Python返回字符串列表（如果其他列表中存在任何子字符串）

Question

假设你有这样的公司信息：

companies = [['zmpEVqsbCUO1aXStxHkSVA', 'palms-car-wash'],
['5T0vKfIJWP1xTnxA7fJ17w', 'meat-and-bread'],
['C0d5kzUx6C19mLcxQyhxCA', 'alamo-drafthouse-cinema-'],
['ch1ercqwoNLpQLxpTb90KQ', 'boston-tea-stop']]

假设您想要排除某些业务，如果列表（以下）的任何字符串/子字符串出现在上面列表的某些信息中：

no_interest = ['museum', 'cinema', 'car']

我已经这样做了，（我们只在每个条目的第2列中查看）：

# KEEPING ONLY RESULTS WHERE WE DO NOT FIND THE SUBSTRINGS
[x for x in companies if (no_interest[0] not in x[1]) & (no_interest[1] not in x[1]) & (no_interest[2] not in x[1])]

# RETURN
[['5T0vKfIJWP1xTnxA7fJ17w', 'meat-and-bread'],
 ['ch1ercqwoNLpQLxpTb90KQ', 'boston-tea-stop']]

即使我希望它使用“OR”语句而不是“AND”（&；）语句，它似乎也能工作，对我来说，AND是一个累积运算符，如果所有条件都满足，它也应该工作（“博物馆”、“电影院”和“汽车”在同一字符串中）

所以我有两个问题：

• Why is the 'AND' statement acting like a 'OR'?
• How can we make this code more pythonic and more efficient?

我们在这里只检查3个子字符串，但它越来越多地与我们正在寻找的数千个事件有关，最好不要重复这些条件，而是有一个更像all()或any()的语句，返回结果而不是布尔值

Answer 1

下面是另一个使用正则表达式的例子，但是（正如亨利·埃克的回答）它假设在任何“无兴趣”元素中都没有干扰正则表达式的特殊字符

import regex as re
pattern = re.compile("|".join(no_interest))
out = [c for c in companies if ((pattern.search(c[0]) == None) and (pattern.search(c[1]) == None))]

Answer 2

Why is the 'AND' statement acting like a 'OR'?

见：DeMorgan's Laws

How can we make this code more pythonic and more efficient?

更像Python：

一种选择是在单独的列表中使用all：

companies = [['zmpEVqsbCUO1aXStxHkSVA', 'palms-car-wash'],
             ['5T0vKfIJWP1xTnxA7fJ17w', 'meat-and-bread'],
             ['C0d5kzUx6C19mLcxQyhxCA', 'alamo-drafthouse-cinema-'],
             ['ch1ercqwoNLpQLxpTb90KQ', 'boston-tea-stop']]

no_interest = ['museum', 'cinema', 'car']

out = [x for x in companies if all([ni not in x[1] for ni in no_interest])]
print(out)

或与not{a4}一起：

out = [x for x in companies if not any([ni in x[1] for ni in no_interest])]

[['5T0vKfIJWP1xTnxA7fJ17w', 'meat-and-bread'],
 ['ch1ercqwoNLpQLxpTb90KQ', 'boston-tea-stop']]

---

更有效率：

使用类似pandas的库：

import pandas as pd

companies = [['zmpEVqsbCUO1aXStxHkSVA', 'palms-car-wash'],
             ['5T0vKfIJWP1xTnxA7fJ17w', 'meat-and-bread'],
             ['C0d5kzUx6C19mLcxQyhxCA', 'alamo-drafthouse-cinema-'],
             ['ch1ercqwoNLpQLxpTb90KQ', 'boston-tea-stop']]

df = pd.DataFrame(data=companies, columns=['id', 'val'])

no_interest = ['museum', 'cinema', 'car']

out = df[~df['val'].str.contains('|'.join(no_interest))]
print(out)

输出为数据帧

                       id              val
1  5T0vKfIJWP1xTnxA7fJ17w   meat-and-bread
3  ch1ercqwoNLpQLxpTb90KQ  boston-tea-stop

输出为列表

print(out.to_numpy().tolist())

[['5T0vKfIJWP1xTnxA7fJ17w', 'meat-and-bread'],
 ['ch1ercqwoNLpQLxpTb90KQ', 'boston-tea-stop']]