Question

我需要一个函数来计算每个字母出现在字符串中的次数。它必须将大写字母和小写字母统计为相同的字母。我有点做了，但它并不完全漂亮。

def lmao(x):

    aye=x.count("a")
    Aye=x.count("A")
    bye=x.count("b")
    Bye=x.count("B")
    cye=x.count("c")
    Cye=x.count("C")
    dye=x.count("d")
    Dye=x.count("D")
    Eye=x.count("E")
    eye=x.count("e")
    Fye=x.count("F")
    fye=x.count("f")
    Gye=x.count("G")
    gye=x.count("g")
    Hye=x.count("H")
    hye=x.count("h")
    Iye=x.count("I")
    iye=x.count("i")
    Jye=x.count("J")
    jye=x.count("j")
    Kye=x.count("K")
    kye=x.count("k")
    Lye=x.count("L")
    lye=x.count("l")
    Mye=x.count("M")
    mye=x.count("m")
    Nye=x.count("N")
    nye=x.count("n")
    Oye=x.count("O")
    oye=x.count("o")
    Pye=x.count("P")
    pye=x.count("P")
    Qye=x.count("Q")
    qye=x.count("q")
    rye=x.count("r")
    Rye=x.count("R")
    sye=x.count("s")
    Sye=x.count("S")
    tye=x.count("t")
    Tye=x.count("T")
    uye=x.count("u")
    Uye=x.count("U")
    Vye=x.count("V")
    vye=x.count("v")
    Wye=x.count("W")
    wye=x.count("w")
    Xye=x.count("X")
    xye=x.count("x")
    Yye=x.count("Y")
    yye=x.count("y")
    Zye=x.count("Z")
    zye=x.count("z")
    killme=(aye+Aye,bye+Bye,cye+Cye,Dye+dye,Eye+eye,Fye+fye,Gye+gye,Hye+hye,Iye+iye,jye+Jye,Kye+kye,Lye+lye,Mye+mye,Nye+nye,Oye+oye,Pye+pye,Qye+qye,rye+Rye,sye+Sye,Tye+tye,uye+Uye,Vye+vye,Wye+wye,xye+Xye,Yye+yye,Zye+zye)
    return killme

所以是的，那就是我想出的灾难。有没有办法缩短这个过程？

Answer 1

使用UIImageView https://docs.python.org/2/library/collections.html#counter-objects

collections.Counter

会给你所有出现的每个字母，忽略大小写

Answer 2

准确回复您的要求：

import string
def lmao(x):
    return tuple(x.lower().count(c) for c in string.lowercase)

Answer 3

这可能有所帮助：

def counter(text):
    count_list = []
    for char in "abcdefghijklmnopqrstuvwxyz":
        count_list.append(text.lower().count(char))
    return tuple(count_list)

print(counter("Helllo123"))

输出：

(0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 3, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)

Answer 4

将计数器与字典理解结合使用，以获得每个字母的计数。

import string
from collections import Counter

对于alpha排序的元组，

def count_letter_tuples(sentence):
    return tuple(Counter(sentence.lower()).get(c, 0) for c in string.ascii_lowercase)

>>> count_letter_tuples("Some Big Sentence")
(0, 0, 1, 0, 4, 0, 1, 0, 1, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0)

或者是字典回复：

def count_letters(sentence):
    return {c: Counter(sentence.lower()).get(c, 0) for c in string.ascii_lowercase}

>>> count_letters("Some Big Sentence")
{'a': 0,
 'b': 1,
 'c': 1,
 'd': 0,
 'e': 4,
 'f': 0,
 'g': 1,
 'h': 0,
 'i': 1,
 'j': 0,
 'k': 0,
 'l': 0,
 'm': 1,
 'n': 2,
 'o': 1,
 'p': 0,
 'q': 0,
 'r': 0,
 's': 2,
 't': 1,
 'u': 0,
 'v': 0,
 'w': 0,
 'x': 0,
 'y': 0,
 'z': 0}

计算每个字母在文本样本中出现的次数

4 个答案: