pythonpandasdataframereplacenegative-number

How to replace negative numbers in Pandas Data Frame by zero


I would like to know if there is someway of replacing all DataFrame negative numbers by zeros?


Solution

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

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

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

    In [1]: import pandas as pd
    
    In [2]: df = pd.DataFrame({'a': [0, -1, 2], 'b': [-3, 2, 1],
                               'c': ['foo', 'goo', 'bar']})
    
    In [3]: df
    Out[3]: 
       a  b    c
    0  0 -3  foo
    1 -1  2  goo
    2  2  1  bar
    
    In [4]: num = df._get_numeric_data()
    
    In [5]: num[num < 0] = 0
    
    In [6]: df
    Out[6]: 
       a  b    c
    0  0  0  foo
    1  0  2  goo
    2  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 pd
    
    In [2]: df = pd.DataFrame({'a': pd.to_timedelta([0, -1, 2], 'd'),
       ...:                    'b': pd.to_timedelta([-3, 2, 1], 'd')})
    
    In [3]: df
    Out[3]: 
            a       b
    0  0 days -3 days
    1 -1 days  2 days
    2  2 days  1 days
    
    In [4]: for k, v in df.iteritems():
       ...:     v[v < 0] = 0
       ...:     
    
    In [5]: df
    Out[5]: 
           a      b
    0 0 days 0 days
    1 0 days 2 days
    2 2 days 1 days
    

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

    In [1]: import pandas as pd
    
    In [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)] = 0
    
    In [4]: df
    Out[4]: 
           a      b
    0 0 days 0 days
    1 0 days 2 days
    2 2 days 1 days