j = 0
    while i < len(string):
        sub_string_out = string[i:len(sub_string)+j]
        if sub_string == sub_string_out:
            count += 1
        i += 1
        j += 1
    return count

重叠出现:

def olpcount(string,pattern,case_sensitive=True):
    if case_sensitive != True:
        string  = string.lower()
        pattern = pattern.lower()
    l = len(pattern)
    ct = 0
    for c in range(0,len(string)):
        if string[c:c+l] == pattern:
            ct += 1
    return ct

test = 'my maaather lies over the oceaaan'
print test
print olpcount(test,'a')
print olpcount(test,'aa')
print olpcount(test,'aaa')

结果:

my maaather lies over the oceaaan
6
4
2

import re
d = [m.start() for m in re.finditer(pattern, string)] 
print(d)

这将查找子字符串在字符串中被找到的次数并显示索引。

string="abc"
mainstr="ncnabckjdjkabcxcxccccxcxcabc"
count=0
for i in range(0,len(mainstr)):
    k=0
    while(k<len(string)):
        if(string[k]==mainstr[i+k]):
            k+=1
        else:
            break   
    if(k==len(string)):
        count+=1;   
print(count)

s = 'arunununghhjj'
sb = 'nun'
results = 0
sub_len = len(sb)
for i in range(len(s)):
    if s[i:i+sub_len] == sb:
        results += 1
print results

根据你的真正意思，我提出以下解决方案:

你的意思是一个空格分隔子字符串的列表，并想知道所有子字符串中的子字符串位置编号是什么: S = 'sub1 sub2 sub3' s.split () .index(“sub2”) > > > 1 你的意思是子字符串在字符串中的char-position: s.find(“sub2”) > > > 5 你的意思是su-bstring的(非重叠)外观计数: s.count(“sub2”) > > > 1 s.count(“子”) > > > 3