Question

我想计算一个子串在字符串中“相邻”出现的次数，例如，给定字符串“8448442844”和子字符串“844”，我想要结果“2”。我知道string.count（substring）函数，在这个例子中，使用这个函数，我会得到结果“3”。但我希望结果为“2”。如何提前感谢，我怎样才能实现目标。

Answer 1

这适用于任何相邻的子字符串：

type = 13

还可以使用a = "8448442844" sol, sub = 0, "844" for i in range(1, a.count(sub)+1): if a.count(str(sub*i)): sol = i else: break print (sol) # 2和Comprehension转换为单行：

sum

Answer 2

使用str.count

>>> nStr = '000123000123'

>>> nStr.count('123')

2

或使用类似的东西

nStr = '000123000123'
pattern = '123'
count =0
flag=True
start=0
while flag:
    a = nStr.find(pattern,start)  # find() returns -1 if the word is not found, 
                                  #start i the starting index from the search starts(default value is 0)
    if a==-1:          #if pattern not found set flag to False
        flag=False
    else:               # if word is found increase count and set starting index to a+1
        count+=1        
        start=a+1
print(count)

Answer 3

你可以这样尝试。也许把它放在def中。它工作得很好

import string

s = "8448442844"
sub = "844"
max_sub = sub
count = 0

while string.find(s, max_sub) != -1:
    max_sub = max_sub + sub
    count +=1

print count

Answer 4

您可以使用str.split，itertools.groupby和max来获取结果：

In [1]: import itertools

In [2]: "8448442844".split("844")
Out[2]: ['', '', '2', '']

In [3]: [len(list(lst)) for key, lst in itertools.groupby("8448442844".split("844")) if key == ""]
Out[3]: [2, 1]

In [4]: max([len(list(lst)) for key, lst in itertools.groupby("8448442844".split("844")) if key == ""])
Out[4]: 2

确定子串在Python中的字符串中相邻出现的次数

4 个答案: