"Curtis Jensen" <cjensen@bioeng.ucsd.edu> wrote in message news:3B6076D4.68B6840C@bioeng.ucsd.edu...
the documenation for the "transpose" function for the "Numeric" module seems to be incoorect, or at least missleading.
The correct place for this is probably the numpy-discussion list, so I'm cc'ing it there. Perhaps the Paul Dubois will see it and have some ideas for clearing of the docs.
It says that transpose is suppose to return a "new" array. In fact it does not. It returns a pointer to the same array, only with transposed indicies.
It's been a while since I looked at the NumPy docs, so I don't recall what conventions they use here. However, transpose does return a new array, it just points to the same data as the original. If a=transpose(b), "a is b" is false and "a.shape == b.shape" is false, so clearly they are different objects (i.e., different arrays). You know most of this, as you demonstrate down below, I just wanted to make the point that there is a difference between a new array and new data. Note that nearly every operation that can return a reference rather than a copy does so. Even slicing, if b=a[3:5], then 'b' holds a reference to the same data as 'a'. Some functions go so far to return a reference when they can, but otherwise copy. See ravel for example -- it copies the data if its argument is not contiguous, otherwise it uses a reference. Now _that_ can be confusing! Useful though. -tim
consider the example:
Python 1.5.2 (#4, Sep 5 2000, 10:29:12) [C] on irix646 Copyright 1991-1995 Stichting Mathematisch Centrum, Amsterdam
from Numeric import * foo = zeros([5,10]) bar = transpose( foo ) foo array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]) bar array([[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]) foo[0,2] = 1 foo array([[0, 0, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]) bar array([[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [1, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]])
if bar was truely a new array, then changes to foo would not affect bar. In the example above, it does. This way actualy works better for my purposes, however, the documentation is missleading.
-- Curtis Jensen cjensen@bioeng.ucsd.edu http://www-bioeng.ucsd.edu/~cjensen/ FAX (425) 740-1451