在Python中将有序字典转换为新列

2024-10-02 16:34:45 发布

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在我的dataframe中,有些列采用OrderedDictionary格式。如何将它们转换为新列(不知道哪些列包含OrderedDictionary和OrderedDictionary中的元素)

示例列:

inventors
[OrderedDict([('@sequence', '001'), ('@app-type', 'applicant'), ('@designation', 'us-only'), ('addressbook', OrderedDict([('last-name', 'Nahm'), ('first-name', 'Seung Hoon'), ('address', OrderedDict([('city', 'Daejeon'), ('country', 'KR')]))])), ('residence', OrderedDict([('country', 'KR')]))]), OrderedDict([('@sequence', '002'), ('@app-type', 'applicant'), ('@designation', 'us-only'), ('addressbook', OrderedDict([('last-name', 'Jang'), ('first-name', 'Hoon Sik'), ('address', OrderedDict([('city', 'Daegu'), ('country', 'KR')]))])), ('residence', OrderedDict([('country', 'KR')]))])]

我想将其转换为以下数据帧(未写入所有列):

@sequence1  @app_type1   @designation1 @last_name1 @first_name1 ....
001         applicant    us_only        Nahm       Seung Hoon

在本例中,姓和名来自另一个嵌套字典。在数据中,我不知道哪些列包含OrderedDictional,为了简单起见,我只包含了数据集中的一列,即inventors


Tags: 数据nameapponlycountryfirstordereddictlast
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1楼 · 发布于 2024-10-02 16:34:45

你考虑过熊猫from_dict()函数吗?你知道吗

# Create example dict
import collections
inventors = collections.OrderedDict()
inventors['@sequence'] =  "001"
inventors['@app-type'] =  "applicant"
inventors['@designation'] =  "us-only"
inventors['last-name'] =  "Nahm"
inventors['first-name'] =  "Seung Hoon"

# Import pandas to use from_dict function
import pandas as pd

# Use from_dict() function; include orient='index' for now to avoid index error
df = pd.DataFrame.from_dict(inventors, orient='index')

# Transpose for final output
df.T

enter image description here

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