def count_substring(string, sub_string):
    k=len(string)
    m=len(sub_string)
    i=0
    l=0
    count=0
    while l<k:
        if string[l:l+m]==sub_string:
            count=count+1
        l=l+1
    return count

if __name__ == '__main__':
    string = input().strip()
    sub_string = input().strip()

    count = count_substring(string, sub_string)
    print(count)

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)

如果你想计算所有的子字符串(包括重叠的)，那么使用这个方法。

import re
def count_substring(string, sub_string):
    regex = '(?='+sub_string+')'
    # print(regex)
    return len(re.findall(regex,string))

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

2+其他人已经提供了这个解决方案，我甚至投票了其中一个，但我的可能是新手最容易理解的。

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

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)