Conditional Replace Pandas
I have a DataFrame, and I want to replace the values in a particular column that exceed a value with zero. I had thought this was a way of achieving this:
df[df.my_channel > 20000].my_channel = 0
If I copy the channel into a new data frame it's simple:
df2 = df.my_channel
df2[df2 > 20000] = 0
This does exactly what I want, but seems not to work with the channel as part of the original DataFrame.
Solution 1:
.ix
indexer works okay for pandas version prior to 0.20.0, but since pandas 0.20.0, the .ix
indexer is deprecated, so you should avoid using it. Instead, you can use .loc
or iloc
indexers. You can solve this problem by:
mask = df.my_channel > 20000
column_name = 'my_channel'
df.loc[mask, column_name] = 0
Or, in one line,
df.loc[df.my_channel > 20000, 'my_channel'] = 0
mask
helps you to select the rows in which df.my_channel > 20000
is True
, while df.loc[mask, column_name] = 0
sets the value 0 to the selected rows where mask
holds in the column which name is column_name
.
Update:
In this case, you should use loc
because if you use iloc
, you will get a NotImplementedError
telling you that iLocation based boolean indexing on an integer type is not available.
Solution 2:
Try
df.loc[df.my_channel > 20000, 'my_channel'] = 0
Note: Since v0.20.0, ix
has been deprecated in favour of loc
/ iloc
.
Solution 3:
np.where
function works as follows:
df['X'] = np.where(df['Y']>=50, 'yes', 'no')
In your case you would want:
import numpy as np
df['my_channel'] = np.where(df.my_channel > 20000, 0, df.my_channel)
Solution 4:
The reason your original dataframe does not update is because chained indexing may cause you to modify a copy rather than a view of your dataframe. The docs give this advice:
When setting values in a pandas object, care must be taken to avoid what is called chained indexing.
You have a few alternatives:-
loc
+ Boolean indexing
loc
may be used for setting values and supports Boolean masks:
df.loc[df['my_channel'] > 20000, 'my_channel'] = 0
mask
+ Boolean indexing
You can assign to your series:
df['my_channel'] = df['my_channel'].mask(df['my_channel'] > 20000, 0)
Or you can update your series in place:
df['my_channel'].mask(df['my_channel'] > 20000, 0, inplace=True)
np.where
+ Boolean indexing
You can use NumPy by assigning your original series when your condition is not satisfied; however, the first two solutions are cleaner since they explicitly change only specified values.
df['my_channel'] = np.where(df['my_channel'] > 20000, 0, df['my_channel'])
Solution 5:
I would use lambda
function on a Series
of a DataFrame
like this:
f = lambda x: 0 if x>100 else 1
df['my_column'] = df['my_column'].map(f)
I do not assert that this is an efficient way, but it works fine.