在 Pandas 中搜索多个字符串而不预先定义要使用的字符串数量

Searching Multiple Strings in pandas without predefining number of strings to use

提问by user3314418

I'm wondering if there's a more general way to do the below? I'm wondering if there's a way to create the st function so that I can search a non-predefined number of strings?

我想知道是否有更通用的方法来执行以下操作？我想知道是否有办法创建 st 函数，以便我可以搜索非预定义数量的字符串？

So for instance, being able to create a generalized st function, and then type st('Governor', 'Virginia', 'Google)

例如，能够创建一个通用的 st 函数，然后输入 st('Governor', 'Virginia', 'Google)

here's my current function, but it predefines two words you can use. (df is a pandas DataFrame)

这是我当前的函数，但它预定义了两个您可以使用的词。（df 是一个Pandas数据帧）

def search(word1, word2, word3 df):
    """
    allows you to search an intersection of three terms
    """
    return df[df.Name.str.contains(word1) & df.Name.str.contains(word2) & df.Name.str.contains(word3)]

st('Governor', 'Virginia', newauthdf)

回答by unutbu

You could use np.logical_and.reduce:

你可以使用np.logical_and.reduce：

import pandas as pd
import numpy as np
def search(df, *words):  #1
    """
    Return a sub-DataFrame of those rows whose Name column match all the words.
    """
    return df[np.logical_and.reduce([df['Name'].str.contains(word) for word in words])]   # 2


df = pd.DataFrame({'Name':['Virginia Google Governor',
                           'Governor Virginia',
                           'Governor Virginia Google']})
print(search(df, 'Governor', 'Virginia', 'Google'))

prints

印刷

                       Name
0  Virginia Google Governor
2  Governor Virginia Google

---

---

1. The *in def search(df, *words)allows searchto accept an unlimited number of positional arguments. It will collect all the arguments (after the first) and place them in a list called words.
2. np.logical_and.reduce([X,Y,Z])is equivalent to X & Y & Z. It allows you to handle an arbitrarily long list, however.

1. 将*在def search(df, *words)允许search接受的位置参数的数量不受限制。它将收集所有参数（在第一个之后）并将它们放在一个名为words.
2. np.logical_and.reduce([X,Y,Z])等价于X & Y & Z. 但是，它允许您处理任意长的列表。

回答by behzad.nouri

str.containscan take regex. so you can use '|'.join(words)as the pattern; to be safe map to re.escapeas well:

str.contains可以带正则表达式。所以你可以'|'.join(words)用作模式；也是安全的映射到re.escape：

>>> df
                 Name
0                Test
1            Virginia
2              Google
3  Google in Virginia
4               Apple

[5 rows x 1 columns]
>>> words = ['Governor', 'Virginia', 'Google']

'|'.join(map(re.escape, words))would be the search pattern:

'|'.join(map(re.escape, words))将是搜索模式：

>>> import re
>>> pat = '|'.join(map(re.escape, words))
>>> df.Name.str.contains(pat)
0    False
1     True
2     True
3     True
4    False
Name: Name, dtype: bool