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!

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)

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

#counting occurence of a substring in another string (overlapping/non overlapping)
s = input('enter the main string: ')# e.g. 'bobazcbobobegbobobgbobobhaklpbobawanbobobobob'
p=input('enter the substring: ')# e.g. 'bob'

counter=0
c=0

for i in range(len(s)-len(p)+1):
    for j in range(len(p)):
        if s[i+j]==p[j]:
            if c<len(p):
                c=c+1
                if c==len(p):
                    counter+=1
                    c=0
                    break
                continue
        else:
            break
print('number of occurences of the substring in the main string is: ',counter)

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

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

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

目前涉及方法计数的最佳答案并不能真正计算重叠出现的次数，也不关心空子字符串。 例如:

>>> a = 'caatatab'
>>> b = 'ata'
>>> print(a.count(b)) #overlapping
1
>>>print(a.count('')) #empty string
9

如果我们考虑重叠的子字符串，第一个答案应该是2而不是1。 对于第二个答案，如果空子字符串返回0作为asnwer会更好。

下面的代码处理这些事情。

def num_of_patterns(astr,pattern):
    astr, pattern = astr.strip(), pattern.strip()
    if pattern == '': return 0

    ind, count, start_flag = 0,0,0
    while True:
        try:
            if start_flag == 0:
                ind = astr.index(pattern)
                start_flag = 1
            else:
                ind += 1 + astr[ind+1:].index(pattern)
            count += 1
        except:
            break
    return count

现在当我们运行它时:

>>>num_of_patterns('caatatab', 'ata') #overlapping
2
>>>num_of_patterns('caatatab', '') #empty string
0
>>>num_of_patterns('abcdabcva','ab') #normal
2