Suppose I have interest in a matrix with the following symbolic structure (specified by 3 parameters: sa, so, m): matrix(c(0,sa*m,so,sa),2,2,byrow=T) What I can't figure out is how to construct a series of matrices, where the elements/parameters are rnorm values. I'd like to construct separate matrices, with each matrix in the series using the 'next random parameter value'. While the following works (for generating, say, 5 such random matrices) replicate(5,matrix(c(0,rnorm(1,0.8,0.1)*rnorm(1,1.2,0.1),rnorm(1,0.5,0.1),rnorm(1,0.8,0.1)),2,2,byrow=T)) its inelegant, and a real pain if the matrix gets large (say, 20 x 20). I'm wondering if there is an easier way. I tried > sa <- rnorm(5,0.8,0.1) > so <- rnorm(5,0.5,0.1) > m <- rnorm(5,1.2,0.1) matrix(c(0,sa*m,so,sa),2,2,byrow=T) but that only returns a single matrix, not 5 matrices as I'd like. I also tried several variants of the 'replicate' approach (above), but didn't stumble across anything that seemed to work. So, is there a better way than something like: replicate(5,matrix(c(0,rnorm(1,0.8,0.1)*rnorm(1,1.2,0.1),rnorm(1,0.5,0.1),rnorm(1,0.8,0.1)),2,2,byrow=T)) Many thanks in advance...
building random matrices from vectors of random parameters
9 messages · Peter Langfelder, Duncan Murdoch, Evan Cooch +1 more
I would try something like
n = 5
a <- rnorm(n,0.8,0.1)
so <- rnorm(n,0.5,0.1)
m <- rnorm(n,1.2,0.1)
mats = mapply(function(sa1, so1, m1) matrix(c(0,sa1*m1,so1,sa1),2,2,byrow=T),
a, so, m, SIMPLIFY = FALSE)
mats
[[1]]
[,1] [,2]
[1,] 0.0000000 0.9129962
[2,] 0.4963598 0.7067311
[[2]]
[,1] [,2]
[1,] 0.0000000 1.0150316
[2,] 0.5489887 0.8469046
[[3]]
[,1] [,2]
[1,] 0.0000000 0.9516137
[2,] 0.3724521 0.8306535
[[4]]
[,1] [,2]
[1,] 0.0000000 1.0525355
[2,] 0.8075108 0.8314638
[[5]]
[,1] [,2]
[1,] 0.0000000 0.9400074
[2,] 0.4803386 0.7901753
On Wed, Sep 27, 2017 at 5:47 PM, Evan Cooch <evan.cooch at gmail.com> wrote:
Suppose I have interest in a matrix with the following symbolic structure (specified by 3 parameters: sa, so, m): matrix(c(0,sa*m,so,sa),2,2,byrow=T) What I can't figure out is how to construct a series of matrices, where the elements/parameters are rnorm values. I'd like to construct separate matrices, with each matrix in the series using the 'next random parameter value'. While the following works (for generating, say, 5 such random matrices) replicate(5,matrix(c(0,rnorm(1,0.8,0.1)*rnorm(1,1.2,0.1),rnorm(1,0.5,0.1),rnorm(1,0.8,0.1)),2,2,byrow=T)) its inelegant, and a real pain if the matrix gets large (say, 20 x 20). I'm wondering if there is an easier way. I tried
sa <- rnorm(5,0.8,0.1) so <- rnorm(5,0.5,0.1) m <- rnorm(5,1.2,0.1)
matrix(c(0,sa*m,so,sa),2,2,byrow=T) but that only returns a single matrix, not 5 matrices as I'd like. I also tried several variants of the 'replicate' approach (above), but didn't stumble across anything that seemed to work. So, is there a better way than something like: replicate(5,matrix(c(0,rnorm(1,0.8,0.1)*rnorm(1,1.2,0.1),rnorm(1,0.5,0.1),rnorm(1,0.8,0.1)),2,2,byrow=T)) Many thanks in advance...
