提问人：Simon Harmel 提问时间：9/13/2023 最后编辑：Simon Harmel 更新时间：9/18/2023 访问量：74
在 R 中将向量转换为矩阵时指定变量名称
Specifying the variable names when converting a vector to a matrix in R
问：
我有一个名为 .该向量有 8 个元素，如下所示： .Rsc(L2DA.L2DF= .637, L2DA.L2G= 0.553,...)
我可以使用（见下文）将这个命名的数值向量转换为 8x8 相关矩阵。metafor::vec2mat(Rs)
问题：但是我想知道如何为该相关矩阵分配行名和列名，以便这些名称表示我原始命名的数字向量中的名称？
例如，在矩阵中，元素 [1,1] = 来自我的向量的第一个元素，因此它应该具有 colname 和 rowname ，依此类推。0.637L2DAL2DF
library(metafor)
dat <- read.csv("https://raw.githubusercontent.com/ilzl/i/master/j.csv")
dat$var1.var2 <- apply(dat[c("var1","var2")],1,paste0,collapse=".")
res <- rma(ri~var1.var2+0,1, data=dat)
(Rs = setNames(coef(res),sub("var1.var2","",names(coef(res)))))
(R_matrix = vec2mat(Rs))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 1.0000000 0.637 0.5533333 0.4180000 0.5550000 0.5678947 0.4781481 0.3675000
[2,] 0.6370000 1.000 0.2440000 0.2900000 0.4840000 0.3500000 0.4750000 0.5700000
[3,] 0.5533333 0.244 1.0000000 0.2933333 0.5100000 0.3300000 0.4775000 0.5765714
[4,] 0.4180000 0.290 0.2933333 1.0000000 0.4627778 0.5212121 0.5569565 0.4928571
[5,] 0.5550000 0.484 0.5100000 0.4627778 1.0000000 0.4695652 0.5140625 0.5313793
[6,] 0.5678947 0.350 0.3300000 0.5212121 0.4695652 1.0000000 0.5194118 0.5258333
[7,] 0.4781481 0.475 0.4775000 0.5569565 0.5140625 0.5194118 1.0000000 0.4240000
[8,] 0.3675000 0.570 0.5765714 0.4928571 0.5313793 0.5258333 0.4240000 1.0000000
答：
4赞
r2evans
9/13/2023
#1
首先，屏蔽是一个问题：vec2mat
Rs
# L2DA.L2DF L2DA.L2G L2DA.L2L L2DA.L2M L2DA.L2P L2DA.L2V L2DF.L2G L2DF.L2L L2DF.L2M L2DF.L2P L2DF.L2V L2G.L2L L2G.L2M L2G.L2P L2M.L2L
# 0.6370000 0.5533333 0.4180000 0.5550000 0.5678947 0.4781481 0.3675000 0.2440000 0.2900000 0.4840000 0.3500000 0.4750000 0.5700000 0.2933333 0.5100000
# L2P.L2L L2P.L2M L2R.L2DA L2R.L2DF L2R.L2G L2R.L2L L2R.L2M L2R.L2P L2R.L2V L2V.L2G L2V.L2L L2V.L2M L2V.L2P
# 0.3300000 0.4775000 0.5765714 0.4627778 0.5212121 0.5569565 0.4928571 0.4695652 0.5140625 0.5313793 0.5194118 0.5258333 0.4240000
请注意，有六个值，表明矩阵的第一列（推断应该是 应该跨越 和 ...但是，它也包括以下值：L2DA.*"L2DA"0.6370.478L2DF.L2G
vec2mat(Rs)
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
# [1,] 1.0000000 0.637 0.5533333 0.4180000 0.5550000 0.5678947 0.4781481 0.3675000
# [2,] 0.6370000 1.000 0.2440000 0.2900000 0.4840000 0.3500000 0.4750000 0.5700000
# [3,] 0.5533333 0.244 1.0000000 0.2933333 0.5100000 0.3300000 0.4775000 0.5765714
# [4,] 0.4180000 0.290 0.2933333 1.0000000 0.4627778 0.5212121 0.5569565 0.4928571
# [5,] 0.5550000 0.484 0.5100000 0.4627778 1.0000000 0.4695652 0.5140625 0.5313793
# [6,] 0.5678947 0.350 0.3300000 0.5212121 0.4695652 1.0000000 0.5194118 0.5258333
# [7,] 0.4781481 0.475 0.4775000 0.5569565 0.5140625 0.5194118 1.0000000 0.4240000
# [8,] 0.3675000 0.570 0.5765714 0.4928571 0.5313793 0.5258333 0.4240000 1.0000000
无论出于何种原因，并非所有组合都存在于系数中;至少缺少：L2DA.L2R
nms <- unique(unlist(strsplit(names(Rs), "[.]")))
