Question

使用分层列创建DataFrame的最简单方法是什么？

我目前正在使用名称词典创建一个DataFrame - ＆gt; Series使用：

df = pd.DataFrame(data=serieses)

我想使用相同的列名称，但在列上添加了更多级别的层次结构。目前我希望附加级别的列具有相同的值，让我们说“估计”。

我正在尝试以下但这似乎不起作用：

pd.DataFrame(data=serieses,columns=pd.MultiIndex.from_tuples([(x, "Estimates") for x in serieses.keys()]))

我得到的只是一个包含所有NaN的DataFrame。

例如，我正在寻找的是：

l1               Estimates    
l2  one  two  one  two  one  two  one  two
r1   1    2    3    4    5    6    7    8
r2   1.1  2    3    4    5    6    71   8.2

其中l1和l2是MultiIndex的标签

Answer 1

这似乎有效：

import pandas as pd

data = {'a': [1,2,3,4], 'b': [10,20,30,40],'c': [100,200,300,400]}

df = pd.concat({"Estimates": pd.DataFrame(data)}, axis=1, names=["l1", "l2"])

l1  Estimates         
l2          a   b    c
0           1  10  100
1           2  20  200
2           3  30  300
3           4  40  400

Answer 2

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0.19.1

Answer 3

我不确定，但我认为使用dict作为你的DF的输入和一个MulitIndex不能很好地结合在一起。使用数组作为输入使它工作。

我经常更喜欢dicts作为输入，一种方法是在创建df之后设置列：

import pandas as pd

data = {'a': [1,2,3,4], 'b': [10,20,30,40],'c': [100,200,300,400]}
df = pd.DataFrame(np.array(data.values()).T, index=['r1','r2','r3','r4'])

tups = zip(*[['Estimates']*len(data),data.keys()])

df.columns = pd.MultiIndex.from_tuples(tups, names=['l1','l2'])

l1          Estimates         
l2          a   c    b
r1          1  10  100
r2          2  20  200
r3          3  30  300
r4          4  40  400

或者当使用数组作为df的输入时：

data_arr = np.array([[1,2,3,4],[10,20,30,40],[100,200,300,400]])

tups = zip(*[['Estimates']*data_arr.shape[0],['a','b','c'])
df = pd.DataFrame(data_arr.T, index=['r1','r2','r3','r4'], columns=pd.MultiIndex.from_tuples(tups, names=['l1','l2']))

这给出了相同的结果。

使用分层列创建DataFrame

3 个答案: