[Solved] Removal of constant columns in R

I was using the prcomp function when I received this error

Error in prcomp.default(x, ...) : 
cannot rescale a constant/zero column to unit variance

I know I can scan my data manually but is there any function or command in R that can help me remove these constant variables?
I know this is a very simple task, but I have never been across any function that does this.

Thanks,

Enquirer: Error404

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Solution #1:

The problem here is that your column variance is equal to zero. You can check which column of a data frame is constant this way, for example :

df <- data.frame(x=1:5, y=rep(1,5))
df
#   x y
# 1 1 1
# 2 2 1
# 3 3 1
# 4 4 1
# 5 5 1

# Supply names of columns that have 0 variance
names(df[, sapply(df, function(v) var(v, na.rm=TRUE)==0)])
# [1] "y" 

So if you want to exclude these columns, you can use :

df[,sapply(df, function(v) var(v, na.rm=TRUE)!=0)]

EDIT : In fact it is simpler to use apply instead. Something like this :

df[,apply(df, 2, var, na.rm=TRUE) != 0]
Respondent: juba

Solution #2:

I guess this Q&A is a popular Google search result but the answer is a bit slow for a large matrix, plus I do not have enough reputation to comment on the first answer. Therefore I post a new answer to the question.

For each column of a large matrix, checking whether the maximum is equal to the minimum is sufficient.

df[,!apply(df, MARGIN = 2, function(x) max(x, na.rm = TRUE) == min(x, na.rm = TRUE))]

This is the test. More than 90% of the time is reduced compared to the first answer. It is also faster than the answer from the second comment on the question.

ncol = 1000000
nrow = 10
df <- matrix(sample(1:(ncol*nrow),ncol*nrow,replace = FALSE), ncol = ncol)
df[,sample(1:ncol,70,replace = FALSE)] <- rep(1,times = nrow) # df is a large matrix

time1 <- system.time(df1 <- df[,apply(df, 2, var, na.rm=TRUE) != 0]) # the first method
time2 <- system.time(df2 <- df[,!apply(df, MARGIN = 2, function(x) max(x, na.rm = TRUE) == min(x, na.rm = TRUE))]) # my method
time3 <- system.time(df3 <- df[,apply(df, 2, function(col) { length(unique(col)) > 1 })]) # Keith's method

time1
#   user  system elapsed 
# 22.267   0.194  22.626 
time2
#   user  system elapsed 
#  2.073   0.077   2.155 
time3
#   user  system elapsed 
#  6.702   0.060   6.790
all.equal(df1, df2)
# [1] TRUE
all.equal(df3, df2)
# [1] TRUE
Respondent: raymkchow

Solution #3:

Since this Q&A is a popular Google search result but the answer is a bit slow for a large matrix and @raymkchow version is slow with NAs i propose a new version using exponential search and data.table power.

This a function I implemented in dataPreparation package.

First build an exemple data.table, with more lines than columns (which is usually the case) and 10% of NAs

ncol = 1000
nrow = 100000
df <- matrix(sample(1:(ncol*nrow),ncol*nrow,replace = FALSE), ncol = ncol)
df <- apply (df, 2, function(x) {x[sample( c(1:nrow), floor(nrow/10))] <- NA; x} ) # Add 10% of NAs
df[,sample(1:ncol,70,replace = FALSE)] <- rep(1,times = nrow) # df is a large matrix
df <- as.data.table(df)

Then benchmark all approaches:

time1 <- system.time(df1 <- df[,apply(df, 2, var, na.rm=TRUE) != 0, with = F]) # the first method
time2 <- system.time(df2 <- df[,!apply(df, MARGIN = 2, function(x) max(x, na.rm = TRUE) == min(x, na.rm = TRUE)), with = F]) # raymkchow
time3 <- system.time(df3 <- df[,apply(df, 2, function(col) { length(unique(col)) > 1 }), with = F]) # Keith's method
time4 <- system.time(df4 <- df[,-whichAreConstant(df, verbose=FALSE)]) # My method

The results are the following:

time1 # Variance approch
#   user  system elapsed 
#   2.55    1.45    4.07
time2 # Min = max approach
#   user  system elapsed 
#  2.72      1.5    4.22
time3 # length(unique()) approach
#   user  system elapsed 
#    6.7    2.75    9.53
time4 # Exponential search approach
#   user  system elapsed 
#   0.39    0.07    0.45
all.equal(df1, df2)
# [1] TRUE
all.equal(df3, df2)
# [1] TRUE
all.equal(df4, df2)
# [1] TRUE

dataPreparation:whichAreConstant is 10 times faster than the other approachs.

Plus the more rows you have the more intersting it is to use.

Respondent: Emmanuel-Lin

Solution #4:

If you are after a dplyr solution that returns the non-constant variables in a df, I’d recommend the following. Optionally, you can add %>% colnames() if the column names are desired:

library(dplyr)
df <- data.frame(x = 1:5, y = rep(1,5))
# returns dataframe
var_df <- df %>%
  select_if(function(v) var(v, na.rm=TRUE) != 0)  
var_df %>% colnames() # returns column names
Respondent: Aaron Clark

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

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