返回字符串“code”出現在給定字符串中任意位置的次數

Question

返回字符串"code"出現在給定字符串中任意位置的次數，除了我們接受'd'任何字母，因此"cope"和"cooe"計數。

我使用正則表達式通過以下代碼實現了這一點：

import re


def count_code(str):
    exp = '^co[a-z|A-Z]e$'
    count = 0
    for i in range(len(str) - 1):
        if re.match(exp, str[i:i + 4]):
            count = count + 1

    return count

print count_code('aaacodebbb')  # prints 1
print count_code('codexxcode')  # prints 2
print count_code('cozexxcope')  # prints 2

有沒有其他方法可以在不使用正則表達式的情況下實現這一目標？

Answer 1

一種方法是您可以使用 co*e 制作每個可能的字符串，其中 * 是任何字母

喜歡

x=["co"+i+"e" for i in string.lowercase]

然后迭代

for i in x:
    if i in <your string>:
        count+=<your string>.count(i)

Answer 2

你可以試試這個：

def count_code(str):
    x=["co"+i+"e" for i in str.lower()]
    count = 0
    index = 0
    for i in x:
        if i in str[index:]:
            index = str.find(i)+1
            count+=1
    return count

print count_code('aaacodebbb')  # prints 1
print count_code('codexxcode')  # prints 2
print count_code('cozexxcope')  # prints 2      

Answer 3

這是針對此問題的簡單而干凈的解決方案：

  def count_code(str):
      count = 0
      for i in range(len(str)):
        if str[i:i+2] == "co" and str[i+3:i+4] == "e":
          count+=1
      return count

Answer 4

def count_code(str):
  a = 0
  for i in range(len(str) - 3):
    if str[i:i+2] + str[i+3] == 'coe':
      a += 1
  return a

Answer 5

您也可以嘗試：使用 Python String Method 'count'

 def count_code1(str):
       counts=0
       for i in range(97,123):   #all the lowercase ASCII characters
        count+= str.count('co'+chr(i)+'e')
       return counts 

Answer 6

def count_code(str):
  code = 0
  for c in range(len(str)-1):
    if str[c+1] == 'o' and str[c:c+4:3] == 'ce':
      code+=1
  return code

Answer 7

def count_code(s):
  count=0
  for i in range(len(s)):
    if s[-(i+3):-(i+1)]=='co' and s[-i]=='e':
      count=count+1
  return count  

Answer 8

您可以以一種可以重復使用的方式定義您的邏輯 - 在這種情況下沒有計數或正則表達式

def count_code(str):
    start = 'co' #first 2 letter
    start1 = 'e' #last letter
    counter = 0 #initiate counter
    strlen=len(str) #for each word
    for i,x in enumerate(str):
        if str[i:i+2]==start: 
        #for each letter - is that letter and the next equal to start
            if len(str[i:strlen]) >=4: #is string long enough?
                if str[i+3]==start1: # if so is last letter right?
                    counter+=1
                else:
                    counter
    return counter

Answer 9

這也應該有效：

def count_code(str):
  counter = 0
  for i in range(len(str)-3):
    if str[i:i+2] == 'co' and str[i+3] == 'e':
      counter +=1
  return counter

希望能幫到你！

Answer 10

def count_code(str):
  a=''
  count=0
  for char in ("abcdefghijklmnopqrstuvwxyz"):
    a=char
    count+=str.count("co"+a+"e")
  return (count)

Answer 11

  a = 0
  for i in range(len(str)-3):
     if str[i:i+2] == 'co' and str[i+3] == 'e':
       a +=1
  return a

Answer 12

    def count_code(str):
  count = 0
  for i in range(len(str)-3):
    if str[i]=='c' and str[i+1] == 'o' and str[i+3]=='e':
      count+=1
  return count