函数在所有实例中查找（）

Question

你能修好这个密码吗。我试着用字符串。查找（）查找给定范围（i到j）内的所有病例

def multi_find(seq, subseq, i, j):
list = []
while i != -1 and i < j:
    list.append(seq.find(subseq,i,j))
    i = seq.find(subseq,i,j) + 1
return list


print multi_find(seq, subseq, i, j)

Answer 1

def multi_find(seq, subseq, i, j):
    lst = []
    loc = seq.find(subseq, i, j) # location of subseq

    while loc != -1:
        lst.append(loc)
        i = loc + 1
        loc = seq.find(subseq, i, j)
    return lst

我认为使用i，j作为在seq中搜索的范围，并使用另一个变量loc来保留搜索结果更为清晰。你知道吗

Answer 2

像这样的事情可能更接近你想要做的事情。你知道吗

import sys

def multi_find(seq, subseq, i, j):
    list = []
    for x in xrange(j - i):
        temp = seq.find(subseq, x + i, j)
        if(temp != -1):
            list.append(temp)
            x += sys.getsizeof(subseq)
        else:
            x += 1
    return list

它运行ad返回子字符串中的令牌数，但我不太确定如何增加I