Re: [問題] list對應座標的運算

看板R_Language作者celestialgod (天)時間7年前 (2018/04/24 19:21)推噓2(2推 0噓 3→)

留言5則, 4人參與討論串2/2 (看更多)

※ 引述《ntpuisbest (阿龍)》之銘言： : 我有個list : 長這樣 : https://imgur.com/glWVNGb

: 他是一個模擬得到的VCOV variance covariance矩陣 : 我希望做到對應的座標取平均 : 也就是 : https://imgur.com/tcpNNnY

: 希望回傳的也是一個三成三的list : 不知道怎麼做 : 發現lapply 也沒用QAQ 給四個方法參考XD library(abind) # data l <- replicate(3L, matrix(rnorm(9), 3), FALSE) # method 1 apply(abind(l, along = 3L), 1:2, mean) # [,1] [,2] [,3] # [1,] 0.08595378 -0.9663702 -0.7770976 # [2,] 0.13758227 0.4697197 0.2799617 # [3,] -0.57574027 -0.4079516 -0.9508097 # method 2 apply(array(unlist(l), dim=rep(3,3)), 1:2, mean) # [,1] [,2] [,3] # [1,] 0.08595378 -0.9663702 -0.7770976 # [2,] 0.13758227 0.4697197 0.2799617 # [3,] -0.57574027 -0.4079516 -0.9508097 # method 3 Reduce("+", l) / length(l) # [,1] [,2] [,3] # [1,] 0.08595378 -0.9663702 -0.7770976 # [2,] 0.13758227 0.4697197 0.2799617 # [3,] -0.57574027 -0.4079516 -0.9508097 # method 4 out <- l[[1]] for (i in 2L:length(l)) out <- out + l[[i]] out / length(l) # [,1] [,2] [,3] # [1,] 0.08595378 -0.9663702 -0.7770976 # [2,] 0.13758227 0.4697197 0.2799617 # [3,] -0.57574027 -0.4079516 -0.9508097 我做了一下benchmark.... 正如我推文所說的，for比較快XD # benchmark forFunc <- function(l){ out <- l[[1]] for (i in 2L:length(l)) out <- out + l[[i]] out / length(l) } library(microbenchmark) l <- replicate(3e3, matrix(rnorm(200^2), 200), FALSE) print(object.size(l), units = "Gb") # 0.9 Gb microbenchmark(method1 = apply(abind(l, along = 3L), 1:2, mean), method2 = apply(array(unlist(l), dim=c(nrow(l[[1]]), ncol(l[[1]]), 3)), 1:2, mean), method3 = Reduce("+", l) / length(l), method4 = forFunc(l), times = 20L) # Unit: milliseconds # expr min lq mean median uq max neval # method1 2481.3607 2681.3842 2730.1426 2776.8715 2803.6407 2821.4399 20 # method2 474.9360 485.1195 531.9193 488.4900 582.2409 670.2529 20 # method3 123.0389 124.6572 144.5512 126.9948 132.6468 310.0517 20 # method4 121.3197 123.1581 127.6650 126.5533 131.4164 139.4469 20 記憶體使用方面，abind是裡面最花記憶體的雖然使用上滿簡單的，但不建議使用abind # memory usage library(data.table) library(profmem) memUsageList <- vector("list", 4L) memUsageList[[1]] <- profmem({apply(abind(l, along = 3L), 1:2, mean)}) memUsageList[[2]] <- profmem({ apply(array(unlist(l), dim=c(nrow(l[[1]]), ncol(l[[1]]), 3)), 1:2, mean) }) memUsageList[[3]] <- profmem({Reduce("+", l) / length(l)}) memUsageList[[4]] <- profmem({forFunc(l)}) data.table(methods = paste0("method", 1:4), "memory (Mb)" = sapply(memUsageList, total) / 2^20) # methods memory (Mb) # 1: method1 5076.6124 # 2: method2 918.5794 # 3: method3 915.6992 # 4: method4 915.6418 以上，供你參考 -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 114.38.104.189 ※ 文章網址: https://www.ptt.cc/bbs/R_Language/M.1524568892.A.5C0.html