Here is my code:
train_points <- read.table(\"kaggle_train_points.txt\", sep=\"\\t\")
train_labels <- read.table(\"kaggle_train_labels.txt\", sep=\"\\t
I had the same issue in trying to apply knn on breast cancer diagnosis from wisconsin dataset I found that the issue was linked to the fact that cl argument need to be a vector factor (my mistake was to write cl=labels , I thought this was the vector to be predicted it was in fact a data frame of one column ) so the solution was to use the following syntax : knn (train, test,cl=labels$diagnosis,k=21) diagnosis was the header of the one column data frame labels and it worked well Hope this help !
Try converting the data into a dataframe using as.dataframe(). I was having the same problem & afterwards it worked fine:
train_pointsdf <- as.data.frame(train_points)
train_labelsdf <- as.data.frame(train_labels)
test_pointsdf <- as.data.frame(test_points)
Simply set drop = TRUE
while you're excluding cl from dataframe, it causes to remove dimension from an array which have only one level:
cl = train_labels[,1, drop = TRUE]
knn(train_points, test_points, cl, k = 5)
Without access to the data, it's really hard to help. However, I suspect that train_labels
should be a vector. So try
cl = train_labels[,1]
knn(train_points, test_points, cl, k = 5)
Also double check:
dim(train_points)
dim(test_points)
length(cl)
I have recently encountered a very similar issue.
I wanted to give only a single column as a predictor. In such cases, selecting a column, you have to remember about drop argument and set it to FALSE. The knn()
function accepts only matrices or data frames as train and test arguments. Not vectors.
knn(train = trainSet[, 2, drop = FALSE], test = testSet[, 2, drop = FALSE], cl = trainSet$Direction, k = 5)