Python中展开（扁平化）字典列

huangapple
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英文:

Python flatten a dictionary column

问题

以下是翻译好的部分：

原始数据框：

Python中展开（扁平化）字典列

  1. df['addresses'][0]
  3. [{'addressLine1': '124 Main Street',
  4.   'addressLine2': '',
  5.   'addressLine3': '',
  6.   'city': 'Portland',
  7.   'region': 'ME',
  8.   'postalCode': '04019',
  9.   'country': 'USA'}]
  1. test = pd.json_normalize(result['addresses'][0])
  2. test

Python中展开（扁平化）字典列

到目前为止，一切都正常，但当我使用该函数并将其应用于整个列时，生成的数据框如下所示。

  1. test = pd.json_normalize(result['addresses'])
  2. test

Python中展开（扁平化）字典列

以下是一些列数据：

  1. [[{'addressLine1': '124 Main Street',
  2.    'addressLine2': '',
  3.    'addressLine3': '',
  4.    'city': 'Portland',
  5.    'region': 'ME',
  6.    'postalCode': '04019',
  7.    'country': 'USA'}],
  8.   ...
  9.   ...
  10. ]
英文:

It should be a simple line of code using pd.json_normalize function but it's working only with a single string and it's not batch processing my whole column

Orginial dataframe

Python中展开（扁平化）字典列

  1. df['addresses'][0]
  3. [{'addressLine1': '124 Main Street',
  4.   'addressLine2': '',
  5.   'addressLine3': '',
  6.   'city': 'Portland',
  7.   'region': 'ME',
  8.   'postalCode': '04019',
  9.   'country': 'USA'}]
  1. test = pd.json_normalize(result['addresses'][0])
  2. test

Python中展开（扁平化）字典列

Everything up to this point works, but when I use the function and apply to the whole column, the resulting dataframe turned out to look like this.

  1. test = pd.json_normalize(result['addresses'])
  2. test

Python中展开（扁平化）字典列

Here are some column data:

  1. [[{'addressLine1': '124 Main Street',
  2.    'addressLine2': '',
  3.    'addressLine3': '',
  4.    'city': 'Portland',
  5.    'region': 'ME',
  6.    'postalCode': '04019',
  7.    'country': 'USA'}],
  8.  [{'addressLine1': '1234 Main Street',
  9.    'addressLine2': '',
  10.    'addressLine3': '',
  11.    'city': 'Chattanooga',
  12.    'region': 'TN',
  13.    'postalCode': '37402',
  14.    'country': 'USA'}],
  15.  [{'addressLine1': '1684151 Chair Street',
  16.    'addressLine2': '',
  17.    'addressLine3': '',
  18.    'city': 'Notaplace',
  19.    'region': 'AL',
  20.    'postalCode': '48835',
  21.    'country': 'USA'}],
  22.  [{'addressLine1': '136 Main Street',
  23.    'addressLine2': '',
  24.    'addressLine3': '',
  25.    'city': 'Portland',
  26.    'region': 'ME',
  27.    'postalCode': '22118',
  28.    'country': 'USA'}],
  29.  [{'addressLine1': '123452 HoneyDo LN',
  30.    'addressLine2': '',
  31.    'addressLine3': '',
  32.    'city': 'Portland',
  33.    'region': 'ME',
  34.    'postalCode': '04019',
  35.    'country': 'USA'}],
  36.  [{'addressLine1': '123 Main Street',
  37.    'addressLine2': 'Apt 2B',
  38.    'addressLine3': 'Building B',
  39.    'city': 'Portland',
  40.    'region': 'ME',
  41.    'postalCode': '04019',
  42.    'country': 'USA'}],
  43.  [{'addressLine1': '123 Main Street',
  44.    'addressLine2': 'Apt 2B',
  45.    'addressLine3': 'Building B',
  46.    'city': 'New York City',
  47.    'region': 'NY',
  48.    'postalCode': '10001',
  49.    'country': 'USA'}],
  50.  [{'addressLine1': '123 Main Street',
  51.    'addressLine2': 'Apt 2B',
  52.    'addressLine3': 'Building B',
  53.    'city': 'Portland',
  54.    'region': 'ME',
  55.    'postalCode': '04019',
  56.    'country': 'USA'}],
  57.  [{'addressLine1': '4578 Shiver Me Timbers Road',
  58.    'addressLine2': '',
  59.    'addressLine3': '',
  60.    'city': 'Portland',
  61.    'region': 'ME',
  62.    'postalCode': '04019',
  63.    'country': 'USA'}],
  64.  [{'addressLine1': '124 Main ST',
  65.    'addressLine2': '',
  66.    'addressLine3': '',
  67.    'city': 'PORTLAND',
  68.    'region': 'ME',
  69.    'postalCode': '04019',
  70.    'country': 'USA'}]]

