Python 具有可编辑性的词典到具有可编辑性索引的多索引数据框架
我想要一个来自字典字典的多索引数据帧。内部字典包含相同长度的列表/numpy数组Python 具有可编辑性的词典到具有可编辑性索引的多索引数据框架,python,pandas,multi-index,Python,Pandas,Multi Index,我想要一个来自字典字典的多索引数据帧。内部字典包含相同长度的列表/numpy数组 x={'a':{'x':[0,1,2],'y':[1,2,3]}, 'b':{'x':[4,6,8],'y':[9,8,7]} 某些函数(x) => 你考虑过使用一些逆向工程吗?我的意思是构建一个多索引数据帧,并检查打印(df.to_dict())时的外观。使用本文的第一部分,我们可以得到您想要的输出 将熊猫作为pd导入 数据=[[0,1,2,1,2,3], [4, 6, 8, 9, 8, 7]] df=pd.Da
x={'a':{'x':[0,1,2],'y':[1,2,3]},
'b':{'x':[4,6,8],'y':[9,8,7]}
某些函数(x)
=>
你考虑过使用一些逆向工程吗?我的意思是构建一个多索引数据帧,并检查打印(df.to_dict())
时的外观。使用本文的第一部分,我们可以得到您想要的输出
将熊猫作为pd导入
数据=[[0,1,2,1,2,3],
[4, 6, 8, 9, 8, 7]]
df=pd.DataFrame(数据)
df.colums=pd.MultiIndex.from_乘积([['x','y',[0,1,2]]))
df.index=['a','b']
打印(df.to_dict())
{('x',0):{'a':0,'b':4},
('x',1):{'a':1,'b':6},
('x',2):{'a':2'b':8},
('y',0):{'a':1,'b':9},
('y',1):{'a':2'b':8},
('y',2):{'a':3,'b':7}
因此,如果您可以将数据作为两个列表,您可以使用pd.MultiIndex.from_product
技巧
否则
将熊猫作为pd导入
数据={'a':{'x':[0,1,2],'y':[1,2,3]},
'b':{'x':[4,6,8],'y':[9,8,7]}
df=pd.DataFrame(data).T
#您可以将列表展开为列
df[“x”]适用(pd系列)
#然后对每一列使用这个
#再次是pd.MultiIndex.from_乘积
#为您提供所需的输出
cols=df.columns
df=pd.concat([df[col].apply(pd.Series)用于col中的col),轴=1)
df.columns=pd.MultiIndex.from_乘积([cols[0,1,2]]))
我创建了两种解决此问题的替代方法,并对结果进行了计时。还包括其他答案以及原始函数
from copy import deepcopy
import pandas as pd
from collections import defaultdict
import numpy as np
x = {'a': {'x': [0, 1, 2], 'y': [1 ,2 ,3]},
'b': {'x': [4, 6, 8], 'y': [9, 8, 7]}}
test = deepcopy(x)
for i in range(1000):
test.update({f'a_{i}':test['a']})
test2 = {k:{key: val*300 for key, val in v.items()} for k, v in x.items()}
print(len(test2['a']['x']))
for i in range(1000):
test2.update({f'a_{i}':test2['a']})
def dict_of_dicts_of_collections_to_multiindex_df(dict_of_dicts_of_collections):
x = dict_of_dicts_of_collections
result = {}
for outer_key, intermediate_dict in x.items():
result[outer_key] = {}
for intermediate_key, collection in intermediate_dict.items():
try:
for i, e in enumerate(collection):
result[outer_key][(intermediate_key, i)] = e
except TypeError:
pass
return pd.DataFrame(result).T
def out_from_other_answer(data):
df = pd.DataFrame(data).T
cols = df.columns
df = pd.concat([df[col].apply(pd.Series) for col in cols], axis=1)
#tweaked to avoid hardcoding [0, 1, 2]
df.columns = pd.MultiIndex.from_product([cols, range(len(df.columns)//len(cols))])
return df
def out2(dict_of_dicts):
df = pd.DataFrame(list(dict_of_dicts.values()))
out_df = pd.concat([pd.DataFrame(df[col].values.tolist())
for col in df.columns
],
axis=1,
keys=df.columns,
)
out_df.index = dict_of_dicts.keys()
return out_df
def out3(data):
temp = defaultdict(list)
for d in list(data.values()):
for k, v in d.items():
temp[k].append(v)
out = pd.concat([pd.DataFrame(v) for v in temp.values()], axis=1, keys=temp.keys())
out.index = data.keys()
return out
让我们看看结果。
用小数据x
%timeit dict_of_dicts_of_collections_to_multiindex_df(x)
%timeit out_from_other_answer(x)
%timeit out2(x)
%timeit out3(x)
1.63 ms ± 102 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
4.49 ms ± 492 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.44 ms ± 102 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
1.49 ms ± 98.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
具有更多行但相同列的测试
%timeit dict_of_dicts_of_collections_to_multiindex_df(test)
%timeit out_from_other_answer(test)
%timeit out2(test)
%timeit out3(test)
70.3 ms ± 10.9 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
494 ms ± 40.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
4.81 ms ± 185 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
3.37 ms ± 115 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
随着行数和列数的增加,test2
%timeit dict_of_dicts_of_collections_to_multiindex_df(test2)
%timeit out_from_other_answer(test2)
%timeit out2(test2)
%timeit out3(test2)
1.24 s ± 19.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
1.1 s ± 63.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
590 ms ± 39.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
598 ms ± 44 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
似乎每个解决方案在数据增加的维度方面都受到不同的影响。总的来说,似乎out3是最好的选择。本质上,最好在处理输入数据之前更改其布局
%timeit dict_of_dicts_of_collections_to_multiindex_df(test2)
%timeit out_from_other_answer(test2)
%timeit out2(test2)
%timeit out3(test2)
1.24 s ± 19.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
1.1 s ± 63.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
590 ms ± 39.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
598 ms ± 44 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)