How to replace negative numbers in Pandas Data Frame by zero

If all your columns are numeric, you can use boolean indexing:

In [1]: import pandas as pdIn [2]: df = pd.DataFrame({'a': [0, -1, 2], 'b': [-3, 2, 1]})In [3]: dfOut[3]:    a  b0  0 -31 -1  22  2  1In [4]: df[df < 0] = 0In [5]: dfOut[5]:    a  b0  0  01  0  22  2  1

For the more general case, this answer shows the private method _get_numeric_data:

In [1]: import pandas as pdIn [2]: df = pd.DataFrame({'a': [0, -1, 2], 'b': [-3, 2, 1],                           'c': ['foo', 'goo', 'bar']})In [3]: dfOut[3]:    a  b    c0  0 -3  foo1 -1  2  goo2  2  1  barIn [4]: num = df._get_numeric_data()In [5]: num[num < 0] = 0In [6]: dfOut[6]:    a  b    c0  0  0  foo1  0  2  goo2  2  1  bar

With timedelta type, boolean indexing seems to work on separate columns, but not on the whole dataframe. So you can do:

In [1]: import pandas as pdIn [2]: df = pd.DataFrame({'a': pd.to_timedelta([0, -1, 2], 'd'),   ...:                    'b': pd.to_timedelta([-3, 2, 1], 'd')})In [3]: dfOut[3]:         a       b0  0 days -3 days1 -1 days  2 days2  2 days  1 daysIn [4]: for k, v in df.iteritems():   ...:     v[v < 0] = 0   ...:     In [5]: dfOut[5]:        a      b0 0 days 0 days1 0 days 2 days2 2 days 1 days

Update: comparison with a pd.Timedelta works on the whole DataFrame:

In [1]: import pandas as pdIn [2]: df = pd.DataFrame({'a': pd.to_timedelta([0, -1, 2], 'd'),   ...:                    'b': pd.to_timedelta([-3, 2, 1], 'd')})In [3]: df[df < pd.Timedelta(0)] = 0In [4]: dfOut[4]:        a      b0 0 days 0 days1 0 days 2 days2 2 days 1 days

Another succinct way of doing this is pandas.DataFrame.clip.

For example:

import pandas as pdIn [20]: df = pd.DataFrame({'a': [-1, 100, -2]})In [21]: dfOut[21]:      a0   -11  1002   -2In [22]: df.clip(lower=0)Out[22]:      a0    01  1002    0

There's also df.clip_lower(0).

Perhaps you could use pandas.where(args) like so:

data_frame = data_frame.where(data_frame < 0, 0)