DataFrame bug when setting column after filtering

[pikeas]

In [615]: df = pd.DataFrame([3,4,5],columns=['a'])
In [616]: df['b'] = 'nope'
In [617]: df[df['a'] == 3]['b'] = 'found'
In [618]: df
Out[618]:
   a          b
0  3  nope
1  4  nope
2  5  nope

#Expected
In [618]: df
Out[618]:
   a          b
0  3  found
1  4  nope
2  5  nope

I believe the intermediate filter step creates a new DataFrame, but as far as I know, the outer step (where the value is set) should modify in-place.

Pandas 0.10.

[wesm]

You should do df.ix[df['a'] == 3, 'b'] = 'found'. What you have modifies a copy not the original data structure.

[changhiskhan]

Since df[df['a'] == 3] creates a new DataFrame, doing df[df['a'] == 3]['b'] = 'found' shouldn't modify df inplace right?

[pikeas]

Chang - df[<stuff>]['b'] looks like it should modify inplace, regardless of what is passed as stuff. I've been working with Pandas for a month and a half and still got bit by this. Possibly worth a mention in the docs?

[changhiskhan]

@pikeas that's a good suggestion. We should put that to the list of "gotchas"

[pikeas]

@wesm Your suggestion of using df.ix[<filter>, <col>] = will not work if the filter matches multiple items and the assignment is a Series.

Eg:

df = pd.DataFrame([3,4,5,3,6], columns=['a'])
df['b'] = 'nope'
df2 = df.copy()
#Works
df.ix[df['a'] == 3, 'b'] = ['f1', 'f2']
#Broken
df2.ix[df2['a'] == 3, 'b'] = pd.Series(['f1', 'f2'])

Out[927]:
   a     b
0  3    f1
1  4  nope
2  5  nope
3  3   NaN
4  6  nope