Using where on DataFrame

Question

I can't seem to figure out how to do this in one step. I want to set the even entries to 1 and the odd ones to -1:

df = DataFrame(np.arange(16).reshape(4,4))
x1 = df.where(df%2==0, -1)
x2 = x1.where(df%2!=0, 1)

The docs for pandas.DataFrame.where say:

Return an object of same shape as self and whose corresponding entries are from self where cond is True and otherwise are from other.

So is this the only way?

Answer 1

We could use np.where to use the boolean condition to set the values where true and to set the value to the alternate value when false. This will return a numpy array which if you want as a df you have to pass as an arg to the df ctor:

In [31]:

df = pd.DataFrame(np.where(df%2==0,-1,1))
df
Out[31]:
   0  1  2  3
0 -1  1 -1  1
1 -1  1 -1  1
2 -1  1 -1  1
3 -1  1 -1  1

Answer 2

numpy.choose can be useful here.

from pandas import DataFrame
import numpy as np
df = DataFrame(np.arange(16).reshape(4,4))
np.choose(df%2, [1, -1])
=> 
   0  1  2  3
0  1 -1  1 -1
1  1 -1  1 -1
2  1 -1  1 -1
3  1 -1  1 -1

The second arg is the list of values to replace with. The value at index=0 replaces where values==0, etc.

Answer 3

For the whole dataframe, in one line:

In [34]: df = pd.DataFrame(randint(0,8,size=(3,3)))

In [35]: df
Out[35]: 
   0  1  2
0  3  4  1
1  3  3  1
2  7  7  3

In [37]: df = ((df%2)-0.5)*-2

In [38]: df
Out[38]: 
   0  1  2
0 -1  1 -1
1 -1 -1 -1
2 -1 -1 -1

By column:

df = pd.DataFrame([list('addfg'),list('LKJHU')]).T
df.Odds = ((df.index.values%2)-0.5)*-2

result:

   0  1  Odds
0  a  L     1
1  d  K    -1
2  d  J     1
3  f  H    -1
4  g  U     1