将 pandas.DataFrame 转换为 Python 中的字典列表
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Convert pandas.DataFrame to list of dictionaries in Python
提问by Shankar Pandey
I have a dictionary which is converted from a dataframe as below :
我有一个从数据帧转换而来的字典，如下所示：
a = d.to_json(orient='index')
Dictionary :
字典：
{"0":{"yr":2017,"PKID":"58306, 57011","Subject":"ABC","ID":"T001"},"1":{"yr":2018,"PKID":"1234,54321","Subject":"XYZ","ID":"T002"}}
What I need is it be in a list, so essentially a list of dictionary. So i just add a [] because that is the format to be used in the rest of the code.
我需要的是它在一个列表中，所以本质上是一个字典列表。所以我只添加一个 [] 因为这是在其余代码中使用的格式。
input_dict = [a]
input_dict :
输入字典：
['
{"0":{"yr":2017,"PKID":"58306, 57011","Subject":"ABC","ID":"T001"},"1":{"yr":2018,"PKID":"1234,54321","Subject":"XYZ","ID":"T002"}}
']
I need to get the single quotes removed just after the [ and just before the ]. Also, have the PKID values in form of list.
我需要在 [ 之后和 ] 之前删除单引号。此外，以列表的形式获取 PKID 值。
How can this be achieved ?
如何做到这一点？
Expected Output :
预期输出：
[ {"yr":2017,"PKID":[58306, 57011],"Subject":"ABC","ID":"T001"},"1":{"yr":2018,"PKID":[1234,54321],"Subject":"XYZ","ID":"T002"} ]
NOTE : The PKID column has multiple integer values which have to come as a lift of integers. a string is not acceptable. so we need like "PKID":[58306, 57011] and not "PKID":"[58306, 57011]"
注意：PKID 列有多个整数值，它们必须是整数的提升。字符串是不可接受的。所以我们需要像 "PKID":[58306, 57011] 而不是 "PKID":"[58306, 57011]"
回答by Norrius
pandas.DataFrame.to_jsonreturns a string (JSON string), not a dictionary. Try to_dictinstead:
pandas.DataFrame.to_json返回一个字符串（JSON 字符串），而不是一个字典。试试吧to_dict：
>>> df
col1 col2
0 1 3
1 2 4
>>> [df.to_dict(orient='index')]
[{0: {'col1': 1, 'col2': 3}, 1: {'col1': 2, 'col2': 4}}]
>>> df.to_dict(orient='records')
[{'col1': 1, 'col2': 3}, {'col1': 2, 'col2': 4}]
回答by jpp
Here is one way:
这是一种方法：
from collections import OrderedDict
d = {"0":{"yr":2017,"PKID":"58306, 57011","Subject":"ABC","ID":"T001"},"1":{"yr":2018,"PKID":"1234,54321","Subject":"XYZ","ID":"T002"}}
list(OrderedDict(sorted(d.items())).values())
# [{'ID': 'T001', 'PKID': '58306, 57011', 'Subject': 'ABC', 'yr': 2017},
# {'ID': 'T002', 'PKID': '1234,54321', 'Subject': 'XYZ', 'yr': 2018}]
Note the ordered dictionary is ordered by text string keys, as supplied. You may wish to convert these to integers first before any processing via d = {int(k): v for k, v in d.items()}.
请注意，有序字典按提供的文本字符串键排序。您可能希望在通过d = {int(k): v for k, v in d.items()}.
回答by Ryan
You are converting your dictionary to jsonwhich is a string. Then you wrap your resulting string a list. So, naturally, the result is a string inside of a list.
您正在将字典转换json为字符串。然后将结果字符串包装成一个列表。因此，自然地，结果是列表中的字符串。
Try instead: [d]where dis your raw dictionary (not converted json
试试看：你的原始字典[d]在哪里d（未转换json
回答by Rakesh
You can use a list comprehension
您可以使用列表理解
Ex:
前任：
d = {"0":{"yr":2017,"PKID":"58306, 57011","Subject":"ABC","ID":"T001"},"1":{"yr":2018,"PKID":"1234,54321","Subject":"XYZ","ID":"T002"}}
print [{k: v} for k, v in d.items()]
Output:
输出：
[{'1': {'PKID': '1234,54321', 'yr': 2018, 'ID': 'T002', 'Subject': 'XYZ'}}, {'0': {'PKID': '58306, 57011', 'yr': 2017, 'ID': 'T001', 'Subject': 'ABC'}}]
回答by RoadRunner - MSFT
What about something like this:
像这样的事情怎么样：
from operator import itemgetter
d = {"0":{"yr":2017,"PKID":"58306, 57011","Subject":"ABC","ID":"T001"},"1":
{"yr":2018,"PKID":"1234,54321","Subject":"XYZ","ID":"T002"}}
sorted_d = sorted(d.items(), key=lambda x: int(x[0]))
print(list(map(itemgetter(1), sorted_d)))
Which Outputs:
哪些输出：
[{'yr': 2017, 'PKID': '58306, 57011', 'Subject': 'ABC', 'ID': 'T001'},
{'yr': 2018, 'PKID': '1234,54321', 'Subject': 'XYZ', 'ID': 'T002'}]
