[Numpy-discussion] another matrix operation

josef.pktd@gmai... josef.pktd@gmai...
Mon Mar 15 09:23:24 CDT 2010


On Mon, Mar 15, 2010 at 9:14 AM,  <josef.pktd@gmail.com> wrote:
> On Mon, Mar 15, 2010 at 9:09 AM, Gerardo Berbeglia <gberbeglia@gmail.com> wrote:
>> I would like to do with numpy the following operation.
>>
>> Let A be an n x n matrix and let s be an integer between 1 and n.
>>
>> I would like to have an n x n matrix B = f(A,s) such that
>>
>> - If we only look at the first s columns of B, we will not see any
>> difference with respect to the first s columns of the n x n identity
>> matrix.
>>
>> - For the last n-s columns of B, we will not see any difference with
>> respect to the last n-s columns of A.
>>
>> Example. n = 4, s = 2
>>
>> A= [[2,2,2,2],[3,3,3,3],[4,4,4,4],[5,5,5,5]]
>>
>> B= f(A,s) = [[1,0,2,2],[0,1,3,3],[0,0,4,4],[0,0,5,5]]
>>
>> Is there a way to do this efficiently (without loops)?
>
> something like this ?
>
> B = A.copy()
> B[:s,:] = np.eye[:s,:]

that changes rows,

this saves building one temporary eye

>>> n,s = 4,2
>>> A = np.array([[2,2,2,2],[3,3,3,3],[4,4,4,4],[5,5,5,5]])
>>> B = np.eye(n)
>>> B[:,-(n-s):] = A[:,-(n-s):]
>>> B
array([[ 1.,  0.,  2.,  2.],
       [ 0.,  1.,  3.,  3.],
       [ 0.,  0.,  4.,  4.],
       [ 0.,  0.,  5.,  5.]])

Josef

>
> Josef
>
>>
>> Thanks in advance.
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>


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