Niels Verburg - 1 year ago 104

R Question

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?

Recommended for you: Get network issues from **WhatsUp Gold**. **Not end users.**

Answer Source

As you noted in a comment on my other answer, the problem is that you get specifically `data.frame`

s 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) objectsIndexing 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.

Recommended from our users: **Dynamic Network Monitoring from WhatsUp Gold from IPSwitch**. ** Free Download**