获取句子中字母的频率

问题描述

我正在尝试编写一个可以输入随机句子的代码，并计算字母在此字符串中返回的次数:

I am trying to make a code where I can input a random sentence, and count the frequency of the times a letter returns in this string:

def getfreq(lines):
    """ calculate a list with letter frequencies

    lines - list of lines (character strings)

    both lower and upper case characters are counted.
    """
    totals = 26*[0]
    chars = []
    for line in lines:
       for ch in line:
           chars.append(totals)

    return totals

    # convert totals to frequency
    freqlst = []
    grandtotal = sum(totals)

    for total in totals:
        freq = totals.count(chars)
        freqlst.append(freq)
    return freqlst

到目前为止，我已经实现了将输入的每个字母追加到列表(字符)中.但是现在我需要一种方法来计算角色在该列表中返回的次数，并以此频率来表示.

So far I have achieved to append each letter of the input in the list (chars). But now I need a way to count the amount of times a character returns in that list, and express this in a frequency.

推荐答案

有一个非常方便的功能，collections模块中的rel ="nofollow noreferrer"> Counter ，它将计算序列中对象的频率:

There's a very handy function, Counter, within the collections module which will compute the frequency of objects within a sequence:

import collections
collections.Counter('A long sentence may contain repeated letters')

将产生:

Counter({' ': 6,
         'A': 1,
         'a': 3,
         'c': 2,
         'd': 1,
         'e': 8,
         'g': 1,
         'i': 1,
         'l': 2,
         'm': 1,
         'n': 5,
         'o': 2,
         'p': 1,
         'r': 2,
         's': 2,
         't': 5,
         'y': 1})

在您的情况下，您可能需要连接行，例如在进入Counter之前使用''.join(lines).

In your case, you might want to concatenate your lines, e.g. using ''.join(lines) before passing into the Counter.

如果您想使用原始字典获得相似的结果，则可能需要执行以下操作:

If you want to achieve a similar result using raw dictionaries, you might want to do something like the following:

counts = {}
for c in my_string:
    counts[c] = counts.get(c, 0) + 1

这取决于您的Python版本，可能会比较慢，但是会使用dict的.get()方法返回现有计数或默认值，然后再增加字符串中每个字符的计数.

Depending on your version of Python, this may be slower, but uses the .get() method of dict to either return an existing count or a default value before incrementing the count for each character in your string.

这篇关于获取句子中字母的频率的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！