I would like to apply operators stored in a vector operatorsUsed to series1 and series2 of the data frame df:
operatorsUsed = c('==', '>=', '<=')
series1 = 1:5
series2 = c(1, 3, 2, 4, 5)
df = data.frame(series1,
series2,
stringsAsFactors = FALSE)
I tried combining the parse() and eval() function:
nbrOperators = length(operatorsUsed)
for (j in 1:nbrOperators){
a = df[eval(parse(text = paste0(df$series1, operatorsUsed[j], df$series2))),]
tableCreated = paste0('b', j)
assign(tableCreated, a)
}
But this doesn't work. With parse, I obtain for e.g. j=1
expression(1==1, 2==3, 3==2, 4==4, 5==5)
Which looks promising but then applying eval yields
[1] TRUE
Rather than the looked for
[1] TRUE FALSE FALSE TRUE TRUE
Is there away I can apply operators stored in a vector as text?
We can use lapply with get
lapply(operatorsUsed, function(op) get(op)(df$series1, df$series2))
#[[1]]
#[1] TRUE FALSE FALSE TRUE TRUE
#[[2]]
#[1] TRUE FALSE TRUE TRUE TRUE
#[[3]]
#[1] TRUE TRUE FALSE TRUE TRUE
as #rawr mentioned in the comments, we can also use match.fun(op) instead of get(op) in the lapply
Related
vector_1 = c(4,3,5,1,2)
vector_2 = c(3,1)
output:
[1] FALSE TRUE FALSE TRUE FALSE
how do I get the output just by using basic operators/loops without using the operator %in% or any functions in R?
See match.fun(`%in%`)
match(vector_1,vector_2, nomatch = 0) > 0
Without "functions" is a bit vague, since virtually anything in R is a function. Probably that's an assignment and a for loop is wanted.
res <- logical(length(vector_1))
for (i in seq_along(vector_1)) {
for (j in seq_along(vector_2)) {
if (vector_1[i] == vector_2[j])
res[i] <- TRUE
}
}
res
# [1] FALSE TRUE FALSE TRUE FALSE
However, that's not very R-ish where you rather want to do something like
apply(outer(vector_1, vector_2, `==`), 1, \(x) x[which.max(x)])
# [1] FALSE TRUE FALSE TRUE FALSE
Data:
vector_1 <- c(4, 3, 5, 1, 2)
vector_2 <- c(3, 1)
One way with sapply() -
sapply(vector_1, function(x) any(x == vector_2))
[1] FALSE TRUE FALSE TRUE FALSE
This is weird.
apply( matrix(c(1,NA,2,3,NA,NA,2,4),ncol = 2), 1, function(x) identical(x[1], x[2]) )
#[1] FALSE TRUE TRUE FALSE
apply( data.frame(a = c(1,NA,2,3),b = c(NA,NA,2,4)), 1, function(x) identical(x[1], x[2]) )
#[1] FALSE FALSE FALSE FALSE
apply( as.matrix(data.frame(a = c(1,NA,2,3),b = c(NA,NA,2,4))), 1, function(x) identical(x[1], x[2]) )
#[1] FALSE FALSE FALSE FALSE
This is due to the names attribute as indicated below by joran. I can obtain the result I expected by:
apply( data.frame(a = c(1,NA,2,3),b = c(NA,NA,2,4)), 1, function(x) identical(unname(x[1]), unname(x[2])) )
or:
apply( data.frame(a = c(1,NA,2,3),b = c(NA,NA,2,4)), 1, function(x) identical(x[[1]], x[[2]]) )
Is there a more natural way to approach this? It would seem that there should be an option to ignore attributes, like in all.equal().
Probably
mapply(identical, x$a, x$b)
#[1] FALSE TRUE TRUE FALSE
where x is a data frame.
As an aside, using apply with a data frame is almost always a mistake. It will coerce the data frame to a matrix which often leads to unexpected results.
how can I check to see if vector exists inside a matrix. The vector will be of size 2. I have an approach but I would like something vectorized/faster.
dim(m)
[1] 30 2
x = c(1, -2)
for(j in 1:nrow(m)){
if ( isTRUE(as.vector(x[1]) == as.vector(m[j,1])) && as.vector(x[2] == as.vector(m[j,2]) )) {
print(TRUE)
}
}
note, x=c(1, -2) is not the same as -2, 1 in the matrix.
