我想使用 来计算 AUC auc(roc(predictions, labels))，其中labels是1(x15) 和0(x500) 的数值向量，并且是具有从[二项式]predictions派生的概率的数值向量。glm它应该很简单，但auc(roc(predictions, labels))会给出错误消息“没有足够的不同预测来计算 ROC 曲线下的面积”。我一定在做一些傻事，但我不知道是什么。你能？
代码是
library(AUC)
#read the data, that come from a previous process of a species distribution modelling
prob<-read.csv("prob.csv")
labels<-read.csv("labels.csv")
#prob is
#labels is
roc(prob,labels)
#Gives the error (that I'm NOT interest in)
Error in `[.data.frame`(predictions, pred.order) : undefined columns selected
In addition: Warning messages:
1: In is.na(x) : is.na() applied to non-(list or vector) of type 'NULL'
2: In is.na(e2) : is.na() applied to non-(list or vector) of type 'NULL'
3: In is.na(e2) : is.na() applied to non-(list or vector) of type 'NULL'
#I change the format to numeric vector
prob<-as.numeric(prob[,2])
labels<-as.numeric(labels[,2])
#Verify it is a vector numeric
class(prob)
[1] "numeric"
class(labels)
[1] "numeric"
#call the roc functoin
roc(prob,labels)
Error in roc(modbrapred, pbbra) : # THIS is the error I0m interested in
Not enough distinct predictions to compute area under the ROC curve.
In addition: Warning messages:
1: In is.na(x) : is.na() applied to non-(list or vector) of type 'NULL'
2: In is.na(e2) : is.na() applied to non-(list or vector) of type 'NULL'
3: In is.na(e2) : is.na() applied to non-(list or vector) of type 'NULL'
Data is as follows
labels.csv
"","x"
"1",1
"2",1
"3",1
"4",1
"5",1
"6",1
...
"164",1
"165",1
"166",0
"167",0
"168",0
"169",0
"170",0
"171",0
"172",0
...
"665",0
prob.csv
"","x"
"1",0.977465874525236
"2",0.989692657762578
"3",0.989692657762578
"4",0.988038430564019
"5",0.443188602491041
"6",0.409732585195485
...
"164",0.988607910625475
"165",0.986296936078692
"166",7.13529696560611e-05
"167",0.000419255989134081
"168",0.00295825183558019
"169",0.00182941235784709
"170",4.85601026999172e-09
"171",0.000953106471289961
"172",1.70252014430306e-05
...
"665",8.13413358866349e-08