[英]pandas str.contains match exact substring not working with regex boudry
I have two dataframes, and trying to find out a way to match the exact substring from one dataframe to another dataframe.我有两个数据框,并试图找到一种方法来匹配从一个 dataframe 到另一个 dataframe 的确切 substring。
First DataFrame :首先 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)
Second 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)
Approach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'])
The current output is as below, though I am expecting to match the exact substring, which is not working with the code above.当前的 output 如下所示,尽管我希望与上面的代码不匹配的确切 substring 匹配。
Current Output:当前 Output:
Place Name Site
TS~HOT_MD~h_PB~progra_VV~gogl programmatic-mechanics
FM~uiosv_PB~emo_SZ~1x1_TG~bhv emoteev
Expected Output:预期 Output:
Place Name Site
TS~HOT_MD~h_PB~progra_VV~gogl programmatic me
FM~uiosv_PB~emo_SZ~1x1_TG~bhv emoteev
Data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)
Map the test_datframe
code
and Actual
into dictionary as key
and value
respectively Map 将
test_datframe
code
和Actual
分别作为key
和value
放入字典
keys=test_dataframe['code'].values.tolist()
dicto=dict(zip(test_dataframe.code, test_dataframe.Actual))
dicto
Join the keys separated by |加入由|分隔的键to enable search of either phrases
启用任一短语的搜索
k = '|'.join(r"{}".format(x) for x in dicto.keys())
k
Extract string from datframe meeting any of the phrases in k and map them to to the dictionary从 datframe 中提取符合 k 和 map 中任何短语的字符串到字典
dataframe['Site'] = dataframe['Place Name'].str.extract('('+ k + ')', expand=False).map(dicto)
dataframe
Output Output
Not the most elegant solution, but this does the trick.不是最优雅的解决方案,但这可以解决问题。
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)
Create a column in test_dataframe
with the substring to match:在
test_dataframe
中使用 ZE83AED3DDF4667DEC0DAAAACB2BB3BE0BZ 创建一列以匹配:
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
Define a function to apply to test_dataframe
:定义一个 function 以应用于
test_dataframe
:
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)
Output: Output:
Place Name Site
0 TS~HOT_MD~h_PB~progra_VV~gogl programmatic me
1 FM~uiosv_PB~emo_SZ~1x1_TG~bhv emoteev
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