Suppose I have a data frame such like
dat<-data.frame('0'=c(1,1,0,0,0,0,0,0),
'1'=c(0,0,1,0,1,0,0,0),
'2'=c(0,0,0,1,0,0,1,1),
'3'=c(0,0,0,0,0,1,0,0))
dat
X0 X1 X2 X3
1 1 0 0 0
2 1 0 0 0
3 0 1 0 0
4 0 0 1 0
5 0 1 0 0
6 0 0 0 1
7 0 0 1 0
8 0 0 1 0
I wanted to convert it to a vector like 1,1,2,3,2,4,3,3 where the numbers corresponding the column-th with unit 1. For example, 4 means the col 4th on row number 6th is 1.
Use
max.col(dat)
# [1] 1 1 2 3 2 4 3 3
In base R, we can use apply
apply(dat == 1, 1, which)
#[1] 1 1 2 3 2 4 3 3
Related
I want to create a new column based on some conditions imposed on several columns. For example, here is an example dataset:
a <- data.frame(x=c(1,0,1,0,0), y=c(0,0,0,0,0), z=c(1,1,0,0,0))
a
x y z
1 1 0 1
2 0 0 1
3 1 0 0
4 0 0 0
5 0 0 0
Specifically, if for any particular row 1 is present, then the new column returns 1. If all are 0, then the new column returns 0. So the dataset with the new column will be
x y z w
1 1 0 1 1
2 0 0 1 1
3 1 0 0 1
4 0 0 0 0
5 0 0 0 0
My initial thought was to use %in% but couldn't get the result I want. Thank you for your help!
If your data frame consists of binary values, e.g., only 0 and 1, you can try the code below with rowSums
a$w <- +(rowSums(a)>0)
such that
> a
x y z w
1 1 0 1 1
2 0 0 1 1
3 1 0 0 1
4 0 0 0 0
5 0 0 0 0
We can use rowMaxs from matrixStats
library(matrixStats)
a$w <- rowMaxs(as.matrix(a))
a$w
#[1] 1 1 1 0 0
You can find max of each row :
a$w <- do.call(pmax, a)
a
# x y z w
#1 1 0 1 1
#2 0 0 1 1
#3 1 0 0 1
#4 0 0 0 0
#5 0 0 0 0
which can also be done with apply :
a$w <- apply(a, 1, max)
Suppose I have something like this:
df<-data.frame(group=c(1, 1,2, 2, 2, 4,4,4,4,6,6,6),
binary1=c(1,0,1,0,0,0,0,0,0,0,0,0),
binary2=c(0,1,0,1,0,1,0,0,0,0,1,1),
binary3=c(0,0,0,0,1,0,1,0,0,0,0,0),
binary4=c(0,0,0,0,0,0,0,1,0,0,0,0))
I want to sum along all possible left to right diagonals within groups (i.e group 1, 2 4 and 6) and return the max sum. This is also in a dataframe, so I would like to specify to only sum along binary1-binary4. Anyone know if this is possible?
Here's my desired output:
group binary1 binary2 binary3 binary4 want
1 1 1 0 0 0 2
2 1 0 1 0 0 2
3 2 1 0 0 0 3
4 2 0 1 0 0 3
5 2 0 0 1 0 3
6 4 0 1 0 0 3
7 4 0 0 1 0 3
8 4 0 0 0 1 3
9 4 0 0 0 0 3
10 6 0 0 0 0 1
11 6 0 1 0 0 1
12 6 0 1 0 0 1
I have circled the "diagonals" I would like summed for group 4 in this image as an example:
Here is another solution where we use row and col indices to get all possible combinations of diagonals. Use by to split by group and merge it with original dataframe.
max_diag <- function(x) max(sapply(split(as.matrix(x), row(x) - col(x)), sum))
merge(df, stack(by(df[-1], df$group, max_diag)), by.x = "group", by.y = "ind")
# group binary1 binary2 binary3 binary4 values
#1 1 1 0 0 0 2
#2 1 0 1 0 0 2
#3 2 1 0 0 0 3
#4 2 0 1 0 0 3
#5 2 0 0 1 0 3
#6 4 0 1 0 0 3
#7 4 0 0 1 0 3
#8 4 0 0 0 1 3
#9 4 0 0 0 0 3
#10 6 0 0 0 0 1
#11 6 0 1 0 0 1
#12 6 0 1 0 0 1
You can split the data.frame and sum the diagonal using diag(). Once you have this sum diagonal per group, it's putting them back into the data.frame by calling the group.
