Spearman correlation loop in R - r

Question

A previous post explained how to do a Chi-squared loop in R on all your data-pairs: Chi Square Analysis using for loop in R.
I wanted to use this code to do the same thing for a Spearman correlation.
I've already tried altering a few of the variables and I was able to calculate the pearson correlation variables using this code:
library(plyr)
combos <- combn(ncol(fullngodata),2)
adply(combos, 2, function(x) {
test <- cor.test(fullngodata[, x[1]], fullngodata[, x[2]])
out <- data.frame("Row" = colnames(fullngodata)[x[1]]
, "Column" = colnames(fullngodata[x[2]])
, "cor" = round(test$statistic,3)
, "df"= test$parameter
, "p.value" = round(test$p.value, 3)
)
return(out)
})
But since I work with data on an ordinal scale, I need to use the Spearman correlation.
I thought I could get this data by just adding the method="spearman" command but this does not seem to work. If I use the code:
library(plyr)
combos <- combn(ncol(fullngodata),2)
adply(combos, 2, function(x) {
test <- cor.test(fullngodata[, x[1]], fullngodata[, x[2]], method="spearman")
out <- data.frame("Row" = colnames(fullngodata)[x[1]]
, "Column" = colnames(fullngodata[x[2]])
, "Chi.Square" = round(test$statistic,3)
, "df"= test$parameter
, "p.value" = round(test$p.value, 3)
)
return(out)
})
I get the response:
Error in data.frame(Row = colnames(fullngodata)[x[1]], Column =
colnames(fullngodata[x[2]]), :
arguments imply differing number of rows: 1, 0
In addition: Warning message:
In cor.test.default(fullngodata[, x[1]], fullngodata[, x[2]], method = "spearman") :
Cannot compute exact p-values with ties
what am I doing wrong?

Answer 1

Try rcor.test function in ltm package.
mat <- matrix(rnorm(1000), 100, 10, dimnames = list(NULL, LETTERS[1:10]))
rcor.test(mat, method = "spearman")
A B C D E F G H I J
A ***** -0.035 0.072 0.238 -0.097 0.007 -0.010 -0.031 0.039 -0.090
B 0.726 ***** -0.042 -0.166 0.005 0.025 0.007 -0.231 0.005 0.006
C 0.473 0.679 ***** 0.046 0.074 -0.020 0.091 -0.183 -0.040 -0.084
D 0.017 0.098 0.647 ***** -0.060 -0.151 -0.175 -0.068 0.039 0.181
E 0.338 0.960 0.466 0.553 ***** 0.254 0.055 -0.031 0.072 -0.059
F 0.948 0.805 0.843 0.133 0.011 ***** -0.014 -0.121 0.153 0.048
G 0.923 0.941 0.370 0.081 0.588 0.892 ***** -0.060 -0.050 0.011
H 0.759 0.021 0.069 0.501 0.756 0.230 0.555 ***** -0.053 -0.193
I 0.700 0.963 0.690 0.701 0.476 0.130 0.621 0.597 ***** -0.034
J 0.373 0.955 0.406 0.072 0.561 0.633 0.910 0.055 0.736 *****
upper diagonal part contains correlation coefficient estimates
lower diagonal part contains corresponding p-values

Answer 2

The problem is that cor.test returns a value NULL for parameter when you do the spearman test. From ?cor.test: parameter: the degrees of freedom of the test statistic in the case that it follows a t distribution.
You can see this in the following example:
x <- c(44.4, 45.9, 41.9, 53.3, 44.7, 44.1, 50.7, 45.2, 60.1)
y <- c( 2.6, 3.1, 2.5, 5.0, 3.6, 4.0, 5.2, 2.8, 3.8)
str(cor.test(x, y, method = "spearman"))
List of 8
$ statistic : Named num 48
..- attr(*, "names")= chr "S"
$ parameter : NULL
$ p.value : num 0.0968
$ estimate : Named num 0.6
..- attr(*, "names")= chr "rho"
$ null.value : Named num 0
..- attr(*, "names")= chr "rho"
$ alternative: chr "two.sided"
$ method : chr "Spearman's rank correlation rho"
$ data.name : chr "x and y"
- attr(*, "class")= chr "htest"
Solution: if you remove the following line from your code, it should work:
, "df"= test$parameter