需要帮助搞清楚什么是错我的Python代码

Question

我现在用的是如何看待一个计算机科学家互动版，和练习一个具有以下要求学习的Python：

“分配给一个变量在你的编写一个三重引号的字符串，其中包含您最喜欢的文本段落 - 可能是一首诗，演讲，烘烤蛋糕的指示，一些励志诗句等。

编写一个函数来计算字母字符数z或A到Z），然后跟踪字母'e'的数量。您的函数应打印如下文本的分析：

您的文本包含243个字母字符，其中109个（44.8％）是'e'。“

我写的代码（对我）似乎在做我刚才被问到的代码，但是当我检查他们的解决方案来测试我的代码时，我得到了不同的结果。

我的代码：

text = ''' "If the automobile had followed the same development cycle as the computer, a 
Rolls-Royce would today cost $100, get a million miles per gallon, and explode 
once a year, killing everyone inside." 
-Robert Cringely''' 

lowercase_text = text.lower() 

def charCounter(some_text): 
    e_counter = 0 
    char_counter = 0 

    for char in lowercase_text:  
     if char == 'e': 
      e_counter = e_counter + 1 
     else: 
      char_counter = char_counter + 1 

    return ("Your text contains " + str(char_counter) + " alphabetic characters, of which " + str(e_counter) + " (" + str((e_counter/char_counter) * 100) + "%)" + "are 'e'.") 

我的代码输出：通过笔者的解决方案

def count(p): 
    lows="abcdefghijklmnopqrstuvwxyz" 
    ups="ABCDEFGHIJKLMNOPQRSTUVWXYZ" 

    numberOfe = 0 
    totalChars = 0 
    for achar in p: 
     if achar in lows or achar in ups: 
      totalChars = totalChars + 1 
      if achar == 'e': 
       numberOfe = numberOfe + 1 


    percent_with_e = (numberOfe/totalChars) * 100 
    print("Your text contains", totalChars, "alphabetic characters of which", numberOfe, "(", percent_with_e, "%)", "are 'e'.") 


p = '''"If the automobile had followed the same development cycle as the computer, a 
Rolls-Royce would today cost $100, get a million miles per gallon, and explode 
once a year, killing everyone inside." 
-Robert Cringely''' 

count(p) 

代码输出：由作者提供

Your text contains 188 alphabetic characters, of which 25 (13.297872340425531%)are 'e'. 

解决方案代码

Your text contains 166 alphabetic characters of which 25 (15.060240963855422 %) are 'e'. 

有人可以请解释我做错了什么？我不明白为什么结果会有这种差异。

Answer 1

您的解决方案不检查一个字符是否确实是字母数字并且也计算空白。另外，'e'不会被添加到总字符数中。

的问题是在你的for循环：

for char in lowercase_text:  
    if char == 'e': 
     e_counter = e_counter + 1 
    else: 
     char_counter = char_counter + 1 

它应该是这样的：

for char in lowercase_text:  
    # Check if we have an alphanumeric string and continue the loop if not 
    if not char.isalpha(): 
     continue 
    # Increment the total character counter 
    char_counter += 1 
    # Additionaly, increment the 'e' counter if we have an 'e' 
    if char == 'e': 
     e_counter += 1 

Answer 2

你在计数中包括标点符号，数字和空格，你不应该这样做。

Answer 3

我有没有办法解决的，它已经解决了这个问题（没有downvotes请），我只是想提出一个更Python的方式来解决这个问题（没有任何需要upvotes）：

import string 

text = ''' "If the automobile had followed the same development cycle as the computer, a 
Rolls-Royce would today cost $100, get a million miles per gallon, and explode 
once a year, killing everyone inside." 
-Robert Cringely''' 

def check_character(text, character): 
    text = text.lower() 
    count_sum = len(list(c for c in text if c in string.ascii_lowercase)) 
    count_char = len(list(c for c in text if c == character)) 
    return count_sum, count_char, 100 * count_char/float(count_sum) 

char = 'e' 
result = check_character(text, char) + (char,) 

print("Your text contains {} alphabetic characters of which {} ({:.2f}%) are '{}'.".format(*result))