CODEWITHSUNDEEP
rloopsrocproc-r-package
Niels Verburg
Niels Verburg

Reputation: 13

Create ROC curve with for loop or sapply

After a lot of searching I could not find an answer to my problem. I would like to generate a ROC curve with the pROC pakkage using a for loop or sapply.

My database looks like this (only with 26 colums and 74 rows):

PT Bpt PA mnT1G mnT01
1   1  1   2.3   4.5
1   2  0   1.2   3.2 
2   1  1   5.4   2.1

I can make a ROC curve 'manually':

plot.new()
roc1 <- roc(cor.datT$PA, cor.datT$mT1G, percent=TRUE, partial.auc=c(100, 90), partial.auc.correct=TRUE, 
            partial.auc.focus="sens", ci=TRUE, boot.n=100, ci.alpha=0.9, stratified=FALSE, plot=TRUE, col= 'red')
roc2 <- roc(cor.datT$PA, cor.datT$mT01, plot=TRUE, add=TRUE, percent=roc1$percent, col = 'blue')

For 'automatic' I tried:

First roc curve always mnT1G:

rocT1G <- roc(cor.datT$PA, cor.datT$mnT1G, percent=TRUE, partial.auc=c(100, 90), partial.auc.correct=TRUE, partial.auc.focus="sens", ci=TRUE, boot.n=100, ci.alpha=0.9, stratified=FALSE, plot=TRUE, col= 'red')

Add other roc curves (data$Img are all the image names (like T1G, T01, etc) from another dataframe). I understand they all will be blue :

sapply(unique(data$Img[data$Img != "T1G"]), FUN = function(i) paste("roc",i,sep="") <- roc(cor.datT$PA, cor.datT[paste("mn",i, sep = "")], plot=TRUE, add=TRUE, percent=rocT1G$percent, col = 'blue'), simplify = FALSE)

But I get this error:

Error in roc.default(cor.datT$PA, cor.datT[paste("mn", i, sep = "")], : Predictor must be numeric or ordered.

Same happens with for loop:

for (i in unique(data$Img[data$Img != "T1G"])){
    plot.new()
    rocT1G <- roc(cor.datT$PA, cor.datT$mnT1G, percent=TRUE, partial.auc=c(100, 90), partial.auc.correct=TRUE, partial.auc.focus="sens", ci=TRUE, boot.n=100, ci.alpha=0.9, stratified=FALSE, plot=TRUE, col= 'red')
    paste("roc",i,sep="") <- roc(cor.datT$PA, cor.datT[paste("mn",i, sep = "")], plot=TRUE, add=TRUE, percent=rocT1G$percent, col = 'blue')
}

I checked the columns and they are all numerical. So maybe something goes wrong with the class in my script?

Upvotes: 1

Views: 1551

Answers (2)

Calimo
Calimo

Reputation: 7969

As you noted in a comment on my other answer, the problem is that you get specifically data.frames out of your extraction.

In a data.frame, extracting with a single character returns a data.frame. This is documented in ?Extract.data.frame:

Data frames can be indexed in several modes. When [ and [[ are used with a single vector index (x[i] or x[[i]]), they index the data frame as if it were a list.

And looking at ?Extract:

Recursive (list-like) objects

Indexing by [ is similar to atomic vectors and selects a list of the specified element(s).

This is not so obvious from the text, but in order to extract a column into a vector, you need to use two brackets [[, so

class(cor.datT[[paste("mn",i, sep = "")]])

should be a vector.

Now the following code should run: 

rocT1G <- roc(cor.datT$PA, cor.datT$mnT1G, percent=TRUE, partial.auc=c(100, 90), partial.auc.correct=TRUE, partial.auc.focus="sens", ci=TRUE, boot.n=100, ci.alpha=0.9, stratified=FALSE, plot=TRUE, col= 'red')
for (i in unique(data$Img[data$Img != "T1G"])){
    roc(cor.datT$PA, cor.datT[[paste("mn",i, sep = "")]], plot=TRUE, add=TRUE, percent=rocT1G$percent, col = 'blue')
}

Upvotes: 0

Calimo
Calimo

Reputation: 7969

One of your column is not numeric as you expect. Unfortunately the error message from R doesn't tell you at which iteration of the loop the problem occurs, but you can easily add print statement to your loop to figure out which column yields the problem

for (i in unique(data$Img[data$Img != "T1G"])){
    print(i)
    plot.new()
    rocT1G <- roc(cor.datT$PA, cor.datT$mnT1G, percent=TRUE, partial.auc=c(100, 90), partial.auc.correct=TRUE, partial.auc.focus="sens", ci=TRUE, boot.n=100, ci.alpha=0.9, stratified=FALSE, plot=TRUE, col= 'red')
    paste("roc",i,sep="") <- roc(cor.datT$PA, cor.datT[paste("mn",i, sep = "")], plot=TRUE, add=TRUE, percent=rocT1G$percent, col = 'blue')
}

The last column number printed will be the one. You can see what is going on with the column:

class(cor.datT[paste("mn",i, sep = "")])
str(cor.datT[paste("mn",i, sep = "")])

More generally you can inspect all the columns of your data frame quite quickly with 

str(cor.datT)

Check out for columns of class character or factor in particular.

Upvotes: 0

Related Questions