vectorize conditional assignment in pandas dataframe

Question

If I have a dataframe df with column x and want to create column y based on values of x using this in pseudo code:

<

Answer 1

This is a good use case for pd.cut where you define ranges and based on those ranges you can assign labels:

df['y'] = pd.cut(df['x'], [-np.inf, -2, 2, np.inf], labels=[1, 0, -1], right=False)

Output

   x  y
0  0  0
1 -3  1
2  5 -1
3 -1  0
4  1  0

Answer 2

One simple method would be to assign the default value first and then perform 2 loc calls:

In [66]:

df = pd.DataFrame({'x':[0,-3,5,-1,1]})
df
Out[66]:
   x
0  0
1 -3
2  5
3 -1
4  1

In [69]:

df['y'] = 0
df.loc[df['x'] < -2, 'y'] = 1
df.loc[df['x'] > 2, 'y'] = -1
df
Out[69]:
   x  y
0  0  0
1 -3  1
2  5 -1
3 -1  0
4  1  0

If you wanted to use np.where then you could do it with a nested np.where:

In [77]:

df['y'] = np.where(df['x'] < -2 , 1, np.where(df['x'] > 2, -1, 0))
df
Out[77]:
   x  y
0  0  0
1 -3  1
2  5 -1
3 -1  0
4  1  0

So here we define the first condition as where x is less than -2, return 1, then we have another np.where which tests the other condition where x is greater than 2 and returns -1, otherwise return 0

timings

In [79]:

%timeit df['y'] = np.where(df['x'] < -2 , 1, np.where(df['x'] > 2, -1, 0))

1000 loops, best of 3: 1.79 ms per loop

In [81]:

%%timeit
df['y'] = 0
df.loc[df['x'] < -2, 'y'] = 1
df.loc[df['x'] > 2, 'y'] = -1

100 loops, best of 3: 3.27 ms per loop

So for this sample dataset the np.where method is twice as fast