nms
# [1] "L2DA" "L2DF" "L2G" "L2L" "L2M" "L2P" "L2V" "L2R"
grep("L2DA", names(Rs), value=TRUE)
# [1] "L2DA.L2DF" "L2DA.L2G" "L2DA.L2L" "L2DA.L2M" "L2DA.L2P" "L2DA.L2V" "L2R.L2DA"
setdiff(nms, sub(".*\\.", "", grep("L2DA", names(Rs), value=TRUE)))
# [1] "L2R"
...虽然存在......似乎没有与数据共享的假设。L2R.L2DAvec2mat
虽然我们可能会蛮力匹配，但我认为这个推理步骤会使所有系数都已正确分配给行/列变得不那么清楚。
而不是，让我们自己做，保留名称。vec2mat
tmp <- data.frame(nm = names(Rs), R = Rs) |>
transform(row = sub("\\..*", "", nm), col = sub(".*\\.", "", nm))
head(tmp)
# nm R row col
# L2DA.L2DF L2DA.L2DF 0.6370000 L2DA L2DF
# L2DA.L2G L2DA.L2G 0.5533333 L2DA L2G
# L2DA.L2L L2DA.L2L 0.4180000 L2DA L2L
# L2DA.L2M L2DA.L2M 0.5550000 L2DA L2M
# L2DA.L2P L2DA.L2P 0.5678947 L2DA L2P
# L2DA.L2V L2DA.L2V 0.4781481 L2DA L2V
reshape2::dcast(tmp, row ~ col, value.var = "R")
# row L2DA L2DF L2G L2L L2M L2P L2V
# 1 L2DA NA 0.6370000 0.5533333 0.4180000 0.5550000 0.5678947 0.4781481
# 2 L2DF NA NA 0.3675000 0.2440000 0.2900000 0.4840000 0.3500000
# 3 L2G NA NA NA 0.4750000 0.5700000 0.2933333 NA
# 4 L2M NA NA NA 0.5100000 NA NA NA
# 5 L2P NA NA NA 0.3300000 0.4775000 NA NA
# 6 L2R 0.5765714 0.4627778 0.5212121 0.5569565 0.4928571 0.4695652 0.5140625
# 7 L2V NA NA 0.5313793 0.5194118 0.5258333 0.4240000 NA
这并不奇怪，因为我们知道我们没有两个方向的对。让我们把两者交换，然后行绑定，然后再来一次。tmpdcast
tmp2 <- transform(tmp, row2 = row, row = col) |>
transform(col = row2, row2 = NULL)
head(tmp2)
# nm R row col
# L2DA.L2DF L2DA.L2DF 0.6370000 L2DF L2DA
# L2DA.L2G L2DA.L2G 0.5533333 L2G L2DA
# L2DA.L2L L2DA.L2L 0.4180000 L2L L2DA
# L2DA.L2M L2DA.L2M 0.5550000 L2M L2DA
# L2DA.L2P L2DA.L2P 0.5678947 L2P L2DA
# L2DA.L2V L2DA.L2V 0.4781481 L2V L2DA
out <- rbind(tmp, tmp2) |>
reshape2::dcast(row ~ col, value.var = "R")
out
# row L2DA L2DF L2G L2L L2M L2P L2R L2V
# 1 L2DA NA 0.6370000 0.5533333 0.4180000 0.5550000 0.5678947 0.5765714 0.4781481
# 2 L2DF 0.6370000 NA 0.3675000 0.2440000 0.2900000 0.4840000 0.4627778 0.3500000
# 3 L2G 0.5533333 0.3675000 NA 0.4750000 0.5700000 0.2933333 0.5212121 0.5313793
# 4 L2L 0.4180000 0.2440000 0.4750000 NA 0.5100000 0.3300000 0.5569565 0.5194118
# 5 L2M 0.5550000 0.2900000 0.5700000 0.5100000 NA 0.4775000 0.4928571 0.5258333
# 6 L2P 0.5678947 0.4840000 0.2933333 0.3300000 0.4775000 NA 0.4695652 0.4240000
# 7 L2R 0.5765714 0.4627778 0.5212121 0.5569565 0.4928571 0.4695652 NA 0.5140625
# 8 L2V 0.4781481 0.3500000 0.5313793 0.5194118 0.5258333 0.4240000 0.5140625 NA
如果你需要它作为一个简单的矩阵，那么我们可以这样做：
