I would like to know if there is someway of replacing all DataFrame negative numbers by zeros?
This question is related to
python
pandas
dataframe
replace
negative-number
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
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
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
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
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)
With lambda function
df['column'] = df['column'].apply(lambda x : x if x > 0 else 0)
Source: Stackoverflow.com