Python字符计数程序报IndexError，请求问题排查

问题概述 #

刚学完Python基础，正在做字符统计任务：用字母列表统计句子中对应字符的数量。当前程序存在两个问题：

1. 计数异常（比如句子里只有1个c，却统计出30）
2. 运行到字符c时抛出索引错误：

File "/home/pi/main.py", line 17, in <module>
    if statementLower[l] == alphabet[m]:
IndexError: string index out of range

当前输出仅：

a:44
b:20
c:30

随后触发上述错误。

原代码 #

alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
statement = "This is my task I want to complete. But I am stuck!!!!!!"
statementLower = statement.lower()
m = 0
l=0
count = 0
print(statementLower)

for i in alphabet:
    frequency = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
    #print("i for alphabet")
    #print("-" + i + "-")
    for k in statementLower:
        #print("k statement Lower")
        #print(k)
        if statementLower[l] == alphabet[m]:
            #print(count)
            count = count+1
            frequency[m] = count
            l+l+1
            continue
        else:
            #print("else continue loop")
            l=l+1
            continue
        
    if frequency[m] != 0:
        #print("frequency loop")
        print(str(alphabet[m]+ ":" + str(frequency[m])))
        m=m+1
        count = 0
        l=0
        continue

错误原因分析 #

1. 索引越界：
   内层循环遍历整个字符串时，手动用l递增索引，遍历完一轮后l等于字符串长度。且只有当字母存在时才重置l=0，若字母不存在，l不会重置，下一轮统计时statementLower[l]必然超出索引范围。
2. 计数逻辑混乱：
   • l+l+1是无效语句，应该写成l += 1，导致匹配到目标字符时l不递增，重复统计同一个位置的字符，造成计数异常。
   • 外层循环遍历alphabet元素，却用m控制当前统计字母，逻辑冗余；内层判断应该直接用当前遍历的i，而非alphabet[m]。
   • frequency数组完全多余，单个变量即可记录当前字母计数。

修复后的代码 #

基础遍历版 #

alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
statement = "This is my task I want to complete. But I am stuck!!!!!!"
statementLower = statement.lower()

for char in alphabet:
    count = 0
    for s_char in statementLower:
        if s_char == char:
            count += 1
    if count > 0:
        print(f"{char}: {count}")

简洁版（用内置count()方法） #

alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
statement = "This is my task I want to complete. But I am stuck!!!!!!"
statementLower = statement.lower()

for char in alphabet:
    count = statementLower.count(char)
    if count > 0:
        print(f"{char}: {count}")

正确输出 #

a: 4
b: 1
c: 1
e: 3
h: 1
i: 4
k: 2
m: 2
n: 1
o: 3
p: 1
s: 3
t: 5
u: 2
w: 1
y: 1

内容的提问来源于stack exchange，提问作者seb