[R] struggling with apply

[Michael Ashton]

This is like "Name that Tune." Can anyone do it in FEWER characters? :-)

On May 27, 2020, at 4:32 PM, Mathew Guilfoyle <mrguilfoyle using gmail.com> wrote:

 A bit quicker:

t(pmin(t(somematrix), UB))



On 27 May 2020, at 20:56, Bert Gunter <bgunter.4567 using gmail.com<mailto:bgunter.4567 using gmail.com>> wrote:

Jeff: Check it!

somematrix <- matrix(c(1,4,3,6,3,9,12,8,5,7,11,11),nrow=3,ncol=4)
UB=c(2.5, 5.5, 8.5, 10.5)
apply( somematrix, 2, function( x ) pmin( x, UB ) )
    [,1] [,2] [,3] [,4]
[1,]    1  2.5  2.5  2.5
[2,]    4  3.0  5.5  5.5
[3,]    3  8.5  5.0  8.5
[4,]    1  6.0 10.5  7.0

Not what was wanted.
Am I missing something?

Bert Gunter

"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Wed, May 27, 2020 at 12:38 PM Jeff Newmiller <jdnewmil using dcn.davis.ca.us<mailto:jdnewmil using dcn.davis.ca.us>>
wrote:

Sigh. Transpose?

apply( somematrix, 2, function( x ) pmin( x, UB ) )

On May 27, 2020 11:22:06 AM PDT, Bert Gunter <bgunter.4567 using gmail.com<mailto:bgunter.4567 using gmail.com>>
wrote:
Better, I think (no indexing):

t(apply(somematrix,1,function(x)pmin(x,UB)))


Bert Gunter

"The trouble with having an open mind is that people keep coming along
and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Wed, May 27, 2020 at 10:56 AM Rui Barradas <ruipbarradas using sapo.pt<mailto:ruipbarradas using sapo.pt>>
wrote:

Hello,

Try pmin. And loop by column/UB index with sapply/seq_along.


sapply(seq_along(UB), function(i) pmin(UB[i], somematrix[,i]))
#     [,1] [,2] [,3] [,4]
#[1,]  1.0  5.5  8.5  7.0
#[2,]  2.5  3.0  8.0 10.5
#[3,]  2.5  5.5  5.0 10.5


Hope this helps,

Rui Barradas


Às 18:46 de 27/05/20, Michael Ashton escreveu:
Hi -

I have a matrix of n rows and 4 columns.

I want to cap the value in each column by a different upper bound.
So,
suppose my matrix is

somematrix <- matrix(c(1,4,3,6,3,9,12,8,5,7,11,11),nrow=3,ncol=4)
somematrix
     [,1] [,2] [,3] [,4]
[1,]    1    6   12    7
[2,]    4    3    8   11
[3,]    3    9    5   11

Now I want to have the maximum value in each column described by
UB=c(2.5, 5.5, 8.5, 10.5)

So that the right answer will look like:
     [,1]      [,2]    [,3]   [,4]
[1,]    1      5.5     8.5    7
[2,]    2.5    3        8     10.5
[3,]    2.5   5.5      5    10.5

I've tried a few things, like:
newmatrix <- apply(somematrix,c(1,2),function(x) min(UB,x))

but I can't figure out to apply the relevant element of the UB list
to
the right element of the matrix. When I run the above, for example,
it
takes min(UB,x) over all UB, so I get:

newmatrix
     [,1] [,2] [,3] [,4]
[1,]  1.0  2.5  2.5  2.5
[2,]  2.5  2.5  2.5  2.5
[3,]  2.5  2.5  2.5  2.5

I'm sure there's a simple and elegant solution but I don't know
what it
is!

Thanks in advance,

Mike

Michael Ashton, CFA
Managing Principal

Enduring Investments LLC
W: 973.457.4602
C: 551.655.8006
Schedule a Call: https://calendly.com/m-ashton


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R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.


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______________________________________________
R-help using r-project.org<mailto:R-help using r-project.org> mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

--
Sent from my phone. Please excuse my brevity.


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______________________________________________
R-help using r-project.org<mailto:R-help using r-project.org> mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
and provide commented, minimal, self-contained, reproducible code.


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