Error in do_one(nmeth) : NA/NaN/Inf in foreign function call (arg 1) [closed]

Question

I have a data table ("norm") containing numeric - at least to what I can see - normalized values of the following form:

When I am executing

k <- kmeans(norm,center=3)

I am receving the following error:

Error in do_one(nmeth) : NA/NaN/Inf in foreign function call (arg 1)

Can you help me? Thank you!

Businesswoman

Have you checked if there's NaN/NA/Inf in your data? You can check using is.na() and is.finite() functions – Namhoi

Yes, there is plenty of NAs in my file sheet, but I thought that shouldnt be a problem?! is.finite() returns a lot of TRUEs but also some FALSEs. How can i fix this? – Businesswoman

You would have to remove the NA/Inf/NaN values from your data. See "missing value imputation" methods for details. One simple method is replacing them by row/column mean values. – Namhoi

@UjjwalKumar Thank you! – Businesswoman

Answer 1

kmeans cannot handle data that has NA values.

The mean and variance are then no longer well defined, and you don't know anymore which center is closest.

Lionhearted

Kmeans cannot handle data that has NA, INF and CONTANTS. – Whittemore

Answer 2

Error in do_one(nmeth) : NA/NaN/Inf in foreign function call (arg 1)

This error occurs also due to non numeric values present in the table.

Heterotopia

Answer 3

all of you all who are having " Error in do_one(nmeth) : NA/NaN/Inf in foreign function call (arg 1)" problem instead of

results <- kmeans(iris.features,3)
results

write the following and please be careful of the case in iris write whatever you have used in the beginning

results <- kmeans(na.omit(irisa.features),3) # this helps in omitting NA 
results

Blubberhead

Answer 4

For error stating:

Error in do_one(nmeth) : NA/NaN/Inf in foreign function call (arg 1)

the dataset you have posted above contains scaled entries, the reason must be that you have NA values in your dataset, hence omit them by the following code.

km_cluster <- kmeans(na.omit(MyData), 3)
km_cluster
km_cluster$withinss
km_cluster$tot.withinss/km_cluster$betweenss

Presentiment

Answer 5

You'll also get this error if you pass a grouped dataframe to the kmeans function. I prepared my data using dplyr and had to close with ungroup() to use kmeans after.

Fronia

Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. – Melee

Answer 6

This error can also be caused by scaling data with a column of values that do not vary (ie, are constant). This is because the standard deviation of those columns equals zero, and scaling produces NaN values. The code below fixes this programmatically by excluding columns with no variation.

# Data
df=copy(mtcars)
df$Constant=2

# Kmeans
fit=kmeans(scale(df),centers=2) # Error

# Identify columns with SD=0
tmp=sapply(df,sd)
tmp=names(tmp)[tmp==0]

# Kmeans
tmp1=setdiff(names(df),tmp)
fit=kmeans(scale(df[,tmp1]),centers=2) # Works

Helmer