Mixing
Most efficient way to use values of Matrix as row indices to lookup values in another matrix in R
I am repeatetly drawing large matrices with random values from a Monte Carlo Simulation. As I explore a large parameter space, the simulation will most likely run for several days, therefore I am trying to find most efficient way to shave off as much time as possible. Consider the following code with a 500x18 Matrix as an example.
U = matrix(sample.int(500, size = 500*18, replace = TRUE), nrow = 500, ncol = 18)
X = matrix(nrow= 500, ncol = 18)
Marginals = matrix(runif(500*18, min = 0, max = 1),500,18)
for (i in 1:18){
for (k in 1:500){
X[k,i] = Marginals[U[k,i],i]
}
}
The randomly drawn values in U serve as the row index, while the col index is provided by column of the respective U.
I know that loops are usually not the R away, is there a more efficient way to use e.g. apply here?
By Yogos Suggesiton, the most efficient code can make due without the k loop:
U = matrix(sample.int(500, size = 500*18, replace = TRUE), nrow = 500, ncol = 18)
X = matrix(nrow= 500, ncol = 18)
Marginals = matrix(runif(500*18, min = 0, max = 1),500,18)
for (i in 1:18){
X[, i] <- Marginals[U[, i], i]
}
r matrix
asked Nov 21 '18 at 11:21
Ozaki_sanOzaki_san
133
add a comment |
I am repeatetly drawing large matrices with random values from a Monte Carlo Simulation. As I explore a large parameter space, the simulation will most likely run for several days, therefore I am trying to find most efficient way to shave off as much time as possible. Consider the following code with a 500x18 Matrix as an example.
U = matrix(sample.int(500, size = 500*18, replace = TRUE), nrow = 500, ncol = 18)
X = matrix(nrow= 500, ncol = 18)
Marginals = matrix(runif(500*18, min = 0, max = 1),500,18)
for (i in 1:18){
for (k in 1:500){
X[k,i] = Marginals[U[k,i],i]
}
}
The randomly drawn values in U serve as the row index, while the col index is provided by column of the respective U.
I know that loops are usually not the R away, is there a more efficient way to use e.g. apply here?
By Yogos Suggesiton, the most efficient code can make due without the k loop:
U = matrix(sample.int(500, size = 500*18, replace = TRUE), nrow = 500, ncol = 18)
X = matrix(nrow= 500, ncol = 18)
Marginals = matrix(runif(500*18, min = 0, max = 1),500,18)
for (i in 1:18){
X[, i] <- Marginals[U[, i], i]
}
r matrix
asked Nov 21 '18 at 11:21
Ozaki_sanOzaki_san
133
1
for (i in 1:18) X[, i] <- Marginals[U[, i], i]
– jogo
Nov 21 '18 at 12:00
eventuallyX <- replicate(18, sample(runif(500), repl=TRUE))is equivalent to your code.
– jogo
Nov 21 '18 at 12:12
Thanks Yogo, the actual numbers in U are not uniformly distributed, but drawn by a more complex function (copula). I will edit your suggestion as the solution, thank you very much
– Ozaki_san
Nov 21 '18 at 12:16
1
I suggest to remove the answer from the question. Then please accept the answer which @jogo provided in order to indicate that you have found an acceptable solution.
– Heikki
Nov 21 '18 at 12:37
add a comment |
I am repeatetly drawing large matrices with random values from a Monte Carlo Simulation. As I explore a large parameter space, the simulation will most likely run for several days, therefore I am trying to find most efficient way to shave off as much time as possible. Consider the following code with a 500x18 Matrix as an example.
U = matrix(sample.int(500, size = 500*18, replace = TRUE), nrow = 500, ncol = 18)
X = matrix(nrow= 500, ncol = 18)
Marginals = matrix(runif(500*18, min = 0, max = 1),500,18)
for (i in 1:18){
for (k in 1:500){
X[k,i] = Marginals[U[k,i],i]
}
}
The randomly drawn values in U serve as the row index, while the col index is provided by column of the respective U.
