Removal of constant columns in R
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
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]
The janitor library has the comment remove_constant that can help delete constant columns.
Let's create a synthesis data for illustration:
library(janitor)
test_dat <- data.frame(A=1, B=1:10, C= LETTERS[1:10])
test_dat
This is the test_dat
> test_dat
A B C
1 1 1 A
2 1 2 B
3 1 3 C
4 1 4 D
5 1 5 E
6 1 6 F
7 1 7 G
8 1 8 H
9 1 9 I
10 1 10 J
then the comment remove_constant can help delete the constant column
remove_constant(test_dat)
remove_constant(test_dat, na.rm= TRUE)
Using the above two comments, we will get:
B C
1 1 A
2 2 B
3 3 C
4 4 D
5 5 E
6 6 F
7 7 G
8 8 H
9 9 I
10 10 J
NOTE: use the argument na.rm = TRUE to make sure that any column having one value and NA will also be deleted. For example,
test_dat_with_NA <- data.frame(A=c(1, NA), B=1:10, C= LETTERS[1:10])
test_dat_with_NA
the test_dat_with_NA we get:
A B C
1 1 1 A
2 NA 2 B
3 1 3 C
4 NA 4 D
5 1 5 E
6 NA 6 F
7 1 7 G
8 NA 8 H
9 1 9 I
10 NA 10 J
then the comment
remove_constant(test_dat_with_NA)
could not delete the column A
A B C
1 1 1 A
2 NA 2 B
3 1 3 C
4 NA 4 D
5 1 5 E
6 NA 6 F
7 1 7 G
8 NA 8 H
9 1 9 I
10 NA 10 J
while the comment
remove_constant(test_dat_with_NA, na.rm= TRUE)
could delete the column A with only value 1 and NA:
B C
1 1 A
2 2 B
3 3 C
4 4 D
5 5 E
6 6 F
7 7 G
8 8 H
9 9 I
10 10 J
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 example 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[,-which_are_constant(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:which_are_constant is 10 times faster than the other approaches.
Plus the more rows you have the more interesting it is to use.