If we are comparing the rows of the matrix ('m') with 'x' having the same length as the number of columns of 'm', we can replicate 'x' (x[col(m)]) to make the lengths same, compare (!=), get the rowSums. If the sum is 0 for a particular row, it means that all the values in the vector matches that row of 'm'. Negate (!) to convert 0 to TRUE and all other values as FALSE.
indx1 <- !rowSums(m!=x[col(m)])
Or if we need a solution using apply, we can use identical
indx2 <- apply(m, 1, identical, y=x)
identical(indx1, indx2)
#[1] TRUE
If this to find only a single TRUE/FALSE, we can wrap any to 'indx1' or 'indx2'.
data
x <- c(1, -2)
set.seed(24)
m <- matrix(sample(c(1,-2,3,4), 30*2, replace=TRUE), ncol=2)
Try
m<-matrix(rnorm(60),30)
x<-m[8,]
m[9,]<-c(x[2],x[1]) # to prove 1,-2 not same -2,1
apply(m,1,function(n,x) all(n==x),x=x)
[1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
[24] FALSE FALSE FALSE FALSE FALSE FALSE FALSE
if you need just one T/F use any() you
any(apply(m,1,function(n,x) all(n==x),x=x))
[1] TRUE
if run this code with akrun's data
x <- c(1, -2)
set.seed(24)
m <- matrix(sample(c(1,-2,3,4), 30*2, replace=TRUE), ncol=2)
any(apply(m,1,function(n,x) all(n==x),x=x))
[1] TRUE
I have a matrix A,
A = as.matrix(data.frame(col1 = c(1,1,2,3,1,2), col2 = c(-1,-1,-2,-3,-1,-2), col3 = c(2,6,1,3,2,4)))
And I have a vector v,
v = c(-1, 2)
How can I get a vector of TRUE/FALSE that compares the last two columns of the matrix and returns TRUE if the last two columns match the vector, or false if they don't?
I.e., If I try,
A[,c(2:3)] == v
I obtain,
col2 col3
[1,] TRUE FALSE
[2,] FALSE FALSE
[3,] FALSE FALSE
[4,] FALSE FALSE
[5,] TRUE FALSE
[6,] FALSE FALSE
Which is not what I want, I want both columns to be the same as vector v, more like,
result = c(TRUE, FALSE, FALSE, FALSE, TRUE, FALSE)
Since the first, and 5th rows match the vector v entirely.
Here's a simple alternative
> apply(A[, 2:3], 1, function(x) all(x==v))
[1] TRUE FALSE FALSE FALSE TRUE FALSE
Ooops by looking into R mailing list I found an answer: https://stat.ethz.ch/pipermail/r-help/2010-September/254096.html,
check.equal <- function(x, y)
{
isTRUE(all.equal(y, x, check.attributes=FALSE))
}
result = apply(A[,c(2:3)], 1, check.equal, y=v)
Not sure I need to define a function and do all that, maybe there are easier ways to do it.
Here's another straightforward option:
which(duplicated(rbind(A[, 2:3], v), fromLast=TRUE))
# [1] 1 5
results <- rep(FALSE, nrow(A))
results[which(duplicated(rbind(A[, 2:3], v), fromLast=TRUE))] <- TRUE
results
# [1] TRUE FALSE FALSE FALSE TRUE FALSE
Alternatively, as one line:
duplicated(rbind(A[, 2:3], v), fromLast=TRUE)[-(nrow(A)+1)]
# [1] TRUE FALSE FALSE FALSE TRUE FALSE
A dirty one:
result <- c()
for(n in 1:nrow(A)){result[n] <-(sum(A[n,-1]==v)==2)}
> result
[1] TRUE FALSE FALSE FALSE TRUE FALSE
I have a logical vector, for which I wish to insert new elements at particular indexes. I've come up with a clumsy solution below, but is there a neater way?
probes <- rep(TRUE, 15)
ind <- c(5, 10)
probes.2 <- logical(length(probes)+length(ind))
probes.ind <- ind + 1:length(ind)
probes.original <- (1:length(probes.2))[-probes.ind]
probes.2[probes.ind] <- FALSE
probes.2[probes.original] <- probes
print(probes)
gives
[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
and
print(probes.2)
gives
[1] TRUE TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE TRUE FALSE
[13] TRUE TRUE TRUE TRUE TRUE
So it works but is ugly looking - any suggestions?
These are all very creative approaches. I think working with indexes is definitely the way to go (Marek's solution is very nice).
I would just mention that there is a function to do roughly that: append().
probes <- rep(TRUE, 15)
probes <- append(probes, FALSE, after=5)
probes <- append(probes, FALSE, after=11)
Or you could do this recursively with your indexes (you need to grow the "after" value on each iteration):
probes <- rep(TRUE, 15)
ind <- c(5, 10)
for(i in 0:(length(ind)-1))
probes <- append(probes, FALSE, after=(ind[i+1]+i))
Incidentally, this question was also previously asked on R-Help. As Barry says:
"Actually I'd say there were no ways of doing this, since I dont think you can actually insert into a vector - you have to create a new vector that produces the illusion of insertion!"