I know that loops are usually not the R away, is there a more efficient way to use e.g. apply here?
By Yogos Suggesiton, the most efficient code can make due without the k loop:
U = matrix(sample.int(500, size = 500*18, replace = TRUE), nrow = 500, ncol = 18)
X = matrix(nrow= 500, ncol = 18)
Marginals = matrix(runif(500*18, min = 0, max = 1),500,18)
for (i in 1:18){
X[, i] <- Marginals[U[, i], i]
}
r matrix
asked Nov 21 '18 at 11:21
Ozaki_sanOzaki_san
133
I am repeatetly drawing large matrices with random values from a Monte Carlo Simulation. As I explore a large parameter space, the simulation will most likely run for several days, therefore I am trying to find most efficient way to shave off as much time as possible. Consider the following code with a 500x18 Matrix as an example.
U = matrix(sample.int(500, size = 500*18, replace = TRUE), nrow = 500, ncol = 18)
X = matrix(nrow= 500, ncol = 18)
Marginals = matrix(runif(500*18, min = 0, max = 1),500,18)
for (i in 1:18){
for (k in 1:500){
X[k,i] = Marginals[U[k,i],i]
}
}
The randomly drawn values in U serve as the row index, while the col index is provided by column of the respective U.
I know that loops are usually not the R away, is there a more efficient way to use e.g. apply here?
By Yogos Suggesiton, the most efficient code can make due without the k loop:
U = matrix(sample.int(500, size = 500*18, replace = TRUE), nrow = 500, ncol = 18)
X = matrix(nrow= 500, ncol = 18)
Marginals = matrix(runif(500*18, min = 0, max = 1),500,18)
for (i in 1:18){
X[, i] <- Marginals[U[, i], i]
}
r matrix
r matrix
asked Nov 21 '18 at 11:21
Ozaki_sanOzaki_san
133
asked Nov 21 '18 at 11:21
Ozaki_sanOzaki_san
133
edited Nov 21 '18 at 12:18
Ozaki_san
asked Nov 21 '18 at 11:21
Ozaki_sanOzaki_san
133
asked Nov 21 '18 at 11:21
Ozaki_sanOzaki_san
133
asked Nov 21 '18 at 11:21
Ozaki_sanOzaki_san
133
133
1
for (i in 1:18) X[, i] <- Marginals[U[, i], i]
– jogo
Nov 21 '18 at 12:00
eventuallyX <- replicate(18, sample(runif(500), repl=TRUE))is equivalent to your code.
– jogo
Nov 21 '18 at 12:12
Thanks Yogo, the actual numbers in U are not uniformly distributed, but drawn by a more complex function (copula). I will edit your suggestion as the solution, thank you very much
– Ozaki_san
Nov 21 '18 at 12:16
1
I suggest to remove the answer from the question. Then please accept the answer which @jogo provided in order to indicate that you have found an acceptable solution.
– Heikki
Nov 21 '18 at 12:37
add a comment |
1
for (i in 1:18) X[, i] <- Marginals[U[, i], i]
– jogo
Nov 21 '18 at 12:00
eventuallyX <- replicate(18, sample(runif(500), repl=TRUE))is equivalent to your code.
– jogo
Nov 21 '18 at 12:12
Thanks Yogo, the actual numbers in U are not uniformly distributed, but drawn by a more complex function (copula). I will edit your suggestion as the solution, thank you very much
– Ozaki_san
Nov 21 '18 at 12:16
1
I suggest to remove the answer from the question. Then please accept the answer which @jogo provided in order to indicate that you have found an acceptable solution.
