我想要计算AUC使用auc(roc(predictions, labels)),其中labels是1(x15)和0(x500)predictions的数值向量,并且是具有从glm[二项式] 导出的概率的数值向量.它应该非常简单,但auc(roc(predictions, labels))会给出一个错误,说"在ROC曲线下计算区域的预测不够".我必须做些傻事,但我无法发现什么.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
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use*_*623 24
问题是我的"标签"是一个数字向量,但我需要一个因素.所以我改变了
labels <- factor(labels)
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并且roc正常工作
感谢您的专注时间