SVM 的测试集中预测值的数量是否正确？

Question

所以我有一个 nrow = 218 的数据集，我正在通过 [this][https://iamnagdev.com/2018/01/02/sound-analytics-in-r-for-animal-sound-classification-using-vector-machine/] example [git here][https://github.com/nagdevAmruthnath].我已将我的数据分为训练 (nrow = 163; ~75%) 和测试 (nrow = 55; ~25%)。

当我到达“pred

一些假数据：

featuredata_all <- matrix(rexp(218, rate=.1), ncol=23)

部分代码：

library(data.table)

pt1 <- scale(featuredata_all[,1:22],center=T)
pt2 <- as.character(featuredata_all[,23]) #since the label is a string I kept it separate 

ft<-cbind.data.frame(pt1,pt2) #to preserve the label in text
colnames(ft)[23]<- "Cluster"

## 75% of the sample size
smp_size <- floor(0.75 * nrow(ft))

## set the seed to make your partition reproducible
set.seed(123)
train_ind <- sample(seq_len(nrow(ft)), size = smp_size)

train <- ft[train_ind,1:22] #163 reads
test  <- ft[-train_ind,1:22] #55 reads

trainlabel<- ft[train_ind,23] #163 labels
testlabel <- ft[-train_ind,23] #55 labels

#ftID <- cbind(ft, seq.int(nrow(ft))
#colnames(ftID)[24]<- "RowID"
#ftIDtestrows <- ftID[-train_ind,24]

#Support Vector Machine for classification
model_svm <- svm(trainlabel ~ as.matrix(train) )
summary(model_svm)

#Use the predictions on the data
# ---------------- This is where the question is ---------------- #
pred <- predict(model_svm, test)
# ----------------------------------------------------------------#

print(confusionMatrix(pred[1:nrow(test)],testlabel))

#ROC and AUC curves and their plots
#-----------------also------------->  was trying to get this to work as pred doesn't naturally end up with the expected 55 nrow from test set
roc.multi<-multiclass.roc(testlabel, as.numeric(pred[1:55])) 
rs <- roc.multi[['rocs']]
plot.roc(rs[[1]])
sapply(2:length(rs),function(i) lines.roc(rs[[i]],col=i)) ```


 [1]: https://iamnagdev.com/2018/01/02/sound-analytics-in-r-for-animal-sound-classification-using-vector-machine/
 [2]: https://github.com/nagdevAmruthnath

Answer 1

好的，我意识到我是在我的训练数据集上训练模型，然后在我的测试集上对其进行测试。我需要先在重新预测训练集时对其进行测试，然后再将其输入测试集。

 summary(model_svm)
#Use the predictions on the data
pred <- predict(model_svm, train)

model_svm <- svm(trainlabel ~ as.matrix(test) )
 summary(model_svm)
#Use the predictions on the data
pred <- predict(model_svm, test)```

Answer 2

我实际上能够使用以下代码获得 55 行的结果。我所做的一些更改是针对pt2 而不是as.character 我将它变成as.factor 而不是pred <- predict(model_svm, test) 到pred <- predict(model_svm, as.matrix(test))。

# load libraries
library(data.table)
library(e1071)

# create dataset with random values
featuredata_all <- matrix(rnorm(23*218), ncol=23)

# scale features
pt1 <- scale(featuredata_all[,1:22],center=T)

# make column as factor
pt2 <- as.factor(ifelse(featuredata_all[,23]>0, 0,1)) #since the label is a string I kept it separate 

# join data (optional)
ft<-cbind.data.frame(pt1,pt2) #to preserve the label in text
colnames(ft)[23]<- "Cluster"

## 75% of the sample size
smp_size <- floor(0.75 * nrow(ft))

## set the seed to make your partition reproducible
set.seed(123)
train_ind <- sample(seq_len(nrow(ft)), size = smp_size)

# split data to train
train <- ft[train_ind,1:22] #163 reads
test  <- ft[-train_ind,1:22] #55 reads
dim(train)
# [1] 163  22

dim(test)
# [1] 55  22

# split data to test
trainlabel<- ft[train_ind,23] #163 labels
testlabel <- ft[-train_ind,23] #55 labels
length(trainlabel)
[1] 163

length(testlabel)
[1] 55

#Support Vector Machine for classification
model_svm <- svm(x= as.matrix(train), y = trainlabel, probability = T)
summary(model_svm)

# Call:
#   svm.default(x = as.matrix(train), y = trainlabel, probability = T)
# 
# 
# Parameters:
#   SVM-Type:  C-classification 
# SVM-Kernel:  radial 
# cost:  1 
# 
# Number of Support Vectors:  159
# 
# ( 78 81 )
# 
# 
# Number of Classes:  2 
# 
# Levels: 
#   0 1

#Use the predictions on the data
# ---------------- This is where the question is ---------------- #
pred <- predict(model_svm, as.matrix(test))
length(pred)
# [1] 55
# ----------------------------------------------------------------#

print(table(pred[1:nrow(test)],testlabel))
#    testlabel
#    0  1
# 0 14 14
# 1 11 16

希望这会有所帮助。