r - 没有足够的明确预测来计算 roc 下的面积

Question

我正在使用 auc(roc(predictions, labels)) 计算 AUC，其中 labels 是 1 的数值向量> (x15) 和 0 (x500)，而 predictions 是一个数值向量，其概率源自 glm [二项式]。它应该非常简单，但是 auc(roc(predictions, labels)) 给出错误，指出“没有足够的不同预测来计算 ROC 曲线下的面积”。我一定是在做一些愚蠢的事情，但我无法发现是什么。你可以吗？

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

Answer 1

问题是我的“标签”是一个数字向量，但我需要一个因子。于是我就转型了

labels <- factor(labels)

大鹏鸟按其应有的方式工作

感谢您投入的时间