Geek Answers Handbook

Setting values ​​with multiindex in pandas

There are already a couple of SO questions related to this, especially this one , however, none of the answers work for me and quite a few links to documents (especially lexsorting) are broken, so I will ask for another one.

I am trying to do something (seemingly) very simply. Consider the following multi-indexed Dataframe:

import pandas as pd; import random
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'])
df = pd.concat([pd.Series(np.random.randn(8), index=index), pd.Series(np.random.randn(8), index=index)], axis=1)

Now I want to set all the values ​​in the column 0for some value (for example, np.NaN) for observations in the category one. I did not succeed:

df.loc(axis=0)[:, "one"][0] = 1 # setting with copy warning

and

df.loc(axis=0)[:, "one", 0] = 1

which either issues a warning about the length of keys exceeding the length of the index, or one about the lack of lexsorting at sufficient depth.

What is the right way to do this?

+4
1

, loc MultiIndex 0 :

import pandas as pd; 
import random
arrays = [['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux'],
      ['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two']]

#add for testing
np.random.seed(0)
tuples = list(zip(*arrays))
index = pd.MultiIndex.from_tuples(tuples, names=['first', 'second'])
df = pd.concat([pd.Series(np.random.randn(8), index=index), pd.Series(np.random.randn(8), index=index)], axis=1)

print df
                     0         1
first second                    
bar   one     1.764052 -0.103219
      two     0.400157  0.410599
baz   one     0.978738  0.144044
      two     2.240893  1.454274
foo   one     1.867558  0.761038
      two    -0.977278  0.121675
qux   one     0.950088  0.443863
      two    -0.151357  0.333674

df.loc[('bar', "one"), 0] = 1
print df
                     0         1
first second                    
bar   one     1.000000 -0.103219
      two     0.400157  0.410599
baz   one     0.978738  0.144044
      two     2.240893  1.454274
foo   one     1.867558  0.761038
      two    -0.977278  0.121675
qux   one     0.950088  0.443863
      two    -0.151357  0.333674

second one, slice(None):

df.loc[(slice(None), "one"), 0] = 1
print df
                     0         1
first second                    
bar   one     1.000000 -0.103219
      two     0.400157  0.410599
baz   one     1.000000  0.144044
      two     2.240893  1.454274
foo   one     1.000000  0.761038
      two    -0.977278  0.121675
qux   one     1.000000  0.443863
      two    -0.151357  0.333674

Docs.

+4

More articles:


All Articles