這是一個嘗試使用talkstats.com的朋友的建議在R中製作MTMM。我不知道它是否正確,因爲我沒有一個測試(基準)數據集用於使用已知是正確的MTMM。請批評。這是一個MTMM還是隻在對角線上具有可靠性的隨機矩陣?
只要記住dim是方法,r是列名和行名的構造。
require(CTT); require(foreign)
dat22 <-read.csv(url("http://dl.dropbox.com/u/61803503/dat.csv"), header=TRUE,
strip.white = TRUE, sep=",", as.is=FALSE, na.strings= c("999", "NA", " "))
#group items by method(dim) and construct(r)
dim1r1 <- dat2[, c(3, 5, 9, 10)]
dim2r1 <- dat2[, c(4, 13:15)]
dim3r1 <- dat2[, c(1, 6, 7, 11, 12)]
dim4r1 <- dat2[, c(2, 8, 16, 17)]
dim1r2 <- dat2[, c(3, 5, 9, 10)+17]
dim2r2 <- dat2[, c(4, 13:15)+17]
dim3r2 <- dat2[, c(1, 6, 7, 11, 12)+17]
dim4r2 <- dat2[, c(2, 8, 16, 17)+17]
dim1r3 <- dat2[, c(3, 5, 9, 10)+17*2]
dim2r3 <- dat2[, c(4, 13:15)+17*2]
dim3r3 <- dat2[, c(1, 6, 7, 11, 12)+17*2]
dim4r3 <- dat2[, c(2, 8, 16, 17)+17*2]
dim1r4 <- dat2[, c(3, 5, 9, 10)+17*3]
dim2r4 <- dat2[, c(4, 13:15)+17*3]
dim3r4 <- dat2[, c(1, 6, 7, 11, 12)+17*3]
dim4r4 <- dat2[, c(2, 8, 16, 17)+17*3]
#make a list from the above items
#dim1r1 means methid 1 (dim1) and construct 1(r1)
LIST2 <- list(dim1r1, dim1r2, dim1r3, dim1r4, dim2r1, dim2r2, dim2r3, dim2r4,
dim3r1, dim3r2, dim3r3, dim3r4, dim4r1, dim4r2, dim4r3, dim4r4)
#get the sums of the items by method and construct
#and generate correlation amtrix (all in 1 step)
mtmm <- round(cor(sapply(LIST2, function(x) rowSums(x))), digits=3)
#generate and order row and column names
VN <- expand.grid(paste('dim', 1:4, sep=""), paste('r', 1:4, sep=""))
VN <- VN[order(VN$Var1, VN$Var2), ]
varNames <- paste(VN[, 1], VN[, 2], sep="")
rownames(mtmm) <- colnames(mtmm) <-varNames
#blank out the upper triangle
mtmm[upper.tri(mtmm)] <- " "
#add cronbach's alpha intot he diagonal
diag(mtmm) <- sapply(LIST2, function(x) round(reliability(x)$alpha, digits=3))
noquote(mtmm)
產生:
dim1r1 dim1r2 dim1r3 dim1r4 dim2r1 dim2r2 dim2r3 dim2r4 dim3r1 dim3r2 dim3r3 dim3r4 dim4r1 dim4r2 dim4r3 dim4r4
dim1r1 0.737
dim1r2 0.82 0.78
dim1r3 0.825 0.755 0.735
dim1r4 0.828 0.783 0.812 0.791
dim2r1 0.415 0.496 0.484 0.495 0.801
dim2r2 0.432 0.615 0.493 0.479 0.818 0.886
dim2r3 0.425 0.473 0.505 0.459 0.89 0.831 0.843
dim2r4 0.355 0.468 0.413 0.482 0.806 0.826 0.837 0.802
dim3r1 0.544 0.518 0.413 0.494 0.281 0.226 0.184 0.233 0.778
dim3r2 0.517 0.585 0.399 0.461 0.306 0.324 0.26 0.293 0.88 0.782
dim3r3 0.491 0.489 0.392 0.421 0.258 0.229 0.232 0.221 0.875 0.912 0.804
dim3r4 0.487 0.492 0.366 0.475 0.269 0.268 0.209 0.274 0.887 0.89 0.859 0.77
dim4r1 0.341 0.399 0.38 0.357 0.387 0.398 0.355 0.375 0.397 0.417 0.387 0.43 0.489
dim4r2 0.274 0.433 0.326 0.323 0.462 0.535 0.416 0.46 0.343 0.422 0.349 0.432 0.863 0.517
dim4r3 0.268 0.368 0.364 0.306 0.329 0.417 0.333 0.341 0.293 0.376 0.34 0.353 0.863 0.856 0.545
dim4r4 0.301 0.403 0.347 0.395 0.377 0.443 0.371 0.483 0.372 0.441 0.345 0.441 0.86 0.84 0.83 0.52
這可能相當使用ggplot或如Excel等
一個外部程序的一般邏輯可能是使用的方法作爲板進行清洗,並加以,特徵作爲分類軸,然後根據相關係數對相關矩陣中的單元格進行着色(例如,請參閱統計站點上的[本答案](http://stats.stackexchange.com/q/9918/1036)) )。如果您真的重現了您的相關矩陣,那將會更容易。對於上帝的愛,請不要在包含的圖片中使用該顏色方案! – 2012-03-09 12:52:40
安迪提請我注意上面的圖片是誤導性的。我真的不想輸出成漂亮的圖形,只是從數字上說輸出是正確的。它可能是醜陋的(現在實際上是首選)。 – 2012-03-11 18:00:25