pandas 从对象创建数据框
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Create a data frame from object
提问by FooBar
I'm given with a collection of objects and need to save their attributes in a data frame. These objects all have the same attributes, but different values.
我得到了一组对象，需要将它们的属性保存在数据框中。这些对象都具有相同的属性，但值不同。
I was trying
我在尝试
dicts = [x.__dict__ for x in objectCollection]
df = pd.concat(dicts)
but that would give me
但这会给我
TypeError: cannot concatenate a non-NDFrame object
Following, for reproduction purposes, my robject (it contains 4 dictionaries)
以下，出于复制目的，我的r对象（它包含 4 个字典）
In[94]: r
Out[94]:
[{'JB': 0.5261750636924186,
'JBar': 0.5261750636925998,
'U': 0.3294737627050343,
'VB': 0,
'VBar': 0,
'WB': 0.3879376586708586,
'WBar': 0.38793765867087865,
'YB': 0.48616322013005736,
'YBar': 0.4861632201302899,
'pB': 1,
'pBar': 1.0000000000002274,
'theta': 30.452217802750784,
'thetaB': 15.226108901371752,
'thetaBar': 15.226108901379034,
'u': 0.027673559739652746,
'vB': 0.4213606342845696,
'vBar': 0.4213606342847711,
'wB': 0.40542217802756686,
'wBar': 0.4054221780275895},
{'JB': 0.591730026927601,
'JBar': 0.5888459364503311,
'U': 0.3574380470030322,
'VB': 0,
'VBar': 0,
'WB': 0.4133529860815169,
'WBar': 0.42286537327529117,
'YB': 0.4450754942945968,
'YBar': 0.4420654404537785,
'pB': 1,
'pBar': 1.0068090684451363,
'theta': 35.72497708330982,
'thetaB': 17.92309565619036,
'thetaBar': 17.801881427119454,
'u': 0.02414497177678684,
'vB': 0.43275263877136705,
'vBar': 0.4298259246315051,
'wB': 0.43134506957181085,
'wBar': 0.44209406710077587},
{'JB': 0.46774949304709174,
'JBar': 0.783373528458816,
'U': 0.4259229207218348,
'VB': 0,
'VBar': 0,
'WB': 0.47789508661595614,
'WBar': 0.5129644238839255,
'YB': 0.5805290540642507,
'YBar': 0.39605190181885574,
'pB': 1,
'pBar': 1.4657903456561598,
'theta': 39.39305475097706,
'thetaB': 12.929841051766449,
'thetaBar': 26.463213699210606,
'u': 0.02341904411689341,
'vB': 0.302804518015738,
'vBar': 0.6197431690965912,
'wB': 0.4948305475091409,
'wBar': 0.5414095820747569},
{'JB': 0.5246012340492012,
'JBar': 0.8731358801035817,
'U': 0.4675627478175433,
'VB': 0,
'VBar': 0,
'WB': 0.5155637079793887,
'WBar': 0.5645778456068301,
'YB': 0.5312340916010532,
'YBar': 0.3593021864121634,
'pB': 1,
'pBar': 1.4785161674233223,
'theta': 45.929330938192074,
'thetaB': 15.162892309068837,
'thetaBar': 30.766438629123233,
'u': 0.020410094185461556,
'vB': 0.3094760601521056,
'vBar': 0.6279459101716279,
'wB': 0.5334306672268627,
'wBar': 0.5960674456434281}]
回答by kushan_s
The pandas DataFrame constructordoes accept a list of dicts and can parse them into a DataFrame. This works for me
pandas DataFrame 构造函数接受一个dicts列表，并且可以将它们解析为一个 DataFrame。这对我有用
df = pd.DataFrame(r)
This gave me a dataframe with the dict keys as columns and their attributes as rows. Is this what you were trying to achieve?
这给了我一个数据框，其中 dict 键作为列，它们的属性作为行。这就是你想要达到的目标吗？
回答by Ramón J Romero y Vigil
Try a pd.DataFrame construction within the list comprehension:
在列表推导式中尝试 pd.DataFrame 构造：
import pandas as pd
pd.concat([pd.DataFrame(x[1], index=[x[0]]) for x in enumerate(r)])
