Oh sorry. No. of columns in b and size of a must always be same.
I have made an r code to show the output.
a = c(1, 2, 3)
b = matrix(1:30, nrow=10, ncol=3)
c = array(as.double(0), dim=c(10, 3, 10))
for(i in 1:10){
for(j in 1:3){
for(k in 1:i){
if(k==1) c[i, j, k] = b[i, j]
else c[i, j, k] = c[i-1, j, k-1] * b[i, j]
}
}
}
Output:
> c, , 1
[,1] [,2] [,3]
[1,] 1 11 21
[2,] 2 12 22
[3,] 3 13 23
[4,] 4 14 24
[5,] 5 15 25
[6,] 6 16 26
[7,] 7 17 27
[8,] 8 18 28
[9,] 9 19 29
[10,] 10 20 30
, , 2
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 2 132 462
[3,] 6 156 506
[4,] 12 182 552
[5,] 20 210 600
[6,] 30 240 650
[7,] 42 272 702
[8,] 56 306 756
[9,] 72 342 812
[10,] 90 380 870
, , 3
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 0 0 0
[3,] 6 1716 10626
[4,] 24 2184 12144
[5,] 60 2730 13800
[6,] 120 3360 15600
[7,] 210 4080 17550
[8,] 336 4896 19656
[9,] 504 5814 21924
[10,] 720 6840 24360
, , 4
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 0 0 0
[3,] 0 0 0
[4,] 24 24024 255024
[5,] 120 32760 303600
[6,] 360 43680 358800
[7,] 840 57120 421200
[8,] 1680 73440 491400
[9,] 3024 93024 570024
[10,] 5040 116280 657720
, , 5
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 0 0 0
[3,] 0 0 0
[4,] 0 0 0
[5,] 120 360360 6375600
[6,] 720 524160 7893600
[7,] 2520 742560 9687600
[8,] 6720 1028160 11793600
[9,] 15120 1395360 14250600
[10,] 30240 1860480 17100720
, , 6
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 0 0 0
[3,] 0 0 0
[4,] 0 0 0
[5,] 0 0 0
[6,] 720 5765760 165765600
[7,] 5040 8910720 213127200
[8,] 20160 13366080 271252800
[9,] 60480 19535040 342014400
[10,] 151200 27907200 427518000
, , 7
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 0 0 0
[3,] 0 0 0
[4,] 0 0 0
[5,] 0 0 0
[6,] 0 0 0
[7,] 5040 98017920 4475671200
[8,] 40320 160392960 5967561600
[9,] 181440 253955520 7866331200
[10,] 604800 390700800 10260432000
, , 8
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 0 0 0
[3,] 0 0 0
[4,] 0 0 0
[5,] 0 0 0
[6,] 0 0 0
[7,] 0 0 0
[8,] 40320 1764322560 125318793600
[9,] 362880 3047466240 173059286400
[10,] 1814400 5079110400 235989936000
, , 9
[,1] [,2] [,3]
[1,] 0 0 0.000000e+00
[2,] 0 0 0.000000e+00
[3,] 0 0 0.000000e+00
[4,] 0 0 0.000000e+00
[5,] 0 0 0.000000e+00
[6,] 0 0 0.000000e+00
[7,] 0 0 0.000000e+00
[8,] 0 0 0.000000e+00
[9,] 362880 33522128640 3.634245e+12
[10,] 3628800 60949324800 5.191779e+12
, , 10
[,1] [,2] [,3]
[1,] 0 0 0.000000e+00
[2,] 0 0 0.000000e+00
[3,] 0 0 0.000000e+00
[4,] 0 0 0.000000e+00
[5,] 0 0 0.000000e+00
[6,] 0 0 0.000000e+00
[7,] 0 0 0.000000e+00
[8,] 0 0 0.000000e+00
[9,] 0 0 0.000000e+00
[10,] 3628800 670442572800 1.090274e+14
On Wed, Dec 14, 2016 at 7:27 PM, Avraham Adler <avraham.ad...@gmail.com>
wrote:
>
>
> On Wed, Dec 14, 2016 at 1:24 AM Amina Shahzadi <aminashahz...@gmail.com>
> wrote:
>
>> Hello Avraham --Happy to see you
>>
>> My code is trying to produce a cube c which is going to be constructed by
>> a vector a and matrix b.
