'), ('c', 'TEST'), ('d', 'TEST'), ('e', 'TEST')]
或者,使用map
:
new_a = list(map(lambda x: (x, "TEST"), a))
答案 1 :(得分:2)
另一种方法是使用itertools
模块中的zip_longest
,如下所示(但是在列表解析中使用元组的解决方案应该是Ajax1234's):
from itertools import zip_longest
l = ['a', 'b', 'c', 'd', 'e']
res = list(zip_longest(l, ['TEST'], fillvalue='TEST'))
<强>输出:强>
>>> res
[('a', 'TEST'), ('b', 'TEST'), ('c', 'TEST'), ('d', 'TEST'), ('e', 'TEST')]
答案 2 :(得分:1)
您可以使用zip
:
a = ["a","b","c","d","e"]
new_a = zip(a, ["TEST"] * len(a))
输出:
[('a', 'TEST'), ('b', 'TEST'), ('c', 'TEST'), ('d', 'TEST'), ('e', 'TEST')]
答案 3 :(得分:1)
不是很时髦,但与其他常见语言类似:
a = ["a","b","c","d","e"]
b = []
for x in a:
b.append ( (x, 'TEST'))
a = b
print a