Question

我正在尝试替换数据框（C列）中的列表名称：

姓名列表（小示例，该列表太大）：

types

小数据框示例：

Jack
Liam
John
Ethan
George
...

我的脚本：

       A          B                                   C
  French      house                Phone <phone_numbers>
 English      house                 email <adresse_mail>
  French  apartment                      my name is Liam
  French      house                         Hello George
 English  apartment   Ethan, my phone is <phone_numbers>

作为输出，我有整个数据框，没有任何修改...

我期望的是

import re
import pandas as pd
from pandas import Series

df = pd.read_excel('data_frame.xlsx')
data = Series.to_string(df['C'])

first_names = open('names_list.txt', 'r')
names_read = first_names.readlines()

def names(data):

    names_regex = re.compile(r'\b%s\b' % r'\b|\b'.join(map(re.escape, names_read)))
    replace_names = names_regex.sub('<name>', data)

    return replace_names

no_names = names(data)
print(no_names)

Answer 1

'No trade'

您可以根据您的姓名列表定义一个正则表达式。然后，将这些值与列name_list = ['Jack', 'Liam', 'John', 'Ethan'] mydf = pd.DataFrame({'C': ['Phone <phone_numbers>', 'email <adresse_mail>', 'my name is Liam', 'Hello George', 'Ethan, my phone is <phone_numbers>']})匹配，并用C

替换它们

apply lambda

输出

match = mydf.C.str.extractall('(' + '|'.join(name_list) + ')').reset_index().set_index('level_0').rename(columns={0: 'name'})
mydf = pd.concat([mydf, match], axis=1)
condition = mydf.match.notnull()
mydf.loc[condition, 'C'] = mydf[condition].apply(lambda x: x['C'].replace(x['name'], '<name>'), axis=1)

Answer 2

您可以简单地通过遍历给定列的值来替换它们：

import pandas as pd

l = [
['French','house','Phone <phone_numbers>'],
['English','house','email <adresse_mail>'],
['French','apartment','my name is Liam'],
['French','house','Hello George'],
['English','apartment','Ethan, my phone is <phone_numbers>']
]

names = [
'Jack',
'Liam',
'John',
'Ethan',
'George'
]

df = pd.DataFrame(l, columns = list('ABC'))

for i in names:
  df.C = df.C.str.replace(i,'<name>')

print(df)

替换数据框中的列表名称

2 个答案: