使用pd資料框從包含字典串列的列中提取組織

我的資料框有多個列，例如 ID、組織、日期、位置等。我正在嘗試提取“組織”列中的“組織”值。我想要的輸出應該是新列中的多個組織的名稱，用逗號分隔。例如：

我希望輸出類似于：

我嘗試了以下代碼（輸出為 NaN）：

latin_combined['newOrg'] = latin_combined['organizations'].str[0].str['organization']

編輯： df.head(5)['organizations'].to_dict()給我以下輸出：

{0: '[{organization= Vac, character_offset=14199}, {organization=Health, character_offset=1494}]',
 1: '[{organization=Store, character_offset=700}, {organization= Museum, character_offset=1711}]',
 2: '[{organization= Mart, character_offset=8232}, {organization= Rep, character_offset=5517}]',
 3: '[{organization= Lodge, character_offset=3881}, {organization= Hotel, character_offset=5947}]',
 4: '[{organization=Airport, character_offset=3881}, {organization=Landmark, character_offset=5947}]'}

任何建議都會有所幫助。

uj5u.com熱心網友回復：

看來你有一個字串。您可以使用以 , pivotregex分隔的鍵、值對來提取，=如下所示：

(df['organizations'].str.extractall('([^{=,] )= *([^=,}] )') 
  .rename({0:'key', 1:'value'}, axis = 1).reset_index()
  .groupby(['level_0', 'key'])['value'].agg(', '.join).unstack())

key      character_offset       organization
level_0                                     
0             14199, 1494        Vac, Health
1               700, 1711      Store, Museum
2              8232, 5517          Mart, Rep
3              3881, 5947       Lodge, Hotel
4              3881, 5947  Airport, Landmark

資料

d = {0: '[{organization= Vac, character_offset=14199}, {organization=Health, character_offset=1494}]',
 1: '[{organization=Store, character_offset=700}, {organization= Museum, character_offset=1711}]',
 2: '[{organization= Mart, character_offset=8232}, {organization= Rep, character_offset=5517}]',
 3: '[{organization= Lodge, character_offset=3881}, {organization= Hotel, character_offset=5947}]',
 4: '[{organization=Airport, character_offset=3881}, {organization=Landmark, character_offset=5947}]'}

df = pd.Series(d).to_frame('organizations')

uj5u.com熱心網友回復：

1.根據您最近的資料框更新進行了更新：

data = {'Organizations': ['[{organization= Vac, character_offset=14199}, {organization=Health, character_offset=1494}]',
        '[{organization=Store, character_offset=700}, {organization= Museum, character_offset=1711}]',
        '[{organization= Mart, character_offset=8232}, {organization= Rep, character_offset=5517}]',
        '[{organization= Lodge, character_offset=3881}, {organization= Hotel, character_offset=5947}]',
        '[{organization=Airport, character_offset=3881}, {organization=Landmark, character_offset=5947}]']}
df = pd.DataFrame(data) 
df

2.在您想要的列上使用''.join() regexwith ：.apply()

import re
df.Organizations = df.Organizations.apply(lambda x: ', '.join(re.findall(r'{[^=] =\s*([^=,}] )', x)))
df

3.結果：

我個人認為，在將資料放入資料框之前，您應該嘗試更好地廢棄和/或清理資料。

uj5u.com熱心網友回復：

這是你想要做的嗎？

latin_combined['newOrg'] = latin_combined['organizations'].apply(lambda x : x.split(',')[0])

uj5u.com熱心網友回復：

您可以將串列理解與 apply 一起使用：

import pandas as pd

df = pd.DataFrame([[[{'organization':'Glaxosmithkline', 'character_offset':10512}, {'organization':'Vulpes Life Sciences Fund', 'character_offset':13845}]]], columns=['newOrg'])
df['Organizations'] = df['newOrg'].apply(lambda x: [i['organization'] for i in x])

輸出：

uj5u.com熱心網友回復：

你可以這樣做：

df['organizations'].str.extractall(r"organization= *(\w )") \
    .groupby(level=0).agg(', '.join).rename(columns={0:'Organizations'})

       Organizations
0        Vac, Health
1      Store, Museum
2          Mart, Rep
3       Lodge, Hotel
4  Airport, Landmark

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