– Heikki
Nov 21 '18 at 12:37
1
1
for (i in 1:18) X[, i] <- Marginals[U[, i], i]– jogo
Nov 21 '18 at 12:00
for (i in 1:18) X[, i] <- Marginals[U[, i], i]– jogo
Nov 21 '18 at 12:00
eventually
X <- replicate(18, sample(runif(500), repl=TRUE)) is equivalent to your code.– jogo
Nov 21 '18 at 12:12
eventually
X <- replicate(18, sample(runif(500), repl=TRUE)) is equivalent to your code.– jogo
Nov 21 '18 at 12:12
Thanks Yogo, the actual numbers in U are not uniformly distributed, but drawn by a more complex function (copula). I will edit your suggestion as the solution, thank you very much
– Ozaki_san
Nov 21 '18 at 12:16
Thanks Yogo, the actual numbers in U are not uniformly distributed, but drawn by a more complex function (copula). I will edit your suggestion as the solution, thank you very much
– Ozaki_san
Nov 21 '18 at 12:16
1
1
I suggest to remove the answer from the question. Then please accept the answer which @jogo provided in order to indicate that you have found an acceptable solution.
– Heikki
Nov 21 '18 at 12:37
I suggest to remove the answer from the question. Then please accept the answer which @jogo provided in order to indicate that you have found an acceptable solution.
– Heikki
Nov 21 '18 at 12:37
add a comment |
1 Answer
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votes
You can speed up by calculating column by column:
for (i in 1:18) X[, i] <- Marginals[U[, i], i]
Eventually the following is equivalent to your code:
X <- replicate(18, sample(runif(500), repl=TRUE))
(this will not be much faster than my first variant, but the code is more compact)
answered Nov 21 '18 at 12:23
jogojogo
10k92135
add a comment |
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You can speed up by calculating column by column:
for (i in 1:18) X[, i] <- Marginals[U[, i], i]
Eventually the following is equivalent to your code:
X <- replicate(18, sample(runif(500), repl=TRUE))
(this will not be much faster than my first variant, but the code is more compact)
answered Nov 21 '18 at 12:23
jogojogo
10k92135
add a comment |
You can speed up by calculating column by column:
for (i in 1:18) X[, i] <- Marginals[U[, i], i]
Eventually the following is equivalent to your code:
X <- replicate(18, sample(runif(500), repl=TRUE))
(this will not be much faster than my first variant, but the code is more compact)
answered Nov 21 '18 at 12:23
jogojogo
10k92135
add a comment |
You can speed up by calculating column by column:
for (i in 1:18) X[, i] <- Marginals[U[, i], i]
Eventually the following is equivalent to your code:
X <- replicate(18, sample(runif(500), repl=TRUE))
(this will not be much faster than my first variant, but the code is more compact)
answered Nov 21 '18 at 12:23
jogojogo
10k92135
You can speed up by calculating column by column:
for (i in 1:18) X[, i] <- Marginals[U[, i], i]
Eventually the following is equivalent to your code:
X <- replicate(18, sample(runif(500), repl=TRUE))
(this will not be much faster than my first variant, but the code is more compact)
answered Nov 21 '18 at 12:23
jogojogo
10k92135
edited Nov 21 '18 at 12:33
answered Nov 21 '18 at 12:23
jogojogo
10k92135
answered Nov 21 '18 at 12:23
jogojogo
10k92135
answered Nov 21 '18 at 12:23
jogojogo
10k92135
10k92135
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for (i in 1:18) X[, i] <- Marginals[U[, i], i]– jogo
Nov 21 '18 at 12:00
eventually
X <- replicate(18, sample(runif(500), repl=TRUE))is equivalent to your code.– jogo
Nov 21 '18 at 12:12
Thanks Yogo, the actual numbers in U are not uniformly distributed, but drawn by a more complex function (copula). I will edit your suggestion as the solution, thank you very much
– Ozaki_san
Nov 21 '18 at 12:16
1
I suggest to remove the answer from the question. Then please accept the answer which @jogo provided in order to indicate that you have found an acceptable solution.
– Heikki
Nov 21 '18 at 12:37