这里有一个解决方案，适用于非重叠和重叠的情况。为了澄清:重叠子字符串是指其最后一个字符与其第一个字符相同的子字符串。

def substr_count(st, sub):
    # If a non-overlapping substring then just
    # use the standard string `count` method
    # to count the substring occurences
    if sub[0] != sub[-1]:
        return st.count(sub)

    # Otherwise, create a copy of the source string,
    # and starting from the index of the first occurence
    # of the substring, adjust the source string to start
    # from subsequent occurences of the substring and keep
    # keep count of these occurences
    _st = st[::]
    start = _st.index(sub)
    cnt = 0

    while start is not None:
        cnt += 1
        try:
            _st = _st[start + len(sub) - 1:]
            start = _st.index(sub)
        except (ValueError, IndexError):
            return cnt

    return cnt

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。

下面的逻辑将适用于所有字符串和特殊字符

def cnt_substr(inp_str, sub_str):
    inp_join_str = ''.join(inp_str.split())
    sub_join_str = ''.join(sub_str.split())

    return inp_join_str.count(sub_join_str)

print(cnt_substr("the sky is   $blue and not greenthe sky is   $blue and not green", "the sky"))

如果你正在寻找一个适用于所有情况的幂解，这个函数应该工作:

def count_substring(string, sub_string):
    ans = 0
    for i in range(len(string)-(len(sub_string)-1)):
        if sub_string == string[i:len(sub_string)+i]:
            ans += 1
    return ans

下面是Python 3中的解决方案，不区分大小写:

s = 'foo bar foo'.upper()
sb = 'foo'.upper()
results = 0
sub_len = len(sb)
for i in range(len(s)):
    if s[i:i+sub_len] == sb:
        results += 1
print(results)

这将列出字符串中所有出现的(也是重叠的)并对它们进行计数

def num_occ(str1, str2):
    l1, l2 = len(str1), len(str2)
    return len([str1[i:i + l2] for i in range(l1 - l2 + 1) if str1[i:i + l2] == str2])

例子:

str1 ='abcabcd'
str2 = 'bc'

将创建这个列表，但只保存BOLD值:

[ab, bc, ca, ab, bc, cd]

返回:

len([bc, bc])