>> And the number of rows in b and size of a must be same.
>>
>> So we can assume that if a is a vector of size 3, Then b must be 2 x 3 or
>> 3 X 3 etc.
>>
>> Thank you Avraham for quick response. I hope this will make my question
>> more clear.
>>
>> Best regards
>>
>>
>> On Wed, Dec 14, 2016 at 4:46 PM, Avraham Adler <avraham.ad...@gmail.com>
>> wrote:
>>
>> On Tue, Dec 13, 2016 at 9:51 PM, Amina Shahzadi <aminashahz...@gmail.com>
>> wrote:
>>
>> Hello Friends and Prof. Dirk
>>
>> I am pasting here a code which has a for loop depending on another for
>> loop.
>> I am getting zeros for cube c. I tried and searched a lot but did not get
>> an example of this type. Would you please help in this regard?
>>
>>
>> #include <RcppArmadillo.h>
>> using namespace Rcpp;
>> using namespace RcppArmadillo;
>> using namespace arma;
>> //[[Rcpp::depends(RcppArmadillo)]]
>> //[[Rcpp::export]]
>>
>>
>> arma::cube exam(arma::vec a, arma::mat b)
>> {
>> int m = a.size();
>> int n = b.n_rows;
>> arma::cube c = zeros<cube>(n, m, n);
>> for(int i=0; i<n; i++) {
>> for(int j=0; j<m; j++) {
>> for(int k=0; k<i; k++) {
>> if(k==0) {
>> c(i, j ,k) = c(i, j, k) + b(i, j);
>> }
>> else{
>> c(i, j, k) = c(i, j, k) + c(i-1, j, k) *b(i, j);
>> }
>> }
>> }
>> }
>> return c;
>> }
>>
>>
>> Thank You
>> --
>> *Amina Shahzadi*
>>
>>
>>
>>
>> Hello.
>>
>> I haven't run your code, but it strikes me
>>
>> that I cannot see where are you capturing the number of columns of b.
>>
>> It's a bit confusing as I was always taught a matrix has m rows and n
>>
>> columns. Be that as it may, your k==0 loop looks like it's trying to
>>
>> copy over the original matrix to the first slice, but how do you know
>>
>> that b has m columns, which is what you're assuming by letting j loop to
>>
>> m. Unless you are assuming a square matrix?
>>
>> Even if you are, if your matrix is not the same length as your vector, I
>> think there is an issue with your loop boundaries, unless I've
>> misunderstood something, which is certainly possible.
>>
>> For example, assume a is {1, 2, 3} and b is the 2 x 2 of row 1: [1 2] and
>> row 2: [3 4]. Thus m = 3 and n = 2.
>>
>> Step 1: i = j = k = 0: c(0, 0, 0) becomes b(0, 0) or 1.
>>
>> Step 2: i = 0, j = 1, k = 0: c(0, 1, 0) becomes b(0, 1) or 2.
>>
>> Step 3: i = 0, j = 2, k = 0: c(0, 2, 0) becomes b(0, 2) ?!?! There is no
>> b(0, 2), it's only a 2x2 matrix?
>>
>>
>> Similar to your previous questions, instead of posting code, can you
>> please describe in words what it is you are trying to do? That may help.
>>
>> Avi
>>
>>
>>
>>
>>
>>
>> --
>> *Amina Shahzadi*
>>
>>
>> Constructed how? Can you provide a simple set of inputs and the expected
> output?
>
> Avi
> --
> Sent from Gmail Mobile
>
--
*Amina Shahzadi*
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