Python Pandas DataFrame 到字典列表
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Pandas DataFrame to List of Dictionaries
提问by Mohamad Ibrahim
I have the following DataFrame:
我有以下数据帧：
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
我想把它翻译成每行的字典列表
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'}]
采纳答案by ComputerFellow
Edit
编辑
As John Galt mentions in his answer , you should probably instead use df.to_dict('records'). It's faster than transposing manually.
正如 John Galt 在他的回答中提到的，您可能应该使用df.to_dict('records'). 它比手动移调更快。
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
Original answer
原答案
Use df.T.to_dict().values(), like below:
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'}]
回答by Zero
Use df.to_dict('records')-- gives the output without having to transpose externally.
使用df.to_dict('records')-- 无需外部转置即可提供输出。
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by Hossain Muctadir
As an extension to John Galt'sanswer -
作为约翰高尔特答案的延伸-
For the following DataFrame,
对于以下数据帧，
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,
如果您想获得包含索引值的字典列表，您可以执行以下操作，
df.to_dict('index')
Which outputs a dictionary of dictionaries where keys of the parent dictionary are index values. In this particular case,
它输出一个字典字典，其中父字典的键是索引值。在这种特殊情况下，
{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'}}
回答by Joe Rivera
If you are interested in only selecting one column this will work.
如果您只对选择一列感兴趣，这将起作用。
df[["item1"]].to_dict("records")
The below will NOTwork 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.
下面将不工作，并产生一个类型错误：不支持的类型。我相信这是因为它试图将系列转换为字典，而不是将数据帧转换为字典。
df["item1"].to_dict("records")
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.
我需要只选择一列并将其转换为以列名作为键的字典列表，并且在此停留了一段时间，所以我想我会分享。