out2 <- as.matrix(out[,-1])
dimnames(out2) <- list(out$row, colnames(out)[-1])
out2
# L2DA L2DF L2G L2L L2M L2P L2R L2V
# L2DA NA 0.6370000 0.5533333 0.4180000 0.5550000 0.5678947 0.5765714 0.4781481
# L2DF 0.6370000 NA 0.3675000 0.2440000 0.2900000 0.4840000 0.4627778 0.3500000
# L2G 0.5533333 0.3675000 NA 0.4750000 0.5700000 0.2933333 0.5212121 0.5313793
# L2L 0.4180000 0.2440000 0.4750000 NA 0.5100000 0.3300000 0.5569565 0.5194118
# L2M 0.5550000 0.2900000 0.5700000 0.5100000 NA 0.4775000 0.4928571 0.5258333
# L2P 0.5678947 0.4840000 0.2933333 0.3300000 0.4775000 NA 0.4695652 0.4240000
# L2R 0.5765714 0.4627778 0.5212121 0.5569565 0.4928571 0.4695652 NA 0.5140625
# L2V 0.4781481 0.3500000 0.5313793 0.5194118 0.5258333 0.4240000 0.5140625 NA
diag(out2) <- 1
out2
# L2DA L2DF L2G L2L L2M L2P L2R L2V
# L2DA 1.0000000 0.6370000 0.5533333 0.4180000 0.5550000 0.5678947 0.5765714 0.4781481
# L2DF 0.6370000 1.0000000 0.3675000 0.2440000 0.2900000 0.4840000 0.4627778 0.3500000
# L2G 0.5533333 0.3675000 1.0000000 0.4750000 0.5700000 0.2933333 0.5212121 0.5313793
# L2L 0.4180000 0.2440000 0.4750000 1.0000000 0.5100000 0.3300000 0.5569565 0.5194118
# L2M 0.5550000 0.2900000 0.5700000 0.5100000 1.0000000 0.4775000 0.4928571 0.5258333
# L2P 0.5678947 0.4840000 0.2933333 0.3300000 0.4775000 1.0000000 0.4695652 0.4240000
# L2R 0.5765714 0.4627778 0.5212121 0.5569565 0.4928571 0.4695652 1.0000000 0.5140625
# L2V 0.4781481 0.3500000 0.5313793 0.5194118 0.5258333 0.4240000 0.5140625 1.0000000
作为功能：
myfun <- function(Rs) {
tmp <- data.frame(nm = names(Rs), R = Rs) |>
transform(row = sub("\\..*", "", nm), col = sub(".*\\.", "", nm))
tmp2 <- transform(tmp, row2 = row, row = col) |>
transform(col = row2, row2 = NULL)
out <- rbind(tmp, tmp2) |>
tidyr::pivot_wider(id_cols = "row", names_from = "col", values_from = "R")
out <- out[order(match(out$row, colnames(out))),]
out2 <- as.matrix(out[,-1])
dimnames(out2) <- list(out$row, colnames(out)[-1])
diag(out2) <- 1
out2
}
Rs