______________________________________________ R-help at 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.
On 27/09/2017 8:47 PM, Evan Cooch wrote:
Suppose I have interest in a matrix with the following symbolic structure (specified by 3 parameters: sa, so, m): matrix(c(0,sa*m,so,sa),2,2,byrow=T) What I can't figure out is how to construct a series of matrices, where the elements/parameters are rnorm values. I'd like to construct separate matrices, with each matrix in the series using the 'next random parameter value'. While the following works (for generating, say, 5 such random matrices) replicate(5,matrix(c(0,rnorm(1,0.8,0.1)*rnorm(1,1.2,0.1),rnorm(1,0.5,0.1),rnorm(1,0.8,0.1)),2,2,byrow=T)) its inelegant, and a real pain if the matrix gets large (say, 20 x 20). I'm wondering if there is an easier way. I tried
> sa <- rnorm(5,0.8,0.1) > so <- rnorm(5,0.5,0.1) > m <- rnorm(5,1.2,0.1)
matrix(c(0,sa*m,so,sa),2,2,byrow=T) but that only returns a single matrix, not 5 matrices as I'd like. I also tried several variants of the 'replicate' approach (above), but didn't stumble across anything that seemed to work. So, is there a better way than something like: replicate(5,matrix(c(0,rnorm(1,0.8,0.1)*rnorm(1,1.2,0.1),rnorm(1,0.5,0.1),rnorm(1,0.8,0.1)),2,2,byrow=T))
Peter's mapply solution is probably the best. Another that might be a little faster (but more obscure) is to use a 3-index array. I think this is what you'd want, with sa, so, and m as defined above: ms <- array(c(rep(0, 5),sa*m,so,sa), c(5, 2, 2)) Then matrix i will be stored as ms[i,,]. Duncan Murdoch
Thanks for both the mapply and array approaches! However, although
intended to generate the same result, they don't:
# mapply approach
n = 3
sa <- rnorm(n,0.8,0.1)
so <- rnorm(n,0.5,0.1)
m <- rnorm(n,1.2,0.1)
mats = mapply(function(sa1, so1, m1)
matrix(c(0,sa1*m1,so1,sa1),2,2,byrow=T), sa, so, m, SIMPLIFY = FALSE)
print(mats)
[[1]]
????????? [,1]????? [,2]
[1,] 0.0000000 0.8643679
[2,] 0.4731249 0.7750431
[[2]]
????????? [,1]????? [,2]
[1,] 0.0000000 0.8838286
[2,] 0.5895258 0.7880983
[[3]]
????????? [,1]????? [,2]
[1,] 0.0000000 1.1491560
[2,] 0.4947322 0.9744166
Now, the array approach:
# array approach
ms <- array(c(rep(0, 3),sa*m,so,sa), c(3, 2, 2))
for (i in 1:n) { print(ms[i,,])
????????? [,1]????? [,2]
[1,] 0.0000000 0.4731249
[2,] 0.8643679 0.7750431
????????? [,1]????? [,2]
[1,] 0.0000000 0.5895258
[2,] 0.8838286 0.7880983
???????? [,1]????? [,2]
[1,] 0.000000 0.4947322
[2,] 1.149156 0.9744166
These matrices are the transpose of those returned by the mapply
approach. To see if one approach or the other is 'confused', I simply
rerun setting sd=0 for the parameters -- thus, every matrix will be the
same. The correct matrix would be:
???? [,1] [,2]
[1,]? 0.0 0.96
[2,]? 0.5 0.80
In fact, this is what is returned by the mapply approach, while the
array approach returns the transpose. I gather the 'missing step' is to
use aperm, but haven't figured out how to get that to work...yet.
On 9/28/2017 5:11 AM, Duncan Murdoch wrote:
ms <- array(c(rep(0, 5),sa*m,so,sa), c(5, 2, 2))
In fact, this is what is returned by the mapply approach, while the array approach returns the transpose. I gather the 'missing step' is to use aperm, but haven't figured out how to get that to work...yet.
