在 column1 上应用条件，其中字符串仅包含给定的关键字集，而不是包含给定的关键字集

Question

如何在 column1 上应用条件，其中字符串 仅包含 给定一组关键字，而不是 包含 给定一组关键字。 强调“仅”这个词

import pandas as pd
import numpy as np
import re

d = {'_id': [1, 2, 3, 4, 5, 6, 7],
     'column1': ['FullName', 'custfullnm', 'nm123', 'sitenm', 'full12', 'suplnm', 'countryfulln'],
     'column2': ['', '', '', '', '', '', '']}

df = pd.DataFrame(data=d)


key_words = ["full", "nm", "name", "txt", "[0-9]"]

check = f"{'|'.join(key_words)}"

mand_key = "full"

df["column2"] = np.where(
    df.column1.str.contains(mand_key, case=False)
    & (df.column1.str.contains(check, case=False, regex=True)),
    "Full Name",
    "",
)

想要的输出：

_id,column1,column2
1,FullName,Full Name
2,custfullnm,
3,nm123,
4,sitenm,
5,full12,Full Name
6,suplnm,
7,countryfullnm,

只有 FullName, full12 符合条件，原因如下：

FullName is only made of words from given set of keywords 'full' & 'name'
full12 is only made of words from given set of keywords 'full' & a number '12'

而rest不符合条件，原因如下：

custfullnm      contains 'cust' not in given list of keywords though contains 'nm' & 'full'
nm123           dones't contain madate keyword 'full' though contains a number & 'nm'
sitename        contains 'site' not in given list of keywords though contains 'name'
suplnm          contains 'supl' not in given list of keywords though contains 'nm'
countryfullnm       contains 'country' not in given list of keywords though contains 'nm' & 'full'

Answer 1

我自己得到了答案：

key_words = ["full", "nm", "name", "txt", "[0-9]"]
check = f"{'|'.join(key_words)}"
mand_key = "full"
df["column2"] = np.where(
    df.column1.str.contains(mand_key, flags=re.IGNORECASE)
    & df.column1.str.match(check, flags=re.IGNORECASE),"Full Name","",)
df

Answer 2

您可以创建一个模式，该模式将从字符串开头开始匹配任意数量的允许子模式，然后尝试匹配您的强制键，然后再次允许任意数量的指定子模式，直到结束字符串：

import pandas as pd
import numpy as np
import re

d = {'_id': [1, 2, 3, 4, 5, 6, 7, 8],
     'column1': ['FullName', 'custfullnm', 'nm123', 'sitenm', 'full12', 'suplnm', 'countryfulln', np.NaN],
     'column2': ['1', '2', '3', '4', '5', '6', '7', '8']}

df = pd.DataFrame(data=d)

key_words = ["full", "nm", "name", "txt", "[0-9]"]
mand_key = "full"
check = rf'^(?:{"|".join(key_words)})*{mand_key}(?:{"|".join(key_words)})*$'
df["column2"] = np.where(df["column1"].str.contains(check, case=False, na=True),"Full Name","")

输出：

>>> df
   _id       column1    column2
0    1      FullName  Full Name
1    2    custfullnm           
2    3         nm123           
3    4        sitenm           
4    5        full12  Full Name
5    6        suplnm           
6    7  countryfulln           
7    8           NaN  Full Name
  

图案看起来像

^(?:full|nm|name|txt|[0-9])*full(?:full|nm|name|txt|[0-9])*$

见regex demo

详情

• ^ - 字符串开头
• (?:full|nm|name|txt|[0-9])* - 指定子模式的 0 次或多次重复
• full - 强制密钥
• (?:full|nm|name|txt|[0-9])* - 指定子模式的 0 次或多次重复
• $ - 字符串结束。