Pandas Dataframe 将多个键值拆分到不同的列

Question

我有一个 dataframe 列，格式如下：

col1    col2   
 A     [{'Id':42,'prices':['30',’78’]},{'Id': 44,'prices':['20','47',‘89’]}]
 B     [{'Id':47,'prices':['30',’78’]},{'Id':94,'prices':['20']},{'Id':84,'prices':['20','98']}]

如何将其转换为以下内容？

col1    Id            price   
  A     42         ['30',’78’]
  A     44         ['20','47',‘89’]
  B     47         ['30',’78’]
  B     94         ['20']
  B     84         ['20','98']

我正在考虑使用 apply 和 lambda 作为解决方案，但我不确定如何。

编辑：为了重新创建这个 dataframe 我使用以下代码：

data = [['A', "[{'Id':42,'prices':['30','78']},{'Id': 44,'prices':['20','47','89']}]"], 
        ['B', "[{'Id':47,'prices':['30','78']},{'Id':94,'prices':['20']},{'Id':84,'prices':['20','98']}]"]] 

df = pd.DataFrame(data, columns = ['col1', 'col2'])

Answer 1

如果col2列中有列表，则解决方案：

print (type(df['col2'].iat[0]))
<class 'list'>

L = [{**{'col1': a}, **x} for a, b in df[['col1','col2']].to_numpy() for x in b]

df = pd.DataFrame(L)
print (df)
  col1  Id        prices
0    A  42      [30, 78]
1    A  44  [20, 47, 89]
2    B  47      [30, 78]
3    B  94          [20]
4    B  84      [20, 98]

如果有字符串：

print (type(df['col2'].iat[0]))
<class 'str'>

import ast

L = [{**{'col1': a}, **x} for a, b in df[['col1','col2']].to_numpy() for x in ast.literal_eval(b)]
df = pd.DataFrame(L)
print (df)
  col1  Id        prices
0    A  42      [30, 78]
1    A  44  [20, 47, 89]
2    B  47      [30, 78]
3    B  94          [20]
4    B  84      [20, 98]

为了更好地理解，可以使用：

import ast

L = []
for a, b in df[['col1','col2']].to_numpy():
    for x in ast.literal_eval(b):
        d = {'col1': a}
        out = {**d, **x}
        L.append(out)

df = pd.DataFrame(L)
print (df)
  col1  Id        prices
0    A  42      [30, 78]
1    A  44  [20, 47, 89]
2    B  47      [30, 78]
3    B  94          [20]
4    B  84      [20, 98]

Answer 2

将“数据”的第二个参数视为列表。

data= [
  ['A', [{'Id':42,'prices':['30','78']},{'Id': 44,'prices':['20','47','89']}]], 
  ['B', [{'Id':47,'prices':['30','78']}, {'Id':94,'prices':['20']},{'Id':84,'prices': 
        ['20','98']}]]
  ]

t_list = []

for i in range(len(data)):
    for j in range(len(data[i][1])):
        t_list.append((data[i][0], data[i][1][j]['Id'], data[i][1][j]['prices']))

df = pd.DataFrame(t_list, columns=['col1', 'id', 'price'])
print(df)

     col1  id         price
0    A     42      [30, 78]
1    A     44  [20, 47, 89]
2    B     47      [30, 78]
3    B     94          [20]
4    B     84      [20, 98]

Answer 3

您可以在此处将df.explode与pd.Series.apply和df.set_index和df.reset_index

df.set_index('col1').explode('col2')['col2'].apply(pd.Series).reset_index()

  col1  Id        prices
0    A  42      [30, 78]
1    A  44  [20, 47, 89]
2    B  47      [30, 78]
3    B  94          [20]
4    B  84      [20, 98]

当col2为字符串时，使用ast.literal_eval

import ast

data = [['A', "[{'Id':42,'prices':['30','78']},{'Id': 44,'prices':['20','47','89']}]"], 
        ['B', "[{'Id':47,'prices':['30','78']},{'Id':94,'prices':['20']},{'Id':84,'prices':['20','98']}]"]] 

df = pd.DataFrame(data, columns = ['col1', 'col2'])
df['col2'] = df['col2'].map(ast.literal_eval)

df.set_index('col1').explode('col2')['col2'].apply(pd.Series).reset_index()

  col1  Id        prices
0    A  42      [30, 78]
1    A  44  [20, 47, 89]
2    B  47      [30, 78]
3    B  94          [20]
4    B  84      [20, 98]