Question

我想计算字符串s中子串“bob”的出现次数。我为edX课程做了这个练习。

AVCaptureVideoDataOutputSampleBufferDelegate

然而，似乎s [position + 1：position + 2]语句无法正常工作。我如何在“b”后面加上两个字符？

Answer 1

不包括第二个切片索引。这意味着s[position+1:position+2]是位置position + 1的单个字符，并且此子字符串不能等于ob。查看相关的answer。您需要[:position + 3]：

s = 'azcbobobegghakl'
counter = 0
numofiterations = len(s)
position = 0

#loop that goes through the string char by char
for iteration in range(numofiterations - 2):
    if s[position] == "b":                  # search  pos. for starting point
        if s[position+1:position+3] == "ob":    # check if complete
            counter += 1        
    position +=1

print("Number of times bob occurs is: " + str(counter))
# 2

Answer 2

您可以将.find与索引一起使用：

s = 'azcbobobegghakl'

needle = 'bob'

idx = -1; cnt = 0
while True:
    idx = s.find(needle, idx+1)
    if idx >= 0:
        cnt += 1
    else:
        break

print("{} was found {} times.".format(needle, cnt))
# bob was found 2 times.

Answer 3

Eric's answer完全解释了为什么你的方法不起作用（在Python中切片是最终排他的），但让我提出另一个选择：

s = 'azcbobobegghakl'
result = sum(1 for ss in [s[i:] for i in range(0, len(s))] if ss.startswith("bob"))

或只是

{{1}}

如何使用切片表示法计算特定子字符串

3 个答案: