CODEWITHSUNDEEP
pythonpandasdataframereplacenegative-number
Hangon
Hangon

Reputation: 2609

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?

Upvotes: 141

Views: 276615

Answers (9)

Alfredo EP
Alfredo EP

Reputation: 81

Try this:

df.loc[(df < 0).index, :] = 0

To avoid getting a unpredicatable behavior: Returning a view versus a copy

Upvotes: 0

follyroof
follyroof

Reputation: 3540

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

For example:

import pandas as pd

In [20]: df = pd.DataFrame({'a': [-1, 100, -2]})

In [21]: df
Out[21]: 
     a
0   -1
1  100
2   -2

In [22]: df.clip(lower=0)
Out[22]: 
     a
0    0
1  100
2    0

Upvotes: 145

tef2128
tef2128

Reputation: 780

If you have a dataset of mixed data types, also consider moving the non-numerics to the index, updating the data, then removing the index:

df = pd.DataFrame({'a': [0, -1, 2], 'b': [-3, 2, 1],
                       'c': ['foo', 'goo', 'bar']})
df = df.set_index('c')
df[df < 0] = 0
df = df.reset_index()

The approach using _get_numeric_data() didn't work for me for some reason.

Upvotes: 0

Terminator17
Terminator17

Reputation: 860

A slight modification of the answers present.

Let's identify all the numeric columns and create a dataframe with all numeric values. Then replace the negative values with NaN in new dataframe

df_numeric = df.select_dtypes(include=[np.number])
df_numeric = df_numeric.where(lambda x: x > 0, np.nan)

Now, drop the columns where negative values are handled in the main data frame and then concatenate the new column values to the main data frame

numeric_cols = df_numeric.columns.values
df = df.drop(columns=numeric_cols)
df = pd.concat([df, df_numeric], axis = 1)

Upvotes: 1

Wickkiey
Wickkiey

Reputation: 4632

With lambda function

df['column'] = df['column'].apply(lambda x : x if x > 0 else 0)

Upvotes: 15

Michael Conlin
Michael Conlin

Reputation: 909

Another clean option that I have found useful is pandas.DataFrame.mask which will "replace values where the condition is true."

Create the DataFrame:

In [2]: import pandas as pd

In [3]: df = pd.DataFrame({'a': [0, -1, 2], 'b': [-3, 2, 1]})

In [4]: df
Out[4]: 
   a  b
0  0 -3
1 -1  2
2  2  1

Replace negative numbers with 0:

In [5]: df.mask(df < 0, 0)
Out[5]: 
   a  b
0  0  0
1  0  2
2  2  1

Or, replace negative numbers with NaN, which I frequently need:

In [7]: df.mask(df < 0)
Out[7]: 
     a    b
0  0.0  NaN
1  NaN  2.0
2  2.0  1.0

Upvotes: 23

MarKo9
MarKo9

Reputation: 97

If you are dealing with a large df (40m x 700 in my case) it works much faster and memory savvy through iteration on columns with something like.

for col in df.columns:
    df[col][df[col] < 0] = 0

Upvotes: 2

alacy
alacy

Reputation: 5064

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

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

Upvotes: 14

Lev Levitsky
Lev Levitsky

Reputation: 65781

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

Upvotes: 153

Related Questions