2

有没有更好的方法来迭代给定数据集的一组参数？显然，我试图得到一个相关系数表：列是“CI、CVP、平均 PAP、平均 SAP”，行是“ALAT、ASAT、GGT、Bili、LDH、FBG”。对于每个组合，我想获得相关系数和显着性水平（p = ...）。下面你会看到“艰难的道路”。但是有没有更优雅的方式，可能是一个可打印的表格？

attach(Liver)
cor.test(CI, ALAT, method = "spearman")
cor.test(CI, ASAT, method = "spearman")
cor.test(CI, GGT, method = "spearman")
cor.test(CI, Bili, method = "spearman")
cor.test(CI, LDH, method = "spearman")
cor.test(CI, FBG, method = "spearman")

cor.test(CVP, ALAT, method = "spearman")
cor.test(CVP, ASAT, method = "spearman")
cor.test(CVP, GGT, method = "spearman")
cor.test(CVP, Bili, method = "spearman")
cor.test(CVP, LDH, method = "spearman")
cor.test(CVP, FBG, method = "spearman")

cor.test(meanPAP, ALAT, method = "spearman")
cor.test(meanPAP, ASAT, method = "spearman")
cor.test(meanPAP, GGT, method = "spearman")
cor.test(meanPAP, Bili, method = "spearman")
cor.test(meanPAP, LDH, method = "spearman")
cor.test(meanPAP, FBG, method = "spearman")

cor.test(meanSAP, ALAT, method = "spearman")
cor.test(meanSAP, ASAT, method = "spearman")
cor.test(meanSAP, GGT, method = "spearman")
cor.test(meanSAP, Bili, method = "spearman")
cor.test(meanSAP, LDH, method = "spearman")
cor.test(meanSAP, FBG, method = "spearman")

detach("Liver")
4

2 回答 2

10

诀窍是获得所有可能的组合。在这里，我创建了一个data.frame10 列并使用该combn函数获取所有组合。然后它非常简单地获得相关值和 p 值。

set.seed(12)
x <- as.data.frame(matrix(rnorm(100), nrow=10))
combinations <- combn(ncol(x), 2)

out <- apply(combinations, 2, function(idx) {
    t <- cor.test(x[, idx[1]], x[, idx[2]], method = "spearman")
    c(names(x)[idx[1]], names(x)[idx[2]], t$estimate, t$p.value)
})
# more formatting if necessary
out <- as.data.frame(t(out))
names(out) <- c("col.idx1", "col.idx2", "cor", "pval")

Edit: An even more compact code by utilizing FUN argument within combn (as per Greg's suggestion)

set.seed(12)
x <- as.data.frame(matrix(rnorm(100), nrow=10))
out <- as.data.frame(t(combn(ncol(x), 2, function(idx) {
    t <- cor.test(x[, idx[1]], x[, idx[2]], method = "spearman")
    c(names(x)[idx[1]], names(x)[idx[2]], t$estimate, t$p.value)
})))
names(out) <- c("col.idx1", "col.idx2", "cor", "pval")
于 2012-12-28T11:15:32.383 回答
9

There is a function rcor.test() in library ltm that makes table of correlation coefficients and p-values. For example used data iris as do not have your data frame.

library(ltm)
rcor.test(iris[,1:4],method="spearman")


             Sepal.Length Sepal.Width Petal.Length Petal.Width
Sepal.Length  *****       -0.167       0.882        0.834     
Sepal.Width   0.041        *****      -0.310       -0.289     
Petal.Length <0.001       <0.001       *****        0.938     
Petal.Width  <0.001       <0.001      <0.001        *****     

upper diagonal part contains correlation coefficient estimates 
lower diagonal part contains corresponding p-values
于 2012-12-28T11:18:13.150 回答