时间戳

首页 > 解决方案 > 如何从python中的pandas数据框中的列中提取关键字（字符串）

python-3.x - 如何从python中的pandas数据框中的列中提取关键字（字符串）

问题描述

我有一个数据框df，它看起来像这样：

         id                        Type                        agent_id  created_at
0       44525   Stunning 6 bedroom villa in New Delhi               184  2018-03-09
1       44859   Villa for sale in Amritsar                          182  2017-02-19
2       45465   House in Faridabad                                  154  2017-04-17
3       50685   5 Hectre land near New Delhi                        113  2017-09-01
4      130728   Duplex in Mumbai                                    157  2017-02-07
5      130856   Large plot with fantastic views in Mumbai           137  2018-01-16
6      130857   Modern Design Penthouse in Bangalore                199  2017-03-24

我有这个表格数据，我试图通过从列中提取关键字来清理这些数据，从而创建一个带有新列的新数据框。

Apartment  = ['apartment', 'penthouse', 'duplex']
House      = ['house', 'villa', 'country estate']
Plot       = ['plot', 'land']
Location   = ['New Delhi','Mumbai','Bangalore','Amritsar']

因此，所需的数据框应如下所示：

         id      Type        Location    agent_id  created_at
0       44525   House       New Delhi         184  2018-03-09
1       44859   House        Amritsar         182  2017-02-19
2       45465   House       Faridabad         154  2017-04-17
3       50685   Plot        New Delhi         113  2017-09-01
4      130728   Apartment      Mumbai         157  2017-02-07
5      130856   Plot           Mumbai         137  2018-01-16
6      130857   Apartment   Bangalore         199  2017-03-24

所以直到现在我已经尝试过这个：

import pandas as pd
df = pd.read_csv('test_data.csv')

#i can extract these keywords one by one by using for loops but how
#can i do this work in pandas with minimum possible line of code.

for index, values in df.type.iteritems():
  for i in Apartment:
     if i in values:
         print(i)

df_new = pd. Dataframe(df['id'])

有人可以告诉我如何解决这个问题吗？

标签： python-3.xpandaslistdataframekeyword

解决方案

首先通过with for regex创建Location列：str.extract|OR

pat = '|'.join(r"\b{}\b".format(x) for x in Location)
df['Location'] = df['Type'].str.extract('('+ pat + ')', expand=False)

然后从另一个 s 创建字典list，用值交换键，并在循环中通过掩码str.contains和参数设置值case=False：

d = {'Apartment' : Apartment,
     'House' : House,
     'Plot' : Plot}

d1 = {k: oldk for oldk, oldv in d.items() for k in oldv}

for k, v in d1.items():
    df.loc[df['Type'].str.contains(k, case=False), 'Type'] = v

print (df)
       id       Type  agent_id  created_at   Location
0   44525      House       184  2018-03-09  New Delhi
1   44859      House       182  2017-02-19   Amritsar
2   45465      House       154  2017-04-17        NaN
3   50685       Plot       113  2017-09-01  New Delhi
4  130728  Apartment       157  2017-02-07     Mumbai
5  130856       Plot       137  2018-01-16     Mumbai
6  130857  Apartment       199  2017-03-24  Bangalore

推荐阅读