我将把我接受的答案作为“简单而明显的方法”，然而，它不包括重叠的情况。 通过多次检查切片就可以简单地找出这些问题——比如:

sum("GCAAAAAGH"[i:].startswith("AAA") for i in range(len("GCAAAAAGH")))

结果是3。

或者可以通过使用正则表达式来实现，如如何使用regex来查找所有重叠的匹配项中所示——它也可以用于良好的代码高尔夫。

这是我的“手工”计数模式重叠出现在一个字符串，它试图不是非常幼稚(至少它不会在每次交互时创建新的字符串对象):

def find_matches_overlapping(text, pattern):
    lpat = len(pattern) - 1
    matches = []
    text = array("u", text)
    pattern = array("u", pattern)
    indexes = {}
    for i in range(len(text) - lpat):
        if text[i] == pattern[0]:
            indexes[i] = -1
        for index, counter in list(indexes.items()):
            counter += 1
            if text[i] == pattern[counter]:
                if counter == lpat:
                    matches.append(index)
                    del indexes[index]
                else:
                    indexes[index] = counter
            else:
                del indexes[index]
    return matches
            
def count_matches(text, pattern):
    return len(find_matches_overlapping(text, pattern))

在Python 3中，要查找字符串中子字符串的重叠情况，该算法将执行以下操作:

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

我亲自检查了这个算法，它是有效的。

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

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

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

场景1:句子中出现一个单词。 str1 =“这是一个例子，很简单”。单词“is”的出现。让str2 = "is"

count = str1.count(str2)

场景二:句子中出现句式。

string = "ABCDCDC"
substring = "CDC"

def count_substring(string,sub_string):
    len1 = len(string)
    len2 = len(sub_string)
    j =0
    counter = 0
    while(j < len1):
        if(string[j] == sub_string[0]):
            if(string[j:j+len2] == sub_string):
                counter += 1
        j += 1

    return counter

谢谢!