替换列表中的单词并附加到相同的列表

Question

我的列表:

city=['Venango Municiplaity', 'Waterford ship','New York']

预期结果:

city = ['Venango Municiplaity ', 'Waterford ship','New York','Venango','Waterford']

常用的词:

common_words = ['ship','municipality']

扫描"我的列表"中的所有项目并删除常用单词并重新插入到预期结果中显示的相同列表中.

我能够搜索包含常用单词但不确定如何用空白替换它并在"我的列表"中重新插入的项目.

我的代码到目前为止:

for item in city:
    if(any(x in s.lower() for s in item.split(' ') for x in common_words)) :

Answer 1

我已经制作了一个按预期工作的小代码:

city=['Venango Municiplaity', 'Waterford ship','New York']
comwo = ['ship','municipality']
for i, c in enumerate(city):
    for ii in comwo:
        if ii in c:
            city.append(city[i].replace(ii,""))
print(city)

输出:

['Venango Municiplaity', 'Waterford ship', 'New York', 'Waterford ']

注意:

您所做的列表包含错误的拼写.
查看list city的第一个元素VenangoMuniciplaity和common_words的第二个元素municipality

编辑:

因此,如果您还想替换单词后面的空格(如果有的话),那么我已经创建了一个单独的代码:

[city.append(city[i].replace(ii,"")) for ii in comwo for i, c in enumerate(city) if ii in c]

输出:

city=['Village home', 'Villagehome','New York']
comwo = ['home']
for i, c in enumerate(city):
    for ii in comwo:
        if ii in c:
            city.append(city[i].replace(" "+ii,"")) if city[i].replace(" "+ii,"") != city[i] else city.append(city[i].replace(ii,""))
print(city)

Answer 2

我建议你使用以下解决方案,使用re.subwith flags=re.IGNORECASE来删除忽略大小写的常用词:

import re

city = ['Venango Municipality', 'Waterford ship','New York']
common_words = ['ship','municipality']

toAppend = []

for c in city:
    for cw in common_words:
        if cw.lower() in c.lower().split():
            toAppend.append(re.sub(cw, "", c, flags=re.IGNORECASE).strip())

city += toAppend

print(city) # ['Venango Municipality', 'Waterford ship', 'New York', 'Venango', 'Waterford']

这是使用列表理解的单行样式解决方案,简短但可读性较差:

import re

city = ['Venango Municipality', 'Waterford ship','New York']
common_words = ['ship','municipality']

city += [re.sub(cw, "", c, flags=re.IGNORECASE).strip() for c in city for cw in common_words if cw.lower() in c.lower().split()]

print(city) # ['Venango Municipality', 'Waterford ship', 'New York', 'Venango', 'Waterford']

Answer 3

您可以尝试它,创建新列表以保存数据应该添加到您的原始列表,然后连接结果:

In [1]: city=['Venango Municiplaity', 'Waterford ship','New York']

In [2]: common_words = ['ship', 'municiplaity']

In [3]: list_add = []

In [4]: for item in city:
   ...:     item_words = [s.lower() for s in item.split(' ')]
   ...:     if set(common_words) & set(item_words):
   ...:         new_item = [s for s in item.split(' ') if s.lower() not in common_words]
   ...:         list_add.append(" ".join(new_item))
   ...:         

In [5]: city + list_add
Out[5]: ['Venango Municiplaity', 'Waterford ship', 'New York', 'Venango', 'Waterford']