简体   繁体   中英

Setting values on pandas DataFrame with multiindex

The following is a minimal example of what I am trying to do. I have a pandas DataFrame with multiindex as follows

import pandas as pd
import numpy as np

arrays = [['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux'],
          ['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two']]
tuples = list(zip(*arrays))
index = pd.MultiIndex.from_tuples(tuples, names=['first', 'second'])
s = pd.DataFrame(np.random.randn(8,2), index=index)

So the DataFrame I have is

                     0         1
first second                    
bar   one    -3.174428 -0.314160
      two     0.968316  0.278967
baz   one     0.171292 -0.789257
      two     1.420621  0.100964
foo   one    -1.001074 -0.517729
      two    -0.211823  0.951422
qux   one     1.173289  0.313692
      two    -0.159855  0.149710

What I want is to set all the observations with the index "second" equal to two as -1. What I have in mind is using .loc, something as follows:

s.loc[(:,'two')]

but .loc would not accept the ":" operator.

Could someone help here?

Option 1:

In [127]: s.loc[pd.IndexSlice[:, 'two'], :] = -1

In [128]: s
Out[128]:
                     0         1
first second
bar   one    -0.581647  0.225254
      two    -1.000000 -1.000000
baz   one     0.705050 -1.414695
      two    -1.000000 -1.000000
foo   one     0.359795  1.468521
      two    -1.000000 -1.000000
qux   one    -0.481149 -0.241922
      two    -1.000000 -1.000000

Option 2:

In [137]: s.loc[(slice(None),'two'), :] = -11

In [138]: s
Out[138]:
                      0          1
first second
bar   one      2.144487   0.024400
      two    -11.000000 -11.000000
baz   one     -0.177128  -1.088566
      two    -11.000000 -11.000000
foo   one     -0.780979   2.701814
      two    -11.000000 -11.000000
qux   one     -0.981635  -0.202875
      two    -11.000000 -11.000000

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM