修改

Question

保持问题尽可能通用 - 我有一个值列表，需要返回指定值的第n次出现的索引...

list = [0,0,1,1,0,1,0,0,0,1,0,1,1,0,1,0]
occurrence = 5
value = 1

所以我正在寻找值1发生的第5次索引。在上面的示例中，返回的值应为11。

'occurrence'和'值'是需要由用户设置的变量

修改

感谢帮助人员，一直在玩各种选项，想出了这个

valList = [0,0,1,1,0,1,0,0,0,1,0,1,1,0,1,0]
occurrence = 5
value = 1

count = 0
rowIndex = -1

for val in valList:
    if count < occurrence:
        rowIndex += 1
        if val == valCR:
            count += 1

Answer 1

>>> mylist = [0,0,1,1,0,1,0,0,0,1,0,1,1,0,1,0]
>>> occurrence = 5
>>> value = 1
>>> [i for i,elem in enumerate(mylist) if elem==value][occurrence-1]
11

或者，itertools方式：

>>> next(itertools.islice((i for i,elem in enumerate(mylist) if elem==value), occurrence-1, occurrence))
11

Answer 2

这是一个有效的版本：

def nth_index(sequence, n, value):
    i = -1
    for _ in xrange(n):
        i = sequence.index(value, i + 1)
    return i

 lst = [0,0,1,1,0,1,0,0,0,1,0,1,1,0,1,0]
 print nth_index(lst, 5, 1) # 11

此行为类似于内置index，并在找不到值或n太大时引发ValueError：

 print nth_index(lst, 55, 1) # ValueError: 1 is not in list

FTR，时间：

import timeit, itertools

lst = range(1000) * 5
value = 0
occurrence = 5

print timeit.timeit(lambda: [i for i,elem in enumerate(lst) if elem==value][occurrence-1], number=1000)
print timeit.timeit(lambda: next(itertools.islice((i for i,elem in enumerate(lst) if elem==value), occurrence-1, occurrence)), number=1000)
print timeit.timeit(lambda: nth_index(lst, occurrence, value), number=1000)

-

0.52188706398
0.416564941406
0.117225885391

如何返回给定值第n次出现的索引号？

修改

2 个答案: