[PYTHON MATRIX-SIG] identity? and asarray?

[Doug Heisterkamp]

Hi,

The function identity(n) does not return an n by n identity matrix. If 
the current definition is changed from:

def identity(n):
    return resize([1]+n*[n], (n,n))

to

def identity(n):
    return resize([1]+n*[0], (n,n))

then it does what I think the function means. The asarray(a,t) function 
is not returning a new matrix of type t.  To do so, change from:

def asarray(a, typecode=None):
    if type(a) == arraytype and (typecode == None or typecode == a.typecode()): 
        return a
    elif typecode == None:
        return array(a)
    else:
        return a

to

def asarray(a, typecode=None):
    if type(a) == arraytype and (typecode == None or typecode == a.typecode()): 
        return a
    elif typecode == None:
        return array(a)
    else:
        return array(a,typecode)

When should a.asType(t) be use instead of asarray(a,t)?  Is asType faster,
always give a new copy, and/or safe?

Doug Heisterkamp
drh@oasis.unl.edu


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