# L2DA.L2DF L2DA.L2G L2DA.L2L L2DA.L2M L2DA.L2P L2DA.L2V L2DF.L2G L2DF.L2L L2DF.L2M L2DF.L2P L2DF.L2V L2G.L2L L2G.L2M L2G.L2P L2M.L2L
# 0.6370000 0.5533333 0.4180000 0.5550000 0.5678947 0.4781481 0.3675000 0.2440000 0.2900000 0.4840000 0.3500000 0.4750000 0.5700000 0.2933333 0.5100000
# L2P.L2L L2P.L2M L2R.L2DA L2R.L2DF L2R.L2G L2R.L2L L2R.L2M L2R.L2P L2R.L2V L2V.L2G L2V.L2L L2V.L2M L2V.L2P
# 0.3300000 0.4775000 0.5765714 0.4627778 0.5212121 0.5569565 0.4928571 0.4695652 0.5140625 0.5313793 0.5194118 0.5258333 0.4240000
myfun(Rs)
# L2DF L2G L2L L2M L2P L2V L2DA L2R
# L2DF 1.0000000 0.3675000 0.2440000 0.2900000 0.4840000 0.3500000 0.6370000 0.4627778
# L2G 0.3675000 1.0000000 0.4750000 0.5700000 0.2933333 0.5313793 0.5533333 0.5212121
# L2L 0.2440000 0.4750000 1.0000000 0.5100000 0.3300000 0.5194118 0.4180000 0.5569565
# L2M 0.2900000 0.5700000 0.5100000 1.0000000 0.4775000 0.5258333 0.5550000 0.4928571
# L2P 0.4840000 0.2933333 0.3300000 0.4775000 1.0000000 0.4240000 0.5678947 0.4695652
# L2V 0.3500000 0.5313793 0.5194118 0.5258333 0.4240000 1.0000000 0.4781481 0.5140625
# L2DA 0.6370000 0.5533333 0.4180000 0.5550000 0.5678947 0.4781481 1.0000000 0.5765714
# L2R 0.4627778 0.5212121 0.5569565 0.4928571 0.4695652 0.5140625 0.5765714 1.0000000
评论
0赞
r2evans
9/13/2023
（1） 当然，.（2） 当然，.myfun <- function(Rs) { tmp <- ...; tmp2 <- ..., out <- ...; out2 <- ...; out2; }diag(out2) <- 1
0赞
r2evans
9/13/2023
看看我的编辑@SimonHarmel，我为你做了这项工作（虽然它没有什么新颖或令人印象深刻的，只是一个包装）。
0赞
r2evans
9/14/2023
基地 R？我建议不要这样做。-to- 是微不足道的：dcastpivot_wider... |> tidyr::pivot_wider(id_cols = "row", names_from = "col", values_from = "R")
0赞
Simon Harmel
9/14/2023
哼哼，但是输出不匹配，例如，对角线元素在变成 1 之前不再都是 NA？
0赞
r2evans
9/14/2023
@SimonHarmel查看我的更新。
2赞
Robert Hacken
9/18/2023
#2
@r2evans解决方案的更简洁的变体：
# matrix with names of rows and columns in its two columns
row.col <- do.call(rbind, strsplit(names(Rs), '\\.'))
# vector of all names
nam <- sort(unique(c(row.col)))
# empty correlation matrix
corr <- matrix(NA, length(nam), length(nam), dimnames=list(nam, nam))
# fill it
diag(corr) <- 1
corr[rbind(row.col, row.col[, 2:1])] <- Rs
corr
# L2DA L2DF L2G L2L L2M L2P L2R L2V
# L2DA 1.00 0.64 0.55 0.42 0.56 0.57 0.58 0.48
# L2DF 0.64 1.00 0.37 0.24 0.29 0.48 0.46 0.35
# L2G 0.55 0.37 1.00 0.48 0.57 0.29 0.52 0.53
# L2L 0.42 0.24 0.48 1.00 0.51 0.33 0.56 0.52
# L2M 0.56 0.29 0.57 0.51 1.00 0.48 0.49 0.53
# L2P 0.57 0.48 0.29 0.33 0.48 1.00 0.47 0.42
# L2R 0.58 0.46 0.52 0.56 0.49 0.47 1.00 0.51
# L2V 0.48 0.35 0.53 0.52 0.53 0.42 0.51 1.00
评论
0赞
Simon Harmel
9/20/2023
谢谢+1，罗伯特。我在这里问了一个关于你的伟大答案的后续问题。你能看一下吗？
评论