通过for循环使用R中的值填充3D矩阵 [英] Filling a 3D Matrix by for loops with values in R

问题描述

我设置了3个尺寸为365x7x4的二维矩阵.

I set up a 3 dimensinoal matrix of size 365x7x4.

x <- array(rep(1, 365*5*4), dim=c(365, 5, 4))

现在，我将使用for循环来为每个元素填充一个值. 可以说每个元素的值应该是行，列和深度的总和. 我想这比较容易.

Now I would to use a for loop to fill each element with a value. Lets say the value of each element should be sum of row, column and depth. I guess this is relatively easy.

谢谢！最好，F

推荐答案

使用一个简单的示例，以便我们了解正在执行的操作

Using a simpler example so we can see what is being done

arr <- array(seq_len(3*3*3), dim = rep(3,3,3))

以下代码给出了请求的输出:

the following code gives the requested output:

dims <- dim(arr)
ind <- expand.grid(lapply(dims, seq_len))
arr[] <- rowSums(ind)

上面给出了

> arr
, , 1

     [,1] [,2] [,3]
[1,]    3    4    5
[2,]    4    5    6
[3,]    5    6    7

, , 2

     [,1] [,2] [,3]
[1,]    4    5    6
[2,]    5    6    7
[3,]    6    7    8

, , 3

     [,1] [,2] [,3]
[1,]    5    6    7
[2,]    6    7    8
[3,]    7    8    9

> arr[1,1,1]
[1] 3
> arr[1,2,3]
[1] 6
> arr[3,3,3]
[1] 9

更新:使用@TimP的答案"中的示例，我更新了答案"以显示如何以更类似于R的方式完成它.

Update: Using the example in @TimP's Answer here I update the Answer to show how it can be done in a more R-like fashion.

给予

arr <- array(seq_len(3*3*3), dim = rep(3,3,3))

用i + j + k替换arr的元素，除非k > 2，在这种情况下，使用j*k-i代替.

Replace elements of arr with i + j + k unless k > 2, in which case j*k-i is used instead.

dims <- dim(arr)
ind <- expand.grid(lapply(dims, seq_len))
## which k > 2
want <- ind[,3] > 2
arr[!want] <- rowSums(ind[!want, ])
arr[want] <- ind[want, 2] * ind[want, 3] - ind[want, 1]

虽然很想坚持使用像循环这样的熟悉的习语，并且与流行的信念循环相反，R的效率不是 ，但学会以向量化的方式思考将为您带来很多回报语言并将其开始应用于数据分析任务.

Whilst it is tempting to stick with familiar idioms like looping, and contrary to popular belief loops are not inefficient in R, learning to think in a vectorised way will pay off many times over as you learn the language and start applying it to data analysis task.

以下是Fabian示例的一些计时信息:

Here are some timings on Fabian's example:

> x <- array(rep(1, 365*5*4), dim=c(365, 5, 4))
> system.time({
+ for (i in seq_len(dim(x)[1])) {
+     for (j in seq_len(dim(x)[2])) {
+         for (k in seq_len(dim(x)[3])) {
+             val = i+j+k
+             if (k > 2) {
+                 val = j*k-i
+             }
+             x[i,j,k] = val
+         }
+     }
+ }
+ })
   user  system elapsed 
  0.043   0.000   0.044 
> arr <- array(rep(1, 365*5*4), dim=c(365, 5, 4))
> system.time({
+ dims <- dim(arr)
+ ind <- expand.grid(lapply(dims, seq_len))
+ ## which k > 2
+ want <- ind[,3] > 2
+ arr[!want] <- rowSums(ind[!want, ])
+ arr[want] <- ind[want, 2] * ind[want, 3] - ind[want, 1]
+ })
   user  system elapsed 
  0.005   0.000   0.006

还有一个更大的问题(至少对于我的笔记本电脑如此！)

and for a much larger (for my ickle laptop at least!) problem

> x <- array(rep(1, 200*200*200), dim=c(200, 200, 200))
> system.time({
+ for (i in seq_len(dim(x)[1])) {
+     for (j in seq_len(dim(x)[2])) {
+         for (k in seq_len(dim(x)[3])) {
+             val = i+j+k
+             if (k > 2) {
+                 val = j*k-i
+             }
+             x[i,j,k] = val
+         }
+     }
+ }
+ })
   user  system elapsed 
 51.759   0.129  53.090
> arr <- array(rep(1, 200*200*200), dim=c(200, 200, 200))
> system.time({
+     dims <- dim(arr)
+     ind <- expand.grid(lapply(dims, seq_len))
+     ## which k > 2
+     want <- ind[,3] > 2
+     arr[!want] <- rowSums(ind[!want, ])
+     arr[want] <- ind[want, 2] * ind[want, 3] - ind[want, 1]
+ })
   user  system elapsed 
  2.282   1.036   3.397 

，但按照今天的标准，这甚至可能很小.您可以看到，由于该方法需要进行所有的函数调用，因此循环开始变得越来越没有竞争性.

but even that may be modest to small by today's standards. You can see that the looping starts to become ever more uncompetitive because of the all the function calls required by that method.

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