Numeric.transpose (incorrect documentation)

[Curtis Jensen]

the documenation for the "transpose" function for the "Numeric" module
seems to be incoorect, or at least missleading.

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.
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 at bioeng.ucsd.edu
http://www-bioeng.ucsd.edu/~cjensen/
FAX (425) 740-1451