I want to find all values in a Pandas dataframe that contain whitespace (any arbitrary amount) and replace those values with NaNs.
Any ideas how this can be improved?
Basically I want to turn this:
A B C
2000-01-01 -0.532681 foo 0
2000-01-02 1.490752 bar 1
2000-01-03 -1.387326 foo 2
2000-01-04 0.814772 baz
2000-01-05 -0.222552 4
2000-01-06 -1.176781 qux
Into this:
A B C
2000-01-01 -0.532681 foo 0
2000-01-02 1.490752 bar 1
2000-01-03 -1.387326 foo 2
2000-01-04 0.814772 baz NaN
2000-01-05 -0.222552 NaN 4
2000-01-06 -1.176781 qux NaN
I've managed to do it with the code below, but man is it ugly. It's not Pythonic and I'm sure it's not the most efficient use of pandas either. I loop through each column and do boolean replacement against a column mask generated by applying a function that does a regex search of each value, matching on whitespace.
for i in df.columns:
df[i][df[i].apply(lambda i: True if re.search('^\s*$', str(i)) else False)]=None
It could be optimized a bit by only iterating through fields that could contain empty strings:
if df[i].dtype == np.dtype('object')
But that's not much of an improvement
And finally, this code sets the target strings to None, which works with Pandas' functions like fillna(), but it would be nice for completeness if I could actually insert a NaN directly instead of None.
I think df.replace() does the job, since pandas 0.13:
df = pd.DataFrame([
[-0.532681, 'foo', 0],
[1.490752, 'bar', 1],
[-1.387326, 'foo', 2],
[0.814772, 'baz', ' '],
[-0.222552, ' ', 4],
[-1.176781, 'qux', ' '],
], columns='A B C'.split(), index=pd.date_range('2000-01-01','2000-01-06'))
# replace field that's entirely space (or empty) with NaN
print(df.replace(r'^\s*$', np.nan, regex=True))
Produces:
A B C
2000-01-01 -0.532681 foo 0
2000-01-02 1.490752 bar 1
2000-01-03 -1.387326 foo 2
2000-01-04 0.814772 baz NaN
2000-01-05 -0.222552 NaN 4
2000-01-06 -1.176781 qux NaN
As Temak pointed it out, use df.replace(r'^\s+$', np.nan, regex=True) in case your valid data contains white spaces.