user17645 - 1 year ago 71

R Question

Consider the following example in R:

`x1 <- rnorm(100000)`

x2 <- rnorm(100000)

g <- cbind(x1, x2, x1^2, x2^2)

gg <- t(g) %*% g

gginv <- solve(gg)

bigmatrix <- outer(x1, x2, "<=")

Gw <- t(g) %*% bigmatrix

beta <- gginv %*% Gw

w1 <- bigmatrix - g %*% beta

If I try to run such a thing in my computer, it will throw a memory error (because the

`bigmatrix`

Do you know how can I achieve the same, without running into this problem?

Answer Source

This is a least squares problem with 100,000 responses. Your `bigmatrix`

is the response (matrix), `beta`

is the coefficient (matrix), while `w1`

is the residual (matrix).

`bigmatrix`

, as well as `w1`

, if formed explicitly, will each cost

```
(100,000 * 100,000 * 8) / (1024 ^ 3) = 74.5 GB
```

This is far too large.

As estimation for each response is independent, there is really no need to form `bigmatrix`

in one go and try to store it in RAM. We can just form it **tile by tile**, and use an iterative procedure: **form a tile, use a tile, then discard it**. For example, the below considers a tile of dimension `100,000 * 2,000`

, with memory size:

```
(100,000 * 2,000 * 8) / (1024 ^ 3) = 1.5 GB
```

By such iterative procedure, the memory usage is effectively under control.

```
x1 <- rnorm(100000)
x2 <- rnorm(100000)
g <- cbind(x1, x2, x1^2, x2^2)
gg <- crossprod(g) ## don't use `t(g) %*% g`
## we also don't explicitly form `gg` inverse
## initialize `beta` matrix (4 coefficients for each of 100,000 responses)
beta <- matrix(0, 4, 100000)
## we split 100,000 columns into 50 tiles, each with 2000 columns
for (i in 1:50) {
start <- 2000 * (i-1) + 1 ## chunk start
end <- 2000 * i ## chunk end
bigmatrix <- outer(x1, x2[start:end], "<=")
Gw <- crossprod(g, bigmatrix) ## don't use `t(g) %*% bigmatrix`
beta[, start:end] <- solve(gg, Gw)
}
```

Note, don't try to compute the residual matrix `w1`

, as It will cost 74.5 GB. If you need residual matrix in later work, you should still try to break it into tiles and work one by one.

You don't need to worry about the loop here. The computation inside each iteration is costly enough to amortize looping overhead.