熊猫str.contains匹配不使用regex的精确子字符串

Question

我有两个dataframe，并试图找到一种方法来匹配从一个dataframe到另一个dataframe的精确子字符串。

First DataFrame

import pandas as pd
import numpy as np

random_data = {'Place Name':['TS~HOT_MD~h_PB~progra_VV~gogl', 'FM~uiosv_PB~emo_SZ~1x1_TG~bhv'], 
              'Site':['DV360', 'Adikteev']}
        
dataframe = pd.DataFrame(random_data)
print(dataframe)

第二DataFrame

test_data = {'code name': ['PB', 'PB', 'PB'], 
             'Actual':['programmatic me', 'emoteev', 'programmatic-mechanics'],
             'code':['progra', 'emo', 'prog']}

test_dataframe = pd.DataFrame(test_data)

逼近

for k, l, m in zip(test_dataframe.iloc[:, 0], test_dataframe.iloc[:, 1], test_dataframe.iloc[:, 2]):
    dataframe['Site'] = np.select([dataframe['Place Name'].str.contains(r'\b{}~{}\b'.format(k, m), regex=False)], [l],
                                  default=dataframe['Site'])

当前的输出如下所示，尽管我期望匹配精确的子字符串，该子字符串与上面的代码不兼容。

当前产出：

Place Name                        Site
TS~HOT_MD~h_PB~progra_VV~gogl     programmatic-mechanics
FM~uiosv_PB~emo_SZ~1x1_TG~bhv     emoteev

预期产出：

Place Name                        Site
TS~HOT_MD~h_PB~progra_VV~gogl     programmatic me
FM~uiosv_PB~emo_SZ~1x1_TG~bhv     emoteev

Answer 1

数据

import pandas as pd
 import numpy as np
 random_data = {'Place Name':['TS~HOT_MD~h_PB~progra_VV~gogl',
                                     'FM~uiosv_PB~emo_SZ~1x1_TG~bhv'], 'Site':['DV360', 'Adikteev']}
    dataframe = pd.DataFrame(random_data)

    test_data = {'code name': ['PB', 'PB', 'PB'], 'Actual':['programmatic me', 'emoteev', 'programmatic-mechanics'],
                 'code':['progra', 'emo', 'prog']}
    test_dataframe  = pd.DataFrame(test_data)

将test_datframe code和Actual分别映射为key和value字典

keys=test_dataframe['code'].values.tolist()

dicto=dict(zip(test_dataframe.code, test_dataframe.Actual))
dicto

加入由|分隔的键，以便搜索任一短语

k = '|'.join(r"{}".format(x) for x in dicto.keys())
k

从数据帧中提取符合k中任何短语的字符串，并将它们映射到字典中。

dataframe['Site'] = dataframe['Place Name'].str.extract('('+ k + ')', expand=False).map(dicto)
dataframe

输出

​

​

Answer 2

这不是最优雅的解决方案，但这确实有效。

设置数据

import pandas as pd
import numpy as np
random_data = {'Place Name':['TS~HOT_MD~h_PB~progra_VV~gogl',
                                'FM~uiosv_PB~emo_SZ~1x1_TG~bhv'], 'Site':['DV360', 'Adikteev']}

dataframe = pd.DataFrame(random_data)

test_data = {'code name': ['PB', 'PB', 'PB'], 'Actual':['programmatic me', 'emoteev', 'programmatic-mechanics'],
             'code':['progra', 'emo', 'prog']}

test_dataframe = pd.DataFrame(test_data)

解决方案

使用要匹配的子字符串在test_dataframe中创建一个列：

test_dataframe['match_str'] = test_dataframe['code name'] + '~' + test_dataframe.code

print(test_dataframe)

  code name                  Actual    code  match_str
0        PB         programmatic me  progra  PB~progra
1        PB                 emoteev     emo     PB~emo
2        PB  programmatic-mechanics    prog    PB~prog

定义一个应用于test_dataframe的函数

def match_string(row, dataframe):
    ind = row.name
    try:
        if row[-1] in dataframe.loc[ind, 'Place Name']:
            return row[1]
        else:
            return dataframe.loc[ind, 'Site']
    except KeyError:
        # More rows in test_dataframe than there are in dataframe
        pass

# Apply match_string and assign back to dataframe
dataframe['Site'] = test_dataframe.apply(match_string, args=(dataframe,), axis=1)

输出：

                      Place Name             Site
0  TS~HOT_MD~h_PB~progra_VV~gogl  programmatic me
1  FM~uiosv_PB~emo_SZ~1x1_TG~bhv          emoteev