问题描述
我正在使用R中的 rcorr（）函数在两个不同的矩阵之间建立关联：
I´m building a correlation between two different matrices with rcorr() function in R:
res <- rcorr(as.matrix(table1), as.matrix(table2),type="pearson")
似乎工作正常，但是我想避免表内相关-有任何建议吗？
It seems to be working fine, however I want to avoid within table correlations - any suggestion?
推荐答案
考虑使用R的基用于两个集合之间的明显相关，因为Hmisc的 rcorr（）返回所有可能的组合。请注意， rcorr（）的右上象限下面（在左下角重复对角线对称）是 cor（）（四舍五入到小数点后两位）。
Consider using R's base cor() for distinct correlations between two sets as Hmisc's rcorr() returns all possible combinations. Notice below the upper right quadrant of rcorr() (which repeats diagonally symmetrical on lower left) is the entire result of cor() (rounded to two decimal points).
table1 <- matrix(rnorm(25),5)
table2 <- matrix(rnorm(25),5)
res <- rcorr(table1, table2, type="pearson")
res
[,1] [,2] [,3] [,4] [,5] | [,6] [,7] [,8] [,9] [,10]
# [1,] 1.00 -0.55 0.95 -0.16 0.17 |-0.46 0.15 0.10 0.69 0.16
# [2,] -0.55 1.00 -0.55 -0.60 -0.79 |-0.45 -0.66 -0.22 -0.30 0.12
# [3,] 0.95 -0.55 1.00 -0.09 0.30 |-0.35 -0.05 -0.17 0.57 -0.03
# [4,] -0.16 -0.60 -0.09 1.00 0.91 | 0.92 0.53 -0.21 -0.58 -0.71
# [5,] 0.17 -0.79 0.30 0.91 1.00 | 0.78 0.41 -0.31 -0.32 -0.68
# ------------------------------------------------------------------
# [6,] -0.46 -0.45 -0.35 0.92 0.78 | 1.00 0.44 -0.14 -0.62 -0.58
# [7,] 0.15 -0.66 -0.05 0.53 0.41 | 0.44 1.00 0.68 0.13 0.13
# [8,] 0.10 -0.22 -0.17 -0.21 -0.31 |-0.14 0.68 1.00 0.59 0.80
# [9,] 0.69 -0.30 0.57 -0.58 -0.32 |-0.62 0.13 0.59 1.00 0.80
#[10,] 0.16 0.12 -0.03 -0.71 -0.68 |-0.58 0.13 0.80 0.80 1.00
# pvalues to follow ...
res <- cor(table1, table2, method="pearson")
res
# [,1] [,2] [,3] [,4] [,5]
# [1,] -0.4551474 0.15080994 0.1008215 0.6894955 0.16390813
# [2,] -0.4468285 -0.66209106 -0.2154960 -0.2954581 0.11662382
# [3,] -0.3542023 -0.05474287 -0.1720881 0.5669501 -0.02880113
# [4,] 0.9246330 0.53456574 -0.2084105 -0.5807386 -0.71108552
# [5,] 0.7788395 0.40551828 -0.3122606 -0.3209273 -0.67912147
唯一的警告是显着性检验统计信息，包括t-stats和p-values在 cor（）中不可用。但是，可以使用 cor.test（）检索它们，您可以使用 mapply（）迭代地运行它们。下面展示了一个测试对，并针对所有其他列进行了概括。请注意，测试的估计值与 cor（）输出中的值相对应。
The only caveat is significance test statistics including t-stats and p-values are not available with cor(). However, they can be retrieved with cor.test() which you can iteratively run with mapply(). Below demonstrates with one test pairing and generalized for all other columns. Notice the test's estimate corresponds to values in cor() output.
# EXAMPLE OF FIRST COL PAIRING
res <- cor.test(table1[,1], table2[,1], method="pearson")
res
# Pearson's product-moment correlation
# data: table1[, 1] and table2[, 1]
# t = -0.88536, df = 3, p-value = 0.4412
# alternative hypothesis: true correlation is not equal to 0
# 95 percent confidence interval:
# -0.9542314 0.7137222
# sample estimates:
# cor
# -0.4551474
# OBTAIN ALL MATRIX COL COMBINATIONS
tblcols <- expand.grid(1:ncol(table1), 1:ncol(table2))
# MAPPLY COR.TEST ACROSS ALL COLS
cfunc <- function(var1, var2) {
cor.test(table1[,var1], table2[,var2], method="pearson")
}
res <- mapply(function(a,b) {
cfunc(var1 = a, var2 = b)
}, tblcols$Var1, tblcols$Var2)
head(res)
# [,1] [,2] [,3] [,4]
# statistic -0.8853596 -0.8650936 -0.6560274 4.204994
# parameter 3 3 3 3
# p.value 0.4411699 0.4506234 0.5586316 0.02455469
# estimate -0.4551474 -0.4468285 -0.3542023 0.924633
# null.value 0 0 0 0
# alternative "two.sided" "two.sided" "two.sided" "two.sided"
# [,5] [,6] [,7] [,8]
# statistic 2.150733 0.2642326 -1.53021 -0.09495982
# parameter 3 3 3 3
# p.value 0.1206246 0.8087132 0.2234562 0.930334
# estimate 0.7788395 0.1508099 -0.6620911 -0.05474287
# null.value 0 0 0 0
# alternative "two.sided" "two.sided" "two.sided" "two.sided"
# ...
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