ms <- array(c(rep(0, 3),sa*m,so,sa), c(3, 2, 2))
ms_new <- aperm(ms,c(1,3,2));
for (i in 1:n) { print(ms_new[i,,]) }
???? [,1] [,2]
[1,]? 0.0 0.96
[2,]? 0.5 0.80
???? [,1] [,2]
[1,]? 0.0 0.96
[2,]? 0.5 0.80
???? [,1] [,2]
[1,]? 0.0 0.96
[2,]? 0.5 0.80
On 9/28/2017 5:11 AM, Duncan Murdoch wrote:
ms <- array(c(rep(0, 5),sa*m,so,sa), c(5, 2, 2))
On 28/09/2017 9:10 AM, Evan Cooch wrote:
Thanks for both the mapply and array approaches! However, although
intended to generate the same result, they don't:
# mapply approach
n = 3
sa <- rnorm(n,0.8,0.1)
so <- rnorm(n,0.5,0.1)
m <- rnorm(n,1.2,0.1)
mats = mapply(function(sa1, so1, m1)
matrix(c(0,sa1*m1,so1,sa1),2,2,byrow=T), sa, so, m, SIMPLIFY = FALSE)
print(mats)
[[1]]
????????? [,1]????? [,2]
[1,] 0.0000000 0.8643679
[2,] 0.4731249 0.7750431
[[2]]
????????? [,1]????? [,2]
[1,] 0.0000000 0.8838286
[2,] 0.5895258 0.7880983
[[3]]
????????? [,1]????? [,2]
[1,] 0.0000000 1.1491560
[2,] 0.4947322 0.9744166
Now, the array approach:
# array approach
ms <- array(c(rep(0, 3),sa*m,so,sa), c(3, 2, 2))
for (i in 1:n) { print(ms[i,,])
????????? [,1]????? [,2]
[1,] 0.0000000 0.4731249
[2,] 0.8643679 0.7750431
????????? [,1]????? [,2]
[1,] 0.0000000 0.5895258
[2,] 0.8838286 0.7880983
???????? [,1]????? [,2]
[1,] 0.000000 0.4947322
[2,] 1.149156 0.9744166
These matrices are the transpose of those returned by the mapply
approach. To see if one approach or the other is 'confused', I simply
rerun setting sd=0 for the parameters -- thus, every matrix will be the
same. The correct matrix would be:
???? [,1] [,2]
[1,]? 0.0 0.96
[2,]? 0.5 0.80
In fact, this is what is returned by the mapply approach, while the
array approach returns the transpose. I gather the 'missing step' is to
use aperm, but haven't figured out how to get that to work...yet.
On 9/28/2017 5:11 AM, Duncan Murdoch wrote:
ms <- array(c(rep(0, 5),sa*m,so,sa), c(5, 2, 2))
Sorry about that -- I didn't notice the "byrow = T" in your original. Duncan Murdoch
Sure -- thanks -- only took me 3-4 attempts to get aperm to work (as opposed to really thinking hard about how it works ;-)
On 9/28/2017 11:55 AM, Duncan Murdoch wrote:
On 28/09/2017 9:10 AM, Evan Cooch wrote:
Thanks for both the mapply and array approaches! However, although
intended to generate the same result, they don't:
# mapply approach
n = 3
sa <- rnorm(n,0.8,0.1)
so <- rnorm(n,0.5,0.1)
m <- rnorm(n,1.2,0.1)
mats = mapply(function(sa1, so1, m1)
matrix(c(0,sa1*m1,so1,sa1),2,2,byrow=T), sa, so, m, SIMPLIFY = FALSE)
print(mats)
[[1]]
?????????? [,1]????? [,2]
[1,] 0.0000000 0.8643679
[2,] 0.4731249 0.7750431
[[2]]
?????????? [,1]????? [,2]
[1,] 0.0000000 0.8838286
[2,] 0.5895258 0.7880983
[[3]]
?????????? [,1]????? [,2]
[1,] 0.0000000 1.1491560
[2,] 0.4947322 0.9744166
Now, the array approach:
# array approach
ms <- array(c(rep(0, 3),sa*m,so,sa), c(3, 2, 2))
for (i in 1:n) { print(ms[i,,])
?????????? [,1]????? [,2]
[1,] 0.0000000 0.4731249
[2,] 0.8643679 0.7750431
?????????? [,1]????? [,2]
[1,] 0.0000000 0.5895258
[2,] 0.8838286 0.7880983
????????? [,1]????? [,2]
[1,] 0.000000 0.4947322
[2,] 1.149156 0.9744166
These matrices are the transpose of those returned by the mapply
approach. To see if one approach or the other is 'confused', I simply
rerun setting sd=0 for the parameters -- thus, every matrix will be
the same. The correct matrix would be:
????? [,1] [,2]
[1,]? 0.0 0.96
[2,]? 0.5 0.80
In fact, this is what is returned by the mapply approach, while the
array approach returns the transpose. I gather the 'missing step' is
to use aperm, but haven't figured out how to get that to work...yet.
