Your answer lies in the pandas docs: returning-a-view-versus-a-copy.
Whenever an array of labels or a boolean vector are involved
in the indexing operation, the result will be a copy.
With single label / scalar indexing and slicing,
e.g. df.ix[3:6] or df.ix[:, 'A'], a view will be returned.
In your example, bar
is a view of slices of foo
. If you wanted a copy, you could have used the copy
method. Modifying bar
also modifies foo
. pandas does not appear to have a copy-on-write mechanism.
See my code example below to illustrate:
In [1]: import pandas as pd
...: import numpy as np
...: foo = pd.DataFrame(np.random.random((10,5)))
...:
In [2]: pd.__version__
Out[2]: '0.12.0.dev-35312e4'
In [3]: np.__version__
Out[3]: '1.7.1'
In [4]: # DataFrame has copy method
...: foo_copy = foo.copy()
In [5]: bar = foo.iloc[3:5,1:4]
In [6]: bar == foo.iloc[3:5,1:4] == foo_copy.iloc[3:5,1:4]
Out[6]:
1 2 3
3 True True True
4 True True True
In [7]: # Changing the view
...: bar.ix[3,1] = 5
In [8]: # View and DataFrame still equal
...: bar == foo.iloc[3:5,1:4]
Out[8]:
1 2 3
3 True True True
4 True True True
In [9]: # It is now different from a copy of original
...: bar == foo_copy.iloc[3:5,1:4]
Out[9]:
1 2 3
3 False True True
4 True True True