答案1

得分: 1

如果我理解您的意思，您可以使用以下示例将您的数据帧 df 转换为包含 dict 数据的形式：

  1. df = pd.concat([df, df.pop('addresses').str[0].apply(pd.Series)], axis=1)
  2. print(df)

打印结果：

  1.                   addressLine1 addressLine2 addressLine3           city region postalCode country
  2. 0              124 Main Street                                 Portland     ME      04019     USA
  3. 1             1234 Main Street                              Chattanooga     TN      37402     USA
  4. 2         1684151 Chair Street                                Notaplace     AL      48835     USA
  5. 3              136 Main Street                                 Portland     ME      22118     USA
  6. 4            123452 HoneyDo LN                                 Portland     ME      04019     USA
  7. 5              123 Main Street       Apt 2B   Building B       Portland     ME      04019     USA
  8. 6              123 Main Street       Apt 2B   Building B  New York City     NY      10001     USA
  9. 7              123 Main Street       Apt 2B   Building B       Portland     ME      04019     USA
  10. 8  4578 Shiver Me Timbers Road                                 Portland     ME      04019     USA
  11. 9                  124 Main ST                                 PORTLAND     ME      04019     USA
英文:

If I understand you correctly, you can transform your dataframe df with dict data with following example:

  1. df = pd.concat([df, df.pop('addresses').str[0].apply(pd.Series)], axis=1)
  2. print(df)

Prints:

  1.                   addressLine1 addressLine2 addressLine3           city region postalCode country
  2. 0              124 Main Street                                 Portland     ME      04019     USA
  3. 1             1234 Main Street                              Chattanooga     TN      37402     USA
  4. 2         1684151 Chair Street                                Notaplace     AL      48835     USA
  5. 3              136 Main Street                                 Portland     ME      22118     USA
  6. 4            123452 HoneyDo LN                                 Portland     ME      04019     USA
  7. 5              123 Main Street       Apt 2B   Building B       Portland     ME      04019     USA
  8. 6              123 Main Street       Apt 2B   Building B  New York City     NY      10001     USA
  9. 7              123 Main Street       Apt 2B   Building B       Portland     ME      04019     USA
  10. 8  4578 Shiver Me Timbers Road                                 Portland     ME      04019     USA
  11. 9                  124 Main ST                                 PORTLAND     ME      04019     USA

答案2

得分: 1

以下是您要翻译的内容：

"It seems your list has one-element lists as elements.

Lets say your list is address_list then you get the first element in that list and then use json_normalize

  1. pd.json_normalize([e[0] for e in address_list])

If the test data that you posted is actually a column then just use:

  1. pd.json_normalize(result["addresses"].str[0])

Or if you have other columns in addition to addresses in your result dataframe:

  1. pd.concat(
  2.     [result.drop(column="addresses"), pd.json_normalize(result["addresses"].str[0])],
  3.     axis=1
  4. )"
英文:

It seems your list has one-element lists as elements.

Lets say your list is address_list then you get the first element in that list and then use json_normalize

  1. pd.json_normalize([e[0] for e in address_list])

If the test data that you posted is actually a column then just use:

  1. pd.json_normalize(result["addresses"].str[0])

Or if you have other columns in addition to addresses in your result dataframe:

  1. pd.concat(
  2.     [result.drop(column="addresses"), pd.json_normalize(result["addresses"].str[0])],
  3.     axis=1
  4. )

huangapple
  • 本文由 发表于 2023年3月1日 08:42:02
  • 转载请务必保留本文链接：https://go.coder-hub.com/75598632.html
  • dataframe
  • dictionary
  • json
  • parsing
  • python
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