如何从列表python中删除重复项[重复]

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Essentially i have to write a program that counts how many time each letter occurs in a string. I got that working but letters duplicate, for example. If the string was hello it would output:

基本上我必须编写一个程序来计算每个字母出现在字符串中的时间。例如，我得到了它，但字母重复。如果字符串是hello，它将输出：

When i need it to output:

当我需要输出时：

The way i've done it so far is without importing anything and my code looks like:

到目前为止我完成它的方式是没有导入任何东西，我的代码看起来像：

input_text = input('Enter some text: ')
s = input_text.lower()
length_s = len(s)
l = 0
while length_s > 0:
    if s[l].isalpha():
        print(s[l], s.count(s[l]))
        l += 1
        length_s -=1
    else:
        l += 1
        length_s -=1

2 个解决方案

#1

You can use a collections.Counter for your use case, it is a dict subclass for counting hashable objects (example - characters in a string) . Example -

您可以将collections.Counter用作您的用例，它是用于计算可哈希对象的dict子类（示例 - 字符串中的字符）。示例 -

>>> from collections import Counter
>>> c = Counter('hello')
>>> c
Counter({'l': 2, 'h': 1, 'o': 1, 'e': 1})
>>> for i,x in c.items():
...     print('{} - {}'.format(i,x))
...
l - 2
h - 1
o - 1
e - 1

Without using imports, you can use set() to store the characters that have already been seen. (This is done so that order is also maintained) -

不使用导入，您可以使用set（）来存储已经看过的字符。（这样做也是为了维持秩序） -

input_text = input('Enter some text: ')
s = input_text.lower()
seen = set()
for c in s:
     if c not in seen:
         print(c, s.count(c))
         seen.add(c)

#2

If you don't want to import anything you could do something like this to get a dictionary:

如果您不想导入任何内容，可以执行以下操作来获取字典：

s = input_text.lower()
countDict = {}
for char in s:
    if (char not in countDict):
        countDict[char] = 1
    else:
        countDict[char] += 1

#1