Pandas DataFrame到字典列表

问题:Pandas DataFrame到字典列表

我有以下DataFrame:

客户item1 item2 item3
1个苹果牛奶番茄
2水橙土豆
3汁芒果片

我想将其翻译为每行词典列表

rows = [{'customer': 1, 'item1': 'apple', 'item2': 'milk', 'item3': 'tomato'},
    {'customer': 2, 'item1': 'water', 'item2': 'orange', 'item3': 'potato'},
    {'customer': 3, 'item1': 'juice', 'item2': 'mango', 'item3': 'chips'}]

I have the following DataFrame:

customer    item1      item2    item3
1           apple      milk     tomato
2           water      orange   potato
3           juice      mango    chips

which I want to translate it to list of dictionaries per row

rows = [{'customer': 1, 'item1': 'apple', 'item2': 'milk', 'item3': 'tomato'},
    {'customer': 2, 'item1': 'water', 'item2': 'orange', 'item3': 'potato'},
    {'customer': 3, 'item1': 'juice', 'item2': 'mango', 'item3': 'chips'}]

回答 0

编辑

正如John Galt在回答中提到的那样,您可能应该改用df.to_dict('records')。它比手动移调要快。

In [20]: timeit df.T.to_dict().values()
1000 loops, best of 3: 395 µs per loop

In [21]: timeit df.to_dict('records')
10000 loops, best of 3: 53 µs per loop

原始答案

使用df.T.to_dict().values(),如下所示:

In [1]: df
Out[1]:
   customer  item1   item2   item3
0         1  apple    milk  tomato
1         2  water  orange  potato
2         3  juice   mango   chips

In [2]: df.T.to_dict().values()
Out[2]:
[{'customer': 1.0, 'item1': 'apple', 'item2': 'milk', 'item3': 'tomato'},
 {'customer': 2.0, 'item1': 'water', 'item2': 'orange', 'item3': 'potato'},
 {'customer': 3.0, 'item1': 'juice', 'item2': 'mango', 'item3': 'chips'}]

Edit

As John Galt mentions in his answer , you should probably instead use df.to_dict('records'). It’s faster than transposing manually.

In [20]: timeit df.T.to_dict().values()
1000 loops, best of 3: 395 µs per loop

In [21]: timeit df.to_dict('records')
10000 loops, best of 3: 53 µs per loop

Original answer

Use df.T.to_dict().values(), like below:

In [1]: df
Out[1]:
   customer  item1   item2   item3
0         1  apple    milk  tomato
1         2  water  orange  potato
2         3  juice   mango   chips

In [2]: df.T.to_dict().values()
Out[2]:
[{'customer': 1.0, 'item1': 'apple', 'item2': 'milk', 'item3': 'tomato'},
 {'customer': 2.0, 'item1': 'water', 'item2': 'orange', 'item3': 'potato'},
 {'customer': 3.0, 'item1': 'juice', 'item2': 'mango', 'item3': 'chips'}]

回答 1

用途df.to_dict('records')-提供输出,而无需外部转置。

In [2]: df.to_dict('records')
Out[2]:
[{'customer': 1L, 'item1': 'apple', 'item2': 'milk', 'item3': 'tomato'},
 {'customer': 2L, 'item1': 'water', 'item2': 'orange', 'item3': 'potato'},
 {'customer': 3L, 'item1': 'juice', 'item2': 'mango', 'item3': 'chips'}]

Use df.to_dict('records') — gives the output without having to transpose externally.

In [2]: df.to_dict('records')
Out[2]:
[{'customer': 1L, 'item1': 'apple', 'item2': 'milk', 'item3': 'tomato'},
 {'customer': 2L, 'item1': 'water', 'item2': 'orange', 'item3': 'potato'},
 {'customer': 3L, 'item1': 'juice', 'item2': 'mango', 'item3': 'chips'}]

回答 2

作为对John Galt答案的扩展-

对于以下DataFrame,

   customer  item1   item2   item3
0         1  apple    milk  tomato
1         2  water  orange  potato
2         3  juice   mango   chips

如果要获取包含索引值的词典列表,可以执行以下操作:

df.to_dict('index')

输出字典的字典,其中父字典的键是索引值。在这种情况下

{0: {'customer': 1, 'item1': 'apple', 'item2': 'milk', 'item3': 'tomato'},
 1: {'customer': 2, 'item1': 'water', 'item2': 'orange', 'item3': 'potato'},
 2: {'customer': 3, 'item1': 'juice', 'item2': 'mango', 'item3': 'chips'}}

As an extension to John Galt’s answer –

For the following DataFrame,

   customer  item1   item2   item3
0         1  apple    milk  tomato
1         2  water  orange  potato
2         3  juice   mango   chips

If you want to get a list of dictionaries including the index values, you can do something like,

df.to_dict('index')

Which outputs a dictionary of dictionaries where keys of the parent dictionary are index values. In this particular case,

{0: {'customer': 1, 'item1': 'apple', 'item2': 'milk', 'item3': 'tomato'},
 1: {'customer': 2, 'item1': 'water', 'item2': 'orange', 'item3': 'potato'},
 2: {'customer': 3, 'item1': 'juice', 'item2': 'mango', 'item3': 'chips'}}

回答 3

如果您只想选择一列,则可以使用。

df[["item1"]].to_dict("records")

下面将工作,并产生一个类型错误:不支持的类型。我相信这是因为它正在尝试将系列转换为字典,而不是将数据帧转换为字典。

df["item1"].to_dict("records")

我只需要选择一个列,然后将其转换为以列名作为键的字典列表,然后在此卡住一会儿,以至于我想与大家分享。

If you are interested in only selecting one column this will work.

df[["item1"]].to_dict("records")

The below will NOT work and produces a TypeError: unsupported type: . I believe this is because it is trying to convert a series to a dict and not a Data Frame to a dict.

df["item1"].to_dict("records")

I had a requirement to only select one column and convert it to a list of dicts with the column name as the key and was stuck on this for a bit so figured I’d share.