拆分嵌入到数据框熊猫列表值列中的字典

Question

我在数据框 df 中有一个列 authors，其中值首先在字典中，然后将字典添加到列表中。然后将列表存储在列中。如：

[{'family': 'Yaisy',
  'given': 'Lisa',
  'affiliation': [{'name': 'Department of Sciences, Faculty Sciences, University of Science'}]},
 {'family': 'Kite',
  'given': 'Hume',
  'affiliation': [{'name': 'Department of Sciences, Science and Technology'}]},
 {'family': 'Jones',
  'given name': 'Mike',
  'localId': 'aza',
  'affiliation': [{'name': 'Department of Health, Science and Technology'}]},
 {'family': 'abc',
  'given name': 'xyz',
  'affiliation': [{'name': 'Health Sciences, University of Science'}]}]

我想将此列表分成不同的列，其中键作为列名，值作为列值。由于键有重复的名称，我可以为每个列名添加 1,2,3 作为后缀。我尝试了this question中建议的解决方案

df.join(pd.json_normalize(df.authors))

但是，我首先需要将列表更改为一个简单的字典，然后使用上面的解决方案。所以我尝试展平列表以获取字典并将它们存储在同一列中：

df.author = [y for x in df.author for y in x]

但是在这里我得到了值的长度与索引的长度不匹配的错误。

谁能帮我解决这个问题？谢谢！

Answer 1

一种方法是手动构建记录列表并创建数据框：

lst = [{'family': 'Yaisy',
  'given': 'Lisa',
  'affiliation': [{'name': 'Department of Sciences, Faculty Sciences, University of Science'}]},
 {'family': 'Kite',
  'given': 'Hume',
  'affiliation': [{'name': 'Department of Sciences, Science and Technology'}]},
 {'family': 'Jones',
  'given name': 'Mike',
  'localId': 'aza',
  'affiliation': [{'name': 'Department of Health, Science and Technology'}]},
 {'family': 'abc',
  'given name': 'xyz',
  'affiliation': [{'name': 'Health Sciences, University of Science'}]}]

records = []
for item in lst:
    records.append({
        'family': item['family'],
        'given': item.get('given', item.get('given name')),
        **item['affiliation'][0]
    })

df = pd.DataFrame(records)
print(df)

打印：

  family given                                               name
0  Yaisy  Lisa  Department of Sciences, Faculty Sciences, Univ...
1   Kite  Hume     Department of Sciences, Science and Technology
2  Jones  Mike       Department of Health, Science and Technology
3    abc   xyz             Health Sciences, University of Science