如何使用切片符號計算特定的子字符串

Question

我想要計算字符串s中子字符串「bob」的出現次數。我爲edX課程做這個練習。

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

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

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

但是，似乎s [position + 1：position + 2]語句無法正常工作。我如何處理「b」後面的兩個字符？

Answer 1

不包括第二個切片索引。這意味着s[position+1:position+2]是位置position + 1上的單個字符，並且此子字符串不能等於ob。查看相關的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

你可以使用.find與索引：

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解釋了爲什麼完美你的方法沒有工作（在Python切片是最終獨佔），但讓我提出另一種選擇：

s = 'azcbobobegghakl' 
substrings = [s[i:] for i in range(0, len(s))] 
filtered_s = filter(substrings, lambda s: s.startswith("bob")) 
result = len(filtered_s) 

或者乾脆

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