Q:

Replacing blank values (white space) with NaN in pandas

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.

A:

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.

A:

If you want to replace an empty string and records with only spaces, the correct answer is!:
df = df.replace(r'^\s*$', np.nan, regex=True)

The accepted answer
df.replace(r'\s+', np.nan, regex=True)

Does not replace an empty string!, you can try yourself with the given example slightly updated:
df = pd.DataFrame([
    [-0.532681, 'foo', 0],
    [1.490752, 'bar', 1],
    [-1.387326, 'fo o', 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'))

Note, also that 'fo o' is not replaced with Nan, though it contains a space.
Further note, that a simple:
df.replace(r'', np.NaN)

Does not work either - try it out.

A:

How about:
d = d.applymap(lambda x: np.nan if isinstance(x, basestring) and x.isspace() else x)

The applymap function applies a function to every cell of the dataframe.

