pandas str.contains match-exact子字符串不适用于正则表达式boudry

Question

我有两个数据帧，并试图找到一种方法来匹配从一个数据帧到另一个数据帧的确切子字符串

第一个数据帧：

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)

第二个数据帧

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

test_dataframe = pd.DataFrame(test_data)

方法

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'])

当前输出如下所示，尽管我希望匹配精确的子字符串，但它不适用于上面的代码

电流输出：

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

这不是最优雅的解决方案，但它确实起到了作用

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)

溶液

在test_dataframe中创建一列，其中包含要匹配的子字符串：

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

定义要应用于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

Answer 2

资料

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{}和Actual分别映射到字典中作为key和value

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

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

连接由|分隔的键以启用对任一短语的搜索

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

从符合k中任何短语的datframe中提取字符串，并将其映射到字典

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

输出