计算字符串中字母出现的频率 (Python)

Counting the Frequency of Letters in a string (Python)

所以我试图在不使用 python 词典的情况下计算用户输入字符串中字母的频率... 我希望输出如下(以字母H为例)

"The letter H occurred 1 time(s)." 

我遇到的问题是程序按字母顺序打印字母 但我希望程序按照输入中给定的顺序打印字母的频率...

例如,如果我输入 "Hello" 该程序将打印

"The letter e occurred 1 time(s)"
"The letter h occurred 1 time(s)"
"The letter l occurred 2 time(s)"
"The letter o occurred 1 time(s)"

但我希望输出如下

"The letter h occurred 1 time(s)"
"The letter e occurred 1 time(s)"
"The letter l occurred 2 time(s)"
"The letter o occurred 1 time(s)"

这是我目前拥有的代码:

originalinput = input("")
if originalinput == "":
    print("There are no letters.")
else:
  word = str.lower(originalinput)

  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']

  for i in range(0,26): 
    if word.count(Alphabet[i]) > 0:
      print("The letter", Alphabet[i] ,"occured", word.count(Alphabet[i]), "times")

您可以使用 if else 检查输入,如果需要,可以使用自定义消息 raise 检查错误

if original_input == "":
    raise RuntimeError("Empty input")
else:
    # Your code goes there

作为方不,input()就够了,不用加引号""

编辑:此问题已编辑,原问题是检查输入是否为空。

第二次编辑:

如果您希望您的代码在输出中打印字母,您应该遍历单词而不是字母表:

for c in word:
     print("The letter", c ,"occured", word.count(c), "times")

正如@BlueSheepToken 提到的,您可以使用简单的 if else 语句。下面是您的代码,其中包含提到的内容。

from collections import defaultdict

originalinput = input()

if originalinput == "":
    raise RuntimeError("Empty input")
else:

    result = defaultdict(int)

    for char in list(originalinput.lower()):
        result[char] += 1

    for letter, num in result.items():
        print(f'The letter {letter} occurred {num} time(s)')  

    #Hello
    #The letter h occurred 1 time(s)
    #The letter e occurred 1 time(s)
    #The letter l occurred 2 time(s)
    #The letter o occurred 1 time(s)

在这种情况下使用 defaultdict 会有所帮助,因为所有未使用的 key 都将是 0.

我建议使用集合库中的计数器。

    from collections import Counter

    print(Counter('sample text'))

    #Output: Counter({'t': 2, 'e': 2, 'l': 1, 's': 1, 'a': 1, ' ': 1, 'p': 1, 'm': 1, 'x': 1})