Question

我想从一个句子中删除逗号，并将所有其他单词（a-z）分开，并一一打印。

a = input()
b=list(a)                      //to remove punctuations
for item in list(b):           //to prevent "index out of range" error. 
    for j in range(len(l)):
        if(item==','):
            b.remove(item)
            break
c="".join(b)                  //sentence without commas
c=c.split()
print(c)

我的输入是：

The university was founded as a standard academy,and developed to a university of technology by Habib Nafisi.

当我删除逗号时：

... founded as a standard academyand developed to a university...

当我拆分单词时：

The
university
.
.
.
academyand
.
.
.

我该怎么做才能防止这种情况？我已经尝试过替换方法，但是它不起作用。

Answer 1

您的问题似乎是逗号和此处的下一个单词之间没有空格：academy,and您可以通过确保有一个空格来解决此问题，以便在使用b=list(a)时该函数将实际上将每个单词分成列表的不同元素。

Answer 2

假设,和输入¹中的下一个单词之间没有空格，可以用空格替换,，然后执行split：

s = 'The university was founded as a standard academy,and developed to a university of technology by Habib Nafisi.'

print(s.replace(',', ' ').split())
# ['The', 'university', 'was', 'founded', 'as', 'a', 'standard', 'academy', 'and', 'developed', 'to', 'a', 'university', 'of', 'technology', 'by', 'Habib', 'Nafisi.']

或者，您也可以在regex上试一试：

import re

s = 'The university was founded as a standard academy,and developed to a university of technology by Habib Nafisi.'

print(re.split(r' |,', s))

¹ _{注意：即使您在,之后有空格（多个），该方法仍然有效，因为最终您split在空白处。}

Answer 3

这可能就是您想要的，我看到您忘记用空格replace逗号。

stri = """ The university was founded as a standard academy,and developed to a university of technology by Habib Nafisi."""
stri.replace(",", " ")
print(stri.split())

将在列表中为您提供输出：

['The', 'university', 'was', 'founded', 'as', 'a', 'standard', 'academy,and', 'developed', 'to', 'a', 'university', 'of', 'technology', 'by', 'Habib', 'Nafisi.']

Answer 4

如果您认为单词是由空格隔开的一系列字符，则如果不使用替换a，则它们之间将没有空格，它将视为一个单词。

最简单的方法是将逗号替换为一个空格，然后根据空格进行分割：

my_string = "The university was founded as a standard academy,and developed to a university of technology by Habib Nafisi."
list_of_words = my_string.replace(",", " ").split()

句子中用逗号分隔的单词

4 个答案: