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how to drop columns by passing variable name with dplyr?
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I'm interested in simplifying the way that I can remove columns with dplyr (version >= 0.7). Let's say that I have a character vector of names.
drop <- c("disp", "drat", "gear", "am")
Selecting Columns
With the current version version of dplyr, you can perform a selection with:
dplyr::select(mtcars, !! rlang::quo(drop))
Or even easier with base R:
mtcars[, drop]
Removing Columns
Removing columns names is another matter. We could use each unquoted column name to remove them:
dplyr::select(mtcars, -disp, -drat, -gear, -am)
But, if you have a data.frame with several hundred columns, this isn't a great solution. The best solution I know of is to use:
dplyr::select(mtcars, -which(names(mtcars) %in% drop))
which is fairly simple and works for both dplyr and base R. However, I wonder if there's an approach which doesn't involve finding the integer positions for each column name in the data.frame.
Use modify_atand set columns to NULL which will remove them:
mtcars %>% modify_at(drop,~NULL)
# mpg cyl hp wt qsec vs carb
# Mazda RX4 21.0 6 110 2.620 16.46 0 4
# Mazda RX4 Wag 21.0 6 110 2.875 17.02 0 4
# Datsun 710 22.8 4 93 2.320 18.61 1 1
# Hornet 4 Drive 21.4 6 110 3.215 19.44 1 1
# Hornet Sportabout 18.7 8 175 3.440 17.02 0 2
# Valiant 18.1 6 105 3.460 20.22 1 1
# ...
Closer to what you were trying, you could have tried magrittr::extract instead of dplyr::select
extract(mtcars,!names(mtcars) %in% drop) # same output
You can use -one_of(drop) with select:
drop <- c("disp", "drat", "gear", "am")
select(mtcars, -one_of(drop)) %>% names()
# [1] "mpg" "cyl" "hp" "wt" "qsec" "vs" "carb"
one_of evaluates the column names in character vector to integers, similar to which(... %in% ...) does:
one_of(drop, vars = names(mtcars))
# [1] 3 5 10 9
which(names(mtcars) %in% drop)
# [1] 3 5 9 10
Related
I have a dataframe that must have a specific layout. Is there a way for me to make R reject any command I attempt that would change the number or names of the columns?
It is easy to check the format of the data table manually, but I have found no way to make R do it for me automatically every time I execute a piece of code.
regards
This doesn’t offer the level of foolproof safety I think you’re looking for (hard to know without more details), but you could define a function operator that yields modified functions that error if changes to columns are detected:
same_cols <- function(fn) {
function(.data, ...) {
out <- fn(.data, ...)
stopifnot(identical(sort(names(.data)), sort(names(out))))
out
}
}
For example, you could create modified versions of dplyr functions:
library(dplyr)
my_mutate <- same_cols(mutate)
my_summarize <- same_cols(summarize)
which work as usual if columns are preserved:
mtcars %>%
my_mutate(mpg = mpg / 2) %>%
head()
# mpg cyl disp hp drat wt qsec vs am gear carb
# Mazda RX4 10.50 6 160 110 3.90 2.620 16.46 0 1 4 4
# Mazda RX4 Wag 10.50 6 160 110 3.90 2.875 17.02 0 1 4 4
# Datsun 710 11.40 4 108 93 3.85 2.320 18.61 1 1 4 1
# Hornet 4 Drive 10.70 6 258 110 3.08 3.215 19.44 1 0 3 1
# Hornet Sportabout 9.35 8 360 175 3.15 3.440 17.02 0 0 3 2
# Valiant 9.05 6 225 105 2.76 3.460 20.22 1 0 3 1
mtcars %>%
my_summarize(across(everything(), mean))
# mpg cyl disp hp drat wt qsec vs am
# 1 20.09062 6.1875 230.7219 146.6875 3.596563 3.21725 17.84875 0.4375 0.40625
# gear carb
# 1 3.6875 2.8125
But throw errors if changes to columns are made:
mtcars %>%
my_mutate(mpg2 = mpg / 2)
# Error in my_mutate(., mpg2 = mpg/2) :
# identical(sort(names(.data)), sort(names(out))) is not TRUE
mtcars %>%
my_summarize(mpg = mean(mpg))
# Error in my_summarize(., mpg = mean(mpg)) :
# identical(sort(names(.data)), sort(names(out))) is not TRUE
You mention the names and columns need to be the same, also realize that with data.table also names are updated by reference. See the example below.
foo <- data.table(
x = letters[1:5],
y = LETTERS[1:5]
)
colnames <- names(foo)
colnames
# [1] "x" "y"
setnames(foo, colnames, c("a", "b"))
foo[, z := "oops"]
colnames
# [1] "a" "b" "z"
identical(colnames, names(foo))
# [1] TRUE
To check that both the columns and names are unalterated (and in same order here) you can take right away a copy of the names. And after each code run, you can check the current names with the copied names.
foo <- data.table(
x = letters[1:5],
y = LETTERS[1:5]
)
colnames <- copy(names(foo))
setnames(foo, colnames, c("a", "b"))
foo[, z := "oops"]
identical(colnames, names(foo))
[1] FALSE
colnames
# [1] "x" "y"
names(foo)
# [1] "a" "b" "z"
I want to order a data.table by using a set of predefined names available in a list.
For example:
library(data.table)
dt <- as.data.table(mtcars)
list_name <-c("mpg", "disp", "xyz")
#Order columns
setcolorder(dt, list_name) #requirement: if "xyz" column doesn't exist it should ignore and take the rest
The use case case is that there are multiple data.tables that are getting created and all of them have column names from a list of names. There can be missing column names in some data but the data needs to be ordered as per a list.
output:
dt
disp wt mpg cyl hp drat qsec vs am gear carb
1: 160.0 2.620 21.0 6 110 3.90 16.46 0 1 4 4
2: 160.0 2.875 21.0 6 110 3.90 17.02 0 1 4 4
3: 108.0 2.320 22.8 4 93 3.85 18.61 1 1 4 1
An option is to load all of them in a list and then use setcolorder by looping over the list with lapply and use intersect on the names of the dataset while ordering
lst1 <- list(dt, dt)
lst1 <- lapply(lst1, function(x) setcolorder(x, intersect(list_name, names(x)))
If we need to reuse, create a function
f1 <- function(dat, nm1) {
setcolorder(dat, intersect(nm1, names(dat)))
}
f1(dt, list_name)
f1(dt2, list_name)
Within the data.table package in R, is there a way in order to use a character vector to be assigned within the by argument of the calculation?
Here is an example of what would be the desired output from this using mtcars:
mtcars <- data.table(mtcars)
ColSelect <- 'cyl' # One Column Option
mtcars[,.( AveMpg = mean(mpg)), by = .(ColSelect)] # Doesn't work
# Desired Output
cyl AveMpg
1: 6 19.74286
2: 4 26.66364
3: 8 15.10000
I know that this is possible to use assigning column names in j by enclosing the vector around brackets.
ColSelect <- 'AveMpg' # Column to be assigned for average mpg value
mtcars[,(ColSelect):= mean(mpg), by = .(cyl)]
head(mtcars)
mpg cyl disp hp drat wt qsec vs am gear carb AveMpg
1: 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 19.74286
2: 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 19.74286
3: 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 26.66364
4: 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1 19.74286
5: 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2 15.10000
6: 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1 19.74286
Is there a suggestion as to what to put in the by argument in order to achieve this?
From ?data.table in the by section it says that by accepts:
a single character string containing comma separated column names (where spaces are significant since column names may contain spaces
even at the start or end): e.g., DT[, sum(a), by="x,y,z"]
a character vector of column names: e.g., DT[, sum(a), by=c("x", "y")]
So yes, you can use the answer in #cccmir's response. You can also use c() as #akrun mentioned, but that seems slightly extraneous unless you want multiple columns.
The reason you cannot use .() syntax is that in data.table .() is an alias for list(). And according to the same help for by the list() syntax requires an expression of column names - not a character string.
Going off the examples in the by help if you wanted to use multiple variables and pass the names as characters you could do:
mtcars[,.( AveMpg = mean(mpg)), by = "cyl,am"]
mtcars[,.( AveMpg = mean(mpg)), by = c("cyl","am")]
try to use it like this
mtcars <- data.table(mtcars)
ColSelect <- 'cyl' # One Column Option
mtcars[, AveMpg := mean(mpg), by = ColSelect] # Should work
I have a pretty big data.table (500 x 2000), and I need to find out if any of the columns are duplicates, i.e., have the same values for all rows. Is there a way to efficiently do this within the data.table structure?
I have tried a naive two loop approach with all(col1 == col2) for each pair of columns, but it takes too long. I have also tried converting it to a data.frame and using the above approach, and it still takes quite a long time.
My current solution is to convert the data.table to a matrix and use the apply() function as:
similarity.matrix <- apply(m, 2, function(x) colSums(x == m)))/nrow(m)
However, the approach forces the modes of all elements to be the same, and I'd rather not have that happen. What other options do I have?
Here is a sample construction for the data.table:
m = matrix(sample(1:10, size=1000000, replace=TRUE), nrow=500, ncol=2000)
DF = as.data.frame(m)
DT = as.data.table(m)
Following the suggestion of #Haboryme*, you can do this using duplicated to find any duplicated vectors. duplicated usually works rowwise, but you can transpose it with t() just for finding the duplicates.
DF <- DF[ , which( !duplicated( t( DF ) ) ) ]
With a data.table, you may need to add with = FALSE (I think this depends on the version of data.table you're using).
DT <- DT[ , which( !duplicated( t( DT ) ) ), with = FALSE ]
*#Haboryme, if you were going to turn your comment into an answer, please do and I'll remove this one.
Here's a different approach, where you hash each column first and then call duplicated.
library(digest)
dups <- duplicated(sapply(DF, digest))
DF <- DF[,which(!dups)]
Depending on your data this might be a faster way.
I am using mtcars for a reproducible result:
library(data.table)
library(digest)
# Create data
data <- as.data.table(mtcars)
data[, car.name := rownames(mtcars)]
data[, car.name.dup := car.name] # create a duplicated row
data[, car.name.not.dup := car.name] # create a second duplicated row...
data[1, car.name.not.dup := "Moon walker"] # ... but change a value so that it is no longer a duplicated column
data contains now:
> head(data)
mpg cyl disp hp drat wt qsec vs am gear carb car.name car.name.dup car.name.not.dup
1: 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 Mazda RX4 Mazda RX4 Moon walker
2: 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 Mazda RX4 Wag Mazda RX4 Wag Mazda RX4 Wag
3: 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 Datsun 710 Datsun 710 Datsun 710
4: 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1 Hornet 4 Drive Hornet 4 Drive Hornet 4 Drive
5: 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2 Hornet Sportabout Hornet Sportabout Hornet Sportabout
6: 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1 Valiant Valiant Valiant
Now find the duplicated colums:
# create a vector with the checksum for each column (and keep the column names as row names)
col.checksums <- sapply(data, function(x) digest(x, "md5"), USE.NAMES = T)
# make a data table with one row per column name and hash value
dup.cols <- data.table(col.name = names(col.checksums), hash.value = col.checksums)
# self join using the hash values and filter out all column name pairs that were joined to themselves
dup.cols[dup.cols,, on = "hash.value"][col.name != i.col.name,]
Results in:
col.name hash.value i.col.name
1: car.name.dup 58fed3da6bbae3976b5a0fd97840591d car.name
2: car.name 58fed3da6bbae3976b5a0fd97840591d car.name.dup
Note: The result still contains both directions (col1 == col2 and col2 == col1) and should be deduplicated ;-)
If I have a sample data frame like mtcars, and I want to find the difference between mtcars$qsec for all rows, I can do diff(mtcars$qsec). But is there a simple way to make diff(mtcars$qsec) a new column in the original mtcars data frame? I'm finding it difficult because there's one less row in diff(mtcars$qsec) than the rest of mtcars.
> head(mtcars,3)
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
Here are two approaches. Both put an NA in the first row of diff_qsec and put diff(qsec) in the remaining rows:
library(dplyr)
mtcars %>% mutate(diff_qsec = qsec - lag(qsec)) # dplyr has its own version of lag
transform(mtcars, diff_qsec = c(NA, diff(qsec)))
Also, on the general issue of padding see: How can I pad a vector with NA from the front?
You could use the base function within() like so:
mtcars <- within(mtcars, difference <- c(NA,diff(qsec)))
This creates a column called "difference" with the first element NA and the rest calculated by diff(qsec).
You could create more columns at the same time by wrapping commands in {}, such as:
mtcars <- within(mtcars, {difference <- c(NA,diff(qsec))
multiple <- qsec*2})
Note that you must use <- for the assignment and not =.