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

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!

my_string = """Strings are amongst the most popular data types in Python. 
               We can create the strings by enclosing characters in quotes.
               Python treats single quotes the same as double quotes."""

Count = my_string.lower().strip("\n").split(" ").count("string")
Count = my_string.lower().strip("\n").split(" ").count("strings")
print("The number of occurance of word String is : " , Count)
print("The number of occurance of word Strings is : " , Count)

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

这个问题不是很清楚，但我可以回答你表面上的问题。

一个长度为L个字符的字符串S，其中S[1]是字符串的第一个字符，S[L]是最后一个字符，它有以下子字符串:

空字符串”。这里有一个。 对于从1到L的每一个值A，对于从A到L的每一个值B，字符串S[A]..S[B] (包容)。有L + L-1 + L-2 +…1个字符串，对于a 共0.5*L*(L+1)。 注意，第二项包括S[1]..S[L]， 即整个原始字符串S。

所以，在长度为L的字符串中有0.5*L*(L+1) +1个子字符串，在Python中呈现这个表达式，你就有了字符串中出现的子字符串的数量。