On 9/28/2017 5:11 AM, Duncan Murdoch wrote:
ms <- array(c(rep(0, 5),sa*m,so,sa), c(5, 2, 2))
Sorry about that -- I didn't notice the "byrow = T" in your original. Duncan Murdoch
The use of aperm is unnecessary if you call array() properly. ms <- array(c(rep(0, 5),so,sa*m,sa), c(5, 2, 2))
Sent from my phone. Please excuse my brevity.
On September 28, 2017 9:10:26 AM PDT, Evan Cooch <evan.cooch at gmail.com> wrote:
>Sure -- thanks -- only took me 3-4 attempts to get aperm to work (as
>opposed to really thinking hard about how it works ;-)
>
>On 9/28/2017 11:55 AM, Duncan Murdoch wrote:
>> On 28/09/2017 9:10 AM, Evan Cooch wrote:
>>> Thanks for both the mapply and array approaches! However, although
>>> intended to generate the same result, they don't:
>>>
>>> # mapply approach
>>>
>>> n = 3
>>> sa <- rnorm(n,0.8,0.1)
>>> so <- rnorm(n,0.5,0.1)
>>> m <- rnorm(n,1.2,0.1)
>>> mats = mapply(function(sa1, so1, m1)
>>> matrix(c(0,sa1*m1,so1,sa1),2,2,byrow=T), sa, so, m, SIMPLIFY =
>FALSE)
>>>
>>> print(mats)
>>>
>>> [[1]]
>>> ?????????? [,1]????? [,2]
>>> [1,] 0.0000000 0.8643679
>>> [2,] 0.4731249 0.7750431
>>>
>>> [[2]]
>>> ?????????? [,1]????? [,2]
>>> [1,] 0.0000000 0.8838286
>>> [2,] 0.5895258 0.7880983
>>>
>>> [[3]]
>>> ?????????? [,1]????? [,2]
>>> [1,] 0.0000000 1.1491560
>>> [2,] 0.4947322 0.9744166
>>>
>>>
>>> Now, the array approach:
>>>
>>> # array approach
>>>
>>> ms <- array(c(rep(0, 3),sa*m,so,sa), c(3, 2, 2))
>>>
>>> for (i in 1:n) { print(ms[i,,])
>>>
>>> ?????????? [,1]????? [,2]
>>> [1,] 0.0000000 0.4731249
>>> [2,] 0.8643679 0.7750431
>>>
>>> ?????????? [,1]????? [,2]
>>> [1,] 0.0000000 0.5895258
>>> [2,] 0.8838286 0.7880983
>>>
>>> ????????? [,1]????? [,2]
>>> [1,] 0.000000 0.4947322
>>> [2,] 1.149156 0.9744166
>>>
>>>
>>> These matrices are the transpose of those returned by the mapply
>>> approach. To see if one approach or the other is 'confused', I
>simply
>>> rerun setting sd=0 for the parameters -- thus, every matrix will be
>>> the same. The correct matrix would be:
>>>
>>> ????? [,1] [,2]
>>> [1,]? 0.0 0.96
>>> [2,]? 0.5 0.80
>>>
>>>
>>> In fact, this is what is returned by the mapply approach, while the
>>> array approach returns the transpose. I gather the 'missing step' is
>
>>> to use aperm, but haven't figured out how to get that to work...yet.
>>>
>>>
>>> On 9/28/2017 5:11 AM, Duncan Murdoch wrote:
>>>> ms <- array(c(rep(0, 5),sa*m,so,sa), c(5, 2, 2))
>>>
>>
>>
>> Sorry about that -- I didn't notice the "byrow = T" in your original.
>>
>> Duncan Murdoch
>>
>
>
> [[alternative HTML version deleted]]
>
>______________________________________________
>R-help at 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.
Makes sense, although (re-)learning what aperm does wasn't a wasted exercise. Thanks!
On 9/28/2017 1:22 PM, Jeff Newmiller wrote:
The use of aperm is unnecessary if you call array() properly. ms <- array(c(rep(0, 5),so,sa*m,sa), c(5, 2, 2))