Python 熊猫：基于json对象替换DataFrame中的列值

我有一个名为“countries”的json对象，如下所示，包含所有国家的ISO代码列表：

countries = [{"name":"Afghanistan","alpha-2":"AF","country-code":"004"},{"name":"Åland Islands","alpha-2":"AX","country-code":"248"},{"name":"Albania","alpha-2":"AL","country-code":"008"},{"name":"Algeria","alpha-2":"DZ","country-code":"012"}]

我有一个带有“国家”列的熊猫数据框：

Country
--------
Albania
Algeria
Algeria

我想用json对象中的“alpha-2”值替换Country列“name”。结果应该是：

Country
---------
AL
DZ
DZ

我正在尝试这样做，它不会给出任何错误，也不会改变值

df['Country'] = df['Country'].replace(lambda y: (x['alpha-2'] for x in countries) if y in (x['name'] for x in countries) else y)

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您正在访问df['Country']中的列Country，因此如果您有其他字段以及有问题的alpha-2，那么为什么不简单地使用df['Country']=df['alpha-2']，它无论如何都比lambda快

熊猫不建议使用行lambda，原因与pd.Series.apply相同。更好的方法是构造一个映射字典，然后使用矢量化：

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如果您已经将Json转换为pandas dataFrame，并且您拥有如列Country所示的dataFrame，那么您可以简单地使用map函数或使用replace方法，这两种方法都可以在这里使用

df['Country'] = df['Country'].map({'Albania': 'AL', 'Algeria': 'DZ'})

或：

或者，您也可以创建一个字典，一次进行多个替换，如下所示

new_vals = {
  'Albania': 'AL',
  'Algeria': 'DZ',
  }


df['Country'].replace(new_vals)
# df['Country'].replace(new_vals, inplace=True)

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您可以用这种方法创建一个新的{country:country_code}字典模式，使用country_to_country_code={v['name']：v['alpha-2']表示v in countries}，然后只需将您的国家列与这个country_to_country_代码字典映射即可

import pandas as pd
df = pd.DataFrame({"Country":["Albania", "Algeria", "Algeria"]})
countries = [{"name":"Afghanistan","alpha-2":"AF","country-code":"004"},{"name":"Åland Islands","alpha-2":"AX","country-code":"248"},{"name":"Albania","alpha-2":"AL","country-code":"008"},{"name":"Algeria","alpha-2":"DZ","country-code":"012"}]

country_to_country_code= {v['name']:v['alpha-2'] for v in countries}
df.loc[:, 'Country'] = df['Country'].map(country_to_country_code)
print(df)

输出

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您可以将列表转换为数据帧，例如df2，然后进行替换

import pandas as pd

countries = [{"name":"Afghanistan","alpha-2":"AF","country-code":"004"},{"name":"Åland Islands","alpha-2":"AX","country-code":"248"},{"name":"Albania","alpha-2":"AL","country-code":"008"},{"name":"Algeria","alpha-2":"DZ","country-code":"012"}]
df2 = pd.DataFrame(countries)

co = [('Country', ['Afghanistan', 'Algeria', 'Albania'])]  # your original dataframe with country
df1 = pd.DataFrame.from_items(co)



df1['Country'] = df1['Country'].replace(df2.set_index('name')['alpha-2'])

df1 should look like:
[enter image description here][1]

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我认为问题是OP是从json对象开始的，所以他们正在寻找一种方法，如果使用pd.Series.map，则将json转换为单个映射。@jpp，这可能是真的：-，否则映射将是合适的。这在某种意义上是好的，OP希望将json转换为单个映射+1

import pandas as pd
df = pd.DataFrame({"Country":["Albania", "Algeria", "Algeria"]})
countries = [{"name":"Afghanistan","alpha-2":"AF","country-code":"004"},{"name":"Åland Islands","alpha-2":"AX","country-code":"248"},{"name":"Albania","alpha-2":"AL","country-code":"008"},{"name":"Algeria","alpha-2":"DZ","country-code":"012"}]

country_to_country_code= {v['name']:v['alpha-2'] for v in countries}
df.loc[:, 'Country'] = df['Country'].map(country_to_country_code)
print(df)

  Country
0      AL
1      DZ
2      DZ

import pandas as pd

countries = [{"name":"Afghanistan","alpha-2":"AF","country-code":"004"},{"name":"Åland Islands","alpha-2":"AX","country-code":"248"},{"name":"Albania","alpha-2":"AL","country-code":"008"},{"name":"Algeria","alpha-2":"DZ","country-code":"012"}]
df2 = pd.DataFrame(countries)

co = [('Country', ['Afghanistan', 'Algeria', 'Albania'])]  # your original dataframe with country
df1 = pd.DataFrame.from_items(co)



df1['Country'] = df1['Country'].replace(df2.set_index('name')['alpha-2'])

df1 should look like:
[enter image description here][1]