如何找到项目在列表中出现的第n次索引? [英] How to find the index of the nth time an item appears in a list?
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问题描述
鉴于:
x = ['w','e','s','s','s ','z','z','s']
每次出现 s
出现在以下指数中:
第一名:2
第二名:3名
3rd:4
4th:7
如果我做 x.index('s')
我将获得第一个索引。
>>> x = ['w','e','s','s','s','z','z','s']
>>> [i for i,n in enumerate(x)if n =='s'] [0]
2
>>> [i for i,n in enumerate(x)if n =='s'] [1]
3
>>> [i for i,n in enumerate(x)if n =='s'] [2]
4
>>> [i for i,n in enumerate(x)if n =='s'] [3]
7
Given:
x = ['w', 'e', 's', 's', 's', 'z','z', 's']
Each occurrence of s
appears at the following indices:
1st: 2
2nd: 3
3rd: 4
4th: 7
If I do x.index('s')
I will get the 1st index.
How do I get the index of the 4th s
?
解决方案
Using list comprehension and enumerate
:
>>> x = [ 'w', 'e', 's', 's', 's', 'z','z', 's']
>>> [i for i, n in enumerate(x) if n == 's'][0]
2
>>> [i for i, n in enumerate(x) if n == 's'][1]
3
>>> [i for i, n in enumerate(x) if n == 's'][2]
4
>>> [i for i, n in enumerate(x) if n == 's'][3]
7
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