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"))

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

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

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

对于重叠计数，我们可以使用use:

def count_substring(string, sub_string):
    count=0
    beg=0
    while(string.find(sub_string,beg)!=-1) :
        count=count+1
        beg=string.find(sub_string,beg)
        beg=beg+1
    return count

对于非重叠的情况，我们可以使用count()函数:

string.count(sub_string)

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

根据你的真正意思，我提出以下解决方案:

你的意思是一个空格分隔子字符串的列表，并想知道所有子字符串中的子字符串位置编号是什么: S = 'sub1 sub2 sub3' s.split () .index(“sub2”) > > > 1 你的意思是子字符串在字符串中的char-position: s.find(“sub2”) > > > 5 你的意思是su-bstring的(非重叠)外观计数: s.count(“sub2”) > > > 1 s.count(“子”) > > > 3