Home Original page

General Numerical Python question

Mark Jackson mjackson at alumni.caltech.edu
Fri Oct 17 13:47:08 EDT 2003 
mcrider at bigfoot.com (2mc) writes:
> Michael Ressler <ressler at cheetah.jpl.nasa.gov> wrote in message news:<slrnboqrh1.6mk.ressler at cheetah.jpl.nasa.gov>...
> > Another example of thinking things differently is suppose you have a
> > vector where the values are randomly positive or negative. Suppose for
> > reasons known only to you, you want to replace the negative values
> > with the sqrt of their absolute values. With Numeric, no loops are
> > involved.
> > 
> > from Numeric import *
> > a=array([1.,2.,-3.,4.,-5.,6.,-7.,-8.,9.])	# make up an array
> > idx=nonzero(a<0)			# indexes of the negative values
> > sqrs=sqrt(abs(take(a,idx)))		# get the sqrts of neg elements
> > put(a,idx,sqrs)				# put them back into a
> > print a					# works!  
> > 
> > You can make the whole thing a one-liner if you want to get carried
> > away with it. It's too bad "nonzero" isn't called "whereis" or
> > something like that - it would make the idx= line more obvious.
> > 
> > Mike
> 
> I think I'm finally getting a handle on this.  So, my thanks to
> everyone who has so graciously helped me out with their suggestions.
> 
> How would you handle the above if "a" were a 2d array since "nonzero"
> only works on 1d arrays?  Could you have used the "nonzero" function
> on a "vertical" slice of the array (from the perspective of an array
> of rows and columns - a vertical slice being the data in the column)?

I'm very new at this myself (currently porting some Fortran code to
Numeric) but I believe that Numeric.putmask is your friend here:

>>> a=Numeric.array([i*(-1)**i for i in range(20)],Numeric.Float)
>>> b=a.resize((4,5))
>>> b
array([[  0.,  -1.,   2.,  -3.,   4.],
       [ -5.,   6.,  -7.,   8.,  -9.],
       [ 10., -11.,  12., -13.,  14.],
       [-15.,  16., -17.,  18., -19.]])
>>> mask = b<0
>>> mask
array([[0, 1, 0, 1, 0],
       [1, 0, 1, 0, 1],
       [0, 1, 0, 1, 0],
       [1, 0, 1, 0, 1]])
>>> Numeric.putmask(b, mask, Numeric.sqrt(abs(b)))
>>> b
array([[  0.        ,   1.        ,   2.        ,   1.73205081,   4.        ],
       [  2.23606798,   6.        ,   2.64575131,   8.        ,   3.        ],
       [ 10.        ,   3.31662479,  12.        ,   3.60555128,  14.        ],
       [  3.87298335,  16.        ,   4.12310563,  18.        ,   4.35889894]])

-- 
Mark Jackson - http://www.alumni.caltech.edu/~mjackson
	There are two kinds of fool.  One says, "This is old,
	and therefore good."  And one says, "This is new, and
	therefore better."		- Dean William Inge
More information about the Python-list mailing list