On 28-Aug-10 11:27:30, Gabor Grothendieck wrote:
> On Sat, Aug 28, 2010 at 1:32 AM, Cheng Peng <cp...@usm.maine.edu>
> wrote:
>>
>> Sorry for possible misunderstanding:
>>
>> I want to define a matrix (B) based on an existing matrix (A) in a
>> single
>> step and keep A unchanged:
>>
>>> #Existing matrix
>>> A=matrix(1:16,ncol=4)
>>> A
>> [,1] [,2] [,3] [,4]
>> [1,] 1 5 9 13
>> [2,] 2 6 10 14
>> [3,] 3 7 11 15
>> [4,] 4 8 12 16
>>> # New matrix B is defined to be the submatrix after row1 and column1
>>> # are deleted.
>>> B=A[-1,-1] # this single step deletes row1 nad column 1 and
>>> assigns the name to the resulting submatrix.
>>> B # check the new matrix B
>> [,1] [,2] [,3]
>> [1,] 6 10 14
>> [2,] 7 11 15
>> [3,] 8 12 16
>>> A # check the original matrix A
>> [,1] [,2] [,3] [,4]
>> [1,] 1 5 9 13
>> [2,] 2 6 10 14
>> [3,] 3 7 11 15
>> [4,] 4 8 12 16
>>
>> Question: How can I do define a new matrix (D) by adding 2*row1
>> to row3 in A in a single step as what was done in the above example?
>>
>> If you do: _A[3,]=2*A[1,]+A[3,], _the new A is not the original A;
>> if you D=A first, then D[3,]=2*D[1,]+D[3,], you used two step!
>>
>> Hope this clarifies my original question. Thanks again.
>
> Try using replace:
>
> D <- replace(A, row(A) == 3, 2 * A[1,] + A[3,])
That's neat -- and subtle! It's the sort of thing one wouldn't think
of doing until one has seen it done!
?replace:
Replace Values in a Vector
Description:
'replace' replaces the values in 'x' with indices given in
'list' by those given in 'values'. If necessary, the values
in 'values' are recycled.
Usage:
replace(x, list, values)
Arguments:
x: vector
list: an index vector
values: replacement values
Value:
A vector with the values replaced.
Note:
'x' is unchanged: remember to assign the result.
One needs to realise that a matrix A is a vector with a dimension
attribute, hence is accessed "sequentially" like any vector
(though displayed as an array). So, with the above matrix A:
row(A)==3
# [,1] [,2] [,3] [,4]
# [1,] FALSE FALSE FALSE FALSE
# [2,] FALSE FALSE FALSE FALSE
# [3,] TRUE TRUE TRUE TRUE
# [4,] FALSE FALSE FALSE FALSE
A[row(A)==3]
# [1] 3 7 11 15
Then 'replace(A, row(A) == 3, 2 * A[1,] + A[3,])' replaces that
part of the vector A as indexed by TRUE with the given expression
2 * A[1,] + A[3,] = 5 17 29 41
Hence
replace(A, row(A) == 3, 2 * A[1,] + A[3,])
# [,1] [,2] [,3] [,4]
# [1,] 1 5 9 13
# [2,] 2 6 10 14
# [3,] 5 17 29 41
# [4,] 4 8 12 16
is then assigned to D.
Hoping this helps to clarify!
Ted.
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E-Mail: (Ted Harding) <ted.hard...@manchester.ac.uk>
Fax-to-email: +44 (0)870 094 0861
Date: 28-Aug-10 Time: 13:11:08
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