You can do some magic with indexes:
First create vector with output values:
probs <- rep(TRUE, 15)
ind <- c(5, 10)
val <- c( probs, rep(FALSE,length(ind)) )
# > val
# [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
# [13] TRUE TRUE TRUE FALSE FALSE
Now trick. Each old element gets rank, each new element gets half-rank
id <- c( seq_along(probs), ind+0.5 )
# > id
# [1] 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0
# [16] 5.5 10.5
Then use order to sort in proper order:
val[order(id)]
# [1] TRUE TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE TRUE FALSE
# [13] TRUE TRUE TRUE TRUE TRUE
probes <- rep(TRUE, 1000000)
ind <- c(50:100)
val <- rep(FALSE,length(ind))
new.probes <- vector(mode="logical",length(probes)+length(val))
new.probes[-ind] <- probes
new.probes[ind] <- val
Some timings:
My method
user system elapsed
0.03 0.00 0.03
Marek method
user system elapsed
0.18 0.00 0.18
R append with for loop
user system elapsed
1.61 0.48 2.10
How about this:
> probes <- rep(TRUE, 15)
> ind <- c(5, 10)
> probes.ind <- rep(NA, length(probes))
> probes.ind[ind] <- FALSE
> new.probes <- as.vector(rbind(probes, probes.ind))
> new.probes <- new.probes[!is.na(new.probes)]
> new.probes
[1] TRUE TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE TRUE FALSE
[13] TRUE TRUE TRUE TRUE TRUE
That is sorta tricky. Here's one way. It iterates over the list, inserting each time, so it's not too efficient.
probes <- rep(TRUE, 15)
probes.ind <- ind + 0:(length(ind)-1)
for (i in probes.ind) {
probes <- c(probes[1:i], FALSE, probes[(i+1):length(probes)])
}
> probes
[1] TRUE TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE TRUE FALSE
[13] TRUE TRUE TRUE TRUE TRUE
This should even work if ind has repeated elements, although ind does need to be sorted for the probes.ind construction to work.
Or you can do it using the insertRow function from the miscTools package.
probes <- rep(TRUE, 15)
ind <- c(5,10)
for (i in ind){
probes <- as.vector(insertRow(as.matrix(probes), i, FALSE))
}
I came up with a good answer that's easy to understand and fairly fast to run, building off Wojciech's answer above. I'll adapt the method for the example here, but it can be easily generalized to pretty much any data type for an arbitrary pattern of missing points (shown below).
probes <- rep(TRUE, 15)
ind <- c(5,10)
probes.final <- rep(FALSE, length(probes)+length(ind))
probes.final[-ind] <- probes
The data I needed this for is sampled at a regular interval, but many samples are thrown out, and the resulting data file only includes the timestamps and measurements for those retained. I needed to produce a vector containing all the timestamps and a data vector with NAs inserted for timestamps that were tossed. I used the "not in" function stolen from here to make it a bit simpler.
`%notin%` <- Negate(`%in%`)
dat <- rnorm(50000) # Data given
times <- seq(from=554.3, by=0.1, length.out=70000] # "Original" time stamps
times <- times[-sample(2:69999, 20000)] # "Given" times with arbitrary points missing from interior
times.final <- seq(from=times[1], to=times[length(times)], by=0.1)
na.ind <- which(times.final %notin% times)
dat.final <- rep(NA, length(times.final))
dat.final[-na.ind] <- dat
Um, hi, I had the same doubt, but I couldn't understand what people had answered, because I'm still learning the language. So I tried make my own and I suppose it works! I created a vector and I wanted to insert the value 100 after the 3rd, 5th and 6th indexes. This is what I wrote.
vector <- c(0:9)
indexes <- c(6, 3, 5)
indexes <- indexes[order(indexes)]
i <- 1
j <- 0
while(i <= length(indexes)){
vector <- append(vector, 100, after = indexes[i] + j)
i <-i + 1
j <- j + 1
}
vector
The vector "indexes" must be in ascending order for this to work. This is why I put them in order at the third line.
The variable "j" is necessary because at each iteration, the length of the new vector increases and the original values are moved.
In the case you wish to insert the new value next to each other, simply repeat the number of the index. For instance, by assigning indexes <- c(3, 5, 5, 5, 6), you should get vector == 0 1 2 100 3 4 100 100 100 5 100 6 7 8 9