Python 代码中计算字符串中字母 'e' 数量的差异

一位初学者在学习 Python 时遇到一个练习题，要求将一段文本存储在变量中，并编写一个函数计算文本中字母 'e' 的数量。他写出了自己的代码，但结果与给定的解决方案不同。

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'.")

print(charCounter(lowercase_text))

使用以上代码得出的结果是：

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

而给定的解决方案代码如下：

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)

使用以上代码得出的结果是：

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

2. 解决方案

问题出在 ursprünglicher 代码的 for 循环中，它没有检查字符是否为字母数字，并且将空格也计算在内。此外，'e' 也没有计入总字符数。

修复代码如下所示：

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:    
        # 检查字符串是否为字母数字，如果不是，则继续循环
        if not char.isalpha():
            continue
        # 将总字符计数器加 1
        char_counter += 1
        # 如果字符为 'e'，则将 'e' 计数器加 1
        if char == 'e':
            e_counter += 1

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

print(charCounter(lowercase_text))

使用修复后的代码得出的结果是：

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

这与给定的解决方案代码的结果相同。