Question

我正在尝试创建一个程序来计算字符串中字母的出现次数，但是我不知道如何在保留原始字母的情况下使其不区分大小写。

我尝试将每个项目分配给字典，其键为字符串中出现的次数，但是，例如，如果我说abA，它将把A和a计为不同的字母。

import operator

def first_non_repeating_letter(string):
    string = string.lower()
    di = {}
    for i in string:
       di[i] = string.count(i)
    if all(value > 1 for value in di.values()):
       return ""
    else:
       var =  min(di.items(), key=operator.itemgetter(1))[0]
       return var

输出：代替

output = {"a":1 , b:"1" , "A":1}

我想要：

output = {"A/a" : 2, "b":1}

并返回：the repeated letter is A or a

Answer 1

这会将所有小写/大写字符及其计数合并到一个字典中：

from collections import Counter
from itertools import groupby

s = 'AbCxbaBc'

out = {}
for v, g in groupby(sorted(Counter(s).items(), key=lambda k: 2*ord(k[0].upper()) - k[0].islower()), lambda k: k[0].lower()):
    l = [*g]
    out[''.join(i[0] for i in l)] = sum(i[1] for i in l)

print(out)

# non repeated letters:

non_repeating = [k for k, v in out.items() if v==1]
print('Non repeated letters:', non_repeating)

# first non repeated letter:

if non_repeating:
    m = min(map(lambda i: (s.index(i), i), non_repeating))
    print('First non repeated letter:', m[-1])
else:
    print('All letters are repeating!')

打印：

{'aA': 2, 'bB': 3, 'cC': 2, 'x': 1}
Non repeated letters: ['x']
First non repeated letter: x

Answer 2

这应该可以满足您的需求。

import operator

def first_non_repeating_letter(string):
    dict = {}
    lowercase_string = string.lower()
    # first we have to make a dictionary counting how many occurrences there are
    for ch in lowercase_string:
        if ch in dict:
            dict[ch] = dict[ch] + 1
        else:
            dict[ch] = 1

    # check if the number of occurrences is one, then return it
    for ch in lowercase_string:
        if ch in dict and dict[ch] == 1:
            index = lowercase_string.find(ch)
            return string[index]

示例：

输入：“ sTreSs”

输出：“ T”

Answer 3

这可能会帮助

import collections
c = collections.Counter('PaRrOt'.lower())
sorted(c.items(), key=lambda c: c[0])
[('a', 1), ('o', 1), ('p', 1), ('r', 2), ('t', 1)]

来源 https://stackoverflow.com/a/22819467/5202279

编辑：基于OP的评论。

您可以简单地检查该特定字母的字典值是否为1，然后再重复一次。

至于字母的case_case。查看字母是否重复，然后使用正则表达式将其与原始字符串匹配。

Answer 4

您总是可以创建一对映射：一个大小写折叠的字符到其计数，另一个到第一次出现的映射。

from collections import Counter

some_string = 'Some string'
lower_string = some_string.casefold()
count = Counter(lower_string)
firsts = dict(zip(lower_string, some_string))

现在，对于count中的和字符，您可以在firsts中查找其原始外观。

无论大小写如何如何退回第一个未重复的字母？

4 个答案: