通过构造函数创建MultiIndexed数据框 [英] Create MultiIndexed dataframe through constructor

问题描述

给出两个数组:

x
[('010_628', '2543677'), ('010_228', '2543677'), ('015_634', '2543677')]

y 
array([['me', 10228955],
       ['me', 10228955],
       ['me', 10228955]], dtype=object)

当前，此代码为我提供了一个具有元组的平面索引的数据框:

Currently, this code gets me a dataframe with a flat index of tuples:

df = pd.DataFrame(x, index=y, columns=['pm_code',   'sec_pm'])
df
                pm_code   sec_pm
(me, 10228955)  010_628  2543677
(me, 10228955)  010_228  2543677
(me, 10228955)  015_634  2543677

我该如何创建一个像这样的MultiIndex数据框?

How can I instead create a MultiIndex dataframe that looks like this?

                  pm_code   sec_pm
state site_no                     
me    10228955   010_628  2543677
                 010_228  2543677
                 015_634  2543677

我尝试使用pd.MultiIndex.from_tuples，但无法正确执行此操作.感谢您的帮助.

I've tried using pd.MultiIndex.from_tuples but I'm not able to get this right. Thanks for the help.

附录:性能比较

Appendix: Performance Comparisons

# unutbu #1
%timeit pd.DataFrame(x, index=pd.MultiIndex.from_arrays(y.T), columns=['pm_code',   'sec_pm'])
1000 loops, best of 3: 1.25 ms per loop

# unutbu #2
%timeit pd.DataFrame(x, index=pd.MultiIndex.from_tuples(y.tolist()), columns=['pm_code',   'sec_pm'])
1000 loops, best of 3: 1.47 ms per loop

# piRSquared
%timeit pd.DataFrame(x, index=y.T.tolist(), columns=['pm_code', 'sec_pm'])
1000 loops, best of 3: 1.41 ms per loop

# Andrew L
%timeit pd.DataFrame(x, index=[y[:,0], y[:,1]], columns=['pm_code',   'sec_pm'])
1000 loops, best of 3: 1.29 ms per loop

大

x2 = np.repeat(x, 10000, 0)
y2 = np.repeat(x, 10000, 0)

# unutbu #1
%timeit pd.DataFrame(x2, index=pd.MultiIndex.from_arrays(y2.T), columns=['pm_code',   'sec_pm'])
100 loops, best of 3: 17.3 ms per loop

# unutbu #2
%timeit pd.DataFrame(x2, index=pd.MultiIndex.from_tuples(y2.tolist()), columns=['pm_code',   'sec_pm'])
10 loops, best of 3: 30.5 ms per loop

# piRSquared
%timeit pd.DataFrame(x2, index=y2.T.tolist(), columns=['pm_code', 'sec_pm'])
10 loops, best of 3: 37.2 ms per loop

# Andrew L
%timeit pd.DataFrame(x2, index=[y2[:,0], y2[:,1]], columns=['pm_code',   'sec_pm'])
100 loops, best of 3: 22 ms per loop

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推荐答案

您可以使用pd.MultiIndex.from_arrays(y.T):

In [53]: pd.DataFrame(x, index=pd.MultiIndex.from_arrays(y.T), columns=['pm_code',   'sec_pm'])
Out[53]: 
             pm_code   sec_pm
me 10228955  010_628  2543677
   10228955  010_228  2543677
   10228955  015_634  2543677

或pd.MultiIndex.from_tuples(y.tolist()):

In [54]: pd.DataFrame(x, index=pd.MultiIndex.from_tuples(y.tolist()), columns=['pm_code',   'sec_pm'])
Out[54]: 
             pm_code   sec_pm
me 10228955  010_628  2543677
   10228955  010_228  2543677
   10228955  015_634  2543677

这篇关于通过构造函数创建MultiIndexed数据框的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！