For循環創建對稱矩陣

Question

我希望減少時間和內存使用（I以前使用外此，但它消耗更多的內存比我）通過減少迭代以創建一個對稱矩陣，即sol[i, j]相同sol[j, i]。

到目前爲止我的代碼：

# Prepare input 
subss <- list(a = c(1, 2, 4), b = c(1, 2, 3), c = c(4, 5)) 
A <- matrix(runif(25), ncol = 5, nrow = 5) 
# Pre allocate memory 
sol <- matrix(nrow = length(subss), ncol = length(subss), 
      dimnames = list(names(subss), names(subss))) 
x <- 0 
for (i in seq_along(subss)) { 
    # Omit for the subsets I already calculated ? 
    for (j in seq_along(subss)) { 
     x <- x + 1 
     message(x) 

     # The function I use here might result in a NA 
     sol[i, j] <- mean(A[subss[[i]], subss[[j]]]) 
     sol[j, i] <- sol[i, j] # Will overwrite when it shouldn't 
    } 
} 

將使用9次迭代，如何避免他們做到這6次迭代？

我需要計算對稱值，因此this question不適用。此other one也不工作，因爲可能有很多組合，並且在某些時候它不能在內存中分配矢量。

Answer 1

甲for循環通常會比outer慢。嘗試字節編譯循環或在Rcpp中實現它。

subss <- list(a = c(1, 2, 4), b = c(1, 2, 3), c = c(4, 5)) 
set.seed(42) 
A <- matrix(runif(25), ncol = 5, nrow = 5) 

#all combinations of indices 
ij <- combn(seq_along(subss), 2) 

#add all i = j 
ij <- matrix(c(ij, rep(seq_along(subss), each = 2)), nrow = 2) 

#preallocate 
res <- numeric(ncol(ij)) 

#only one loop 
for (k in seq_len(ncol(ij))) { 

    message(k) 

    res[k] <- mean(A[subss[[ij[1, k]]], subss[[ij[2, k]]]]) 
} 
#1 
#2 
#3 
#4 
#5 
#6 

#create symmetric sparse matrix  
library(Matrix) 
sol <- sparseMatrix(i = ij[1,], j = ij[2,], 
        x = res, dims = rep(length(subss), 2), 
        symmetric = TRUE, index1 = TRUE) 
#3 x 3 sparse Matrix of class "dsCMatrix" 
#         
#[1,] 0.7764715 0.6696987 0.7304413 
#[2,] 0.6696987 0.6266553 0.6778936 
#[3,] 0.7304413 0.6778936 0.5161089 

Answer 2

我發現了一種用普通的for循環：

x <- 0 
for (i in seq_along(subss)) { 
    for (j in seq_len(i)) { # or for (j in 1:i) as proposed below 
     x <- x + 1 
     message(x) 

     sol[i, j] <- mean(A[subss[[i]], subss[[j]]]) 
     sol[j, i] <- sol[i, j] 
    } 
} 

Answer 3

for (i in 1:length(subss)) { 
    for (j in 1:i) { 
    message(i, ' ', j, ' - ', mean(A[subss[[i]], subss[[j]]])) # Check iterations and value 
    sol2[i, j] <- sol2[j, i] <- mean(A[subss[[i]], subss[[j]]]) 
    } 
} 

我檢查你的腳本的價值觀和不是對稱的：

1 1 - 0.635455905252861 
1 2 - 0.638608284398086 
1 3 - 0.488700995299344 
2 1 - 0.568414432255344 
2 2 - 0.602851431118324 
2 3 - 0.516099992596234 
3 1 - 0.595461705311512 
3 2 - 0.656920690399905 
3 3 - 0.460815121419728 

礦值（同@ LLOPIS）：

1 2 - 0.638608284398086 
1 3 - 0.488700995299344 
2 2 - 0.602851431118324 
2 3 - 0.516099992596234 
3 2 - 0.656920690399905 
3 3 - 0.460815121419728