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.
.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 maskholds 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.