Question

我试图让这个python代码摆脱与单词相关的标点符号并计算独特的单词。出于某种原因，它仍然在计算“你好”。和“你好”。非常感激任何的帮助。

def word_distribution(words):
            word_dict = {}
            words = words.lower()
            words = words.split()
            for word in words:
                if ord('a') <= ord(word[-1]) <= ord('z'):
                    pass
                elif ord('A') <= ord(word[-1]) <= ord('Z'):
                    pass
                else: 
                    word[:-1]
            word_dict = {word:words.count(word)+1 for word in set(words)}
            return(word_dict)

Answer 1

我不知道你为什么要加1来计算。

def word_distribution(words):
        word_dict = {}
        words = words.lower().split()
        for word in words:
            if ord('a') <= ord(word[-1]) <= ord('z'):
                pass
            elif ord('A') <= ord(word[-1]) <= ord('Z'):
                pass
        word_dict = {word:words.count(word) for word in set(words)}
        return(word_dict)

{'你好'：2，'我的'：1，'名字'：1，'是'：1}

编辑：

作为brianpck，指出：

def word_distribution(words):
        word_dict = {}
        words = words.lower().split()
        word_dict = {word:words.count(word) for word in set(words)}
        return(word_dict)

也会给出相同的结果。

Answer 2

你使它变得太复杂了，正如Sohier Dane在评论中提到的那样，你可以利用其他帖子删除标点并简化脚本：

import string
def word_distribution(words):
    words = words.translate(None, string.punctuation).lower()
    d = {}
    for w in words.split():
        if w not in d.keys():
            d[w] = 1
        else:
            d[w] += 1   
    return d

结果：

>>> x='Hello My Name Is hello.'
>>> print word_distribution(x)  
>>> {'is': 1, 'my': 1, 'hello': 2, 'name': 1}

Answer 3

肯定有更好的方法来实现您的目标，但这个答案会修复您的代码。

字符串是不可变的，列表是可变的。您的代码中没有任何地方正在修改列表。并且words[-1]不会产生任何影响，因为您没有重新分配它，字符串是不可变的

def word_distribution(words):
        word_dict = {}
        words = words.lower()
        words = words.split()
        for word in words:
            index = words.index(word)
            if ord('a') <= ord(word[-1]) <= ord('z'):
                pass
            elif ord('A') <= ord(word[-1]) <= ord('Z'):
                pass
            else: 
                word = word[:-1]
                words[index] = word 

        word_dict = {word:words.count(word) for word in set(words)}
        return(word_dict)

Python在迭代中改变列表项

3 个答案: