将不等的列表列表解包到熊猫数据框中

Question

我在变量mydata有以下列表

[
  [["item1"], ["2016","value1"], ["2017","value2"], ["2018","value3"], ["2019","value4"]],
  [["item2"], ["2016","value1"], ["2017","value2"], ["2018","value3"], ["2019","value4"]],
  [["item3"], ["2016","value1"], ["2017","value2"], ["2018","value3"], ["2019","value4"]],
  [["item4"], ["2016","value1"], ["2017","value2"], ["2018","value3"], ["2019","value4"]],
]

我一直在尝试将其放入如下所示的数据框中：

index     item1  | item2  | item3  | item4
2016      value1 | value1 | value1 | value1
2017      value2 | value2 | value2 | value2
2018      value3 | value3 | value3 | value3
2019      value4 | value4 | value4 | value4

但是，当我尝试DataFrame(mydata) ，它给了我以下结果：

index   0                 | 1                 | 2                 | 3  
item1   ["2016","value1"] | ["2017","value2"] | ["2018","value3"] | ["2019","value4"]
item2   ["2016","value1"] | ["2017","value2"] | ["2018","value3"] | ["2019","value4"]
item3   ["2016","value1"] | ["2017","value2"] | ["2018","value3"] | ["2019","value4"]
item4   ["2016","value1"] | ["2017","value2"] | ["2018","value3"] | ["2019","value4"]

我认为这是因为mydata的性质，其中列表列表分布不均。 但是是否可以按照我指定的方式将其解压缩到数据dataFrame中？ 或者我是否必须创建一个 for 循环来遍历每个元素并将其解压缩到dataFrame ？

Answer 1

首先创建dict一种方法：

df = pd.DataFrame({l[0][0]:dict(l[1:]) for l in mydata})
print(df)

输出：

       item1   item2   item3   item4
2016  value1  value1  value1  value1
2017  value2  value2  value2  value2
2018  value3  value3  value3  value3
2019  value4  value4  value4  value4