Group 4 should be zero? Or am I missing something:
DIAG = by(df[,-1],df$group,function(i)sum(diag(as.matrix(i))))
df$want = DIAG[as.character(df$group)]
If I get your definition correct, we define a function to calculate sum of main diagonal:
main_diag = function(m){
sapply(1:(ncol(m)-1),function(i)sum(diag(m[,i:ncol(m)])))
}
Thanks to #IceCreamToucan for correcting this. Then we consider the max of all main diagonals, and their transpose:
DIAG = by(df[,-1],df$group,function(i){
i = as.matrix(i)
max(main_diag(i),main_diag(t(i)))
})
df$want = DIAG[as.character(df$group)]
group binary1 binary2 binary3 binary4 want
1 1 1 0 0 0 2
2 1 0 1 0 0 2
3 2 1 0 0 0 3
4 2 0 1 0 0 3
5 2 0 0 1 0 3
6 4 0 1 0 0 3
7 4 0 0 1 0 3
8 4 0 0 0 1 3
9 4 0 0 0 0 3
10 6 0 0 0 0 1
11 6 0 1 0 0 1
12 6 0 1 0 0 1
This question already has answers here:
Cumulative sum for positive numbers only [duplicate]
(9 answers)
Closed 6 years ago.
If I have the following vector:
x = c(1,1,1,0,0,0,0,1,1,0,0,1,1,1,0,0,1,1,1,1,0,0,0,0,1,1,1)
how can I calculate the cumulative sum for all of the consecutive 1's, resetting each time I hit a 0?
So, the desired output would look like this:
> y
[1] 1 2 3 0 0 0 0 1 2 0 0 1 2 3 0 0 1 2 3 4 0 0 0 0 1 2 3
This works:
unlist(lapply(rle(x)$lengths, FUN = function(z) 1:z)) * x
# [1] 1 2 3 0 0 0 0 1 2 0 0 1 2 3 0 0 1 2 3 4 0 0 0 0 1 2 3
It relies pretty heavily on your special case of only having 1s and 0s, but for that case it works great! Even better, with #nicola's suggested improvements:
sequence(rle(x)$lengths) * x
# [1] 1 2 3 0 0 0 0 1 2 0 0 1 2 3 0 0 1 2 3 4 0 0 0 0 1 2 3
I read this post about how to split a vector, and use splitAt2 by #Calimo.
So it's like this:
splitAt2 <- function(x, pos) {
out <- list()
pos2 <- c(1, pos, length(x)+1)
for (i in seq_along(pos2[-1])) {
out[[i]] <- x[pos2[i]:(pos2[i+1]-1)]
}
return(out)
}
x = c(1,1,1,0,0,0,0,1,1,0,0,1,1,1,0,0,1,1,1,1,0,0,0,0,1,1,1)
where_split = which(x == 0)
x_split = splitAt2(x, where_split)
unlist(sapply(x_split, cumsum))
# [1] 1 2 3 0 0 0 0 1 2 0 0 1 2 3 0 0 1 2 3 4 0 0 0 0 1 2 3
Here is another option
library(data.table)
ave(x, rleid(x), FUN=seq_along)*x
#[1] 1 2 3 0 0 0 0 1 2 0 0 1 2 3 0 0 1 2 3 4 0 0 0 0 1 2 3
Or without any packages
ave(x, cumsum(c(TRUE, x[-1]!= x[-length(x)])), FUN=seq_along)*x
#[1] 1 2 3 0 0 0 0 1 2 0 0 1 2 3 0 0 1 2 3 4 0 0 0 0 1 2 3
I am trying to replace every row's values in 2 columns with a vector of length 2. It is easier to show you.
First here is a some data.
set.seed(1234)
x<-data.frame(x=sample(c(0:3), 10, replace=T))
x$ab<-0 #column that will be replaced
x$cd<-0 #column that will be replaced
The data looks like this:
x ab cd
1 0 0 0
2 2 0 0
3 2 0 0
4 2 0 0
5 3 0 0
6 2 0 0
7 0 0 0
8 0 0 0
9 2 0 0
10 2 0 0
Every time x=2 or x=3, I want to ab=0 and cd=1.
My attempt is this:
x[with(x, which(x==2|x==3)), c(2:3)] <- c(0,1)
Which does not have the intended results:
x ab cd
1 0 0 0
2 2 0 1
3 2 1 0
4 2 0 1
5 3 1 0
6 2 0 1
7 0 0 0
8 0 0 0
9 2 1 0
10 2 0 1
Can you help me?
The reason it doesn't work as you want is because R stores matrices and arrays in column-major layout. And when you a assign a shorter array to a longer array, R cycles through the shorter array. For example if you have
x<-rep(0,20)
x[1:10]<-c(2,3)
then you end up with
[1] 2 3 2 3 2 3 2 3 2 3 0 0 0 0 0 0 0 0 0 0
What is happening in your case is that the sub-array where x is equal to 2 or 3 is being filled in column-wise by cycling through the vector c(0,1). I don't know of any simple way to change this behavior.
Probably the easiest thing to do here is simply fill in the columns one at a time. Or, you could do something like this:
indices<-with(x, which(x==2|x==3))
x[indices,c(2,3)]<-rep(c(0,1),each=length(indices))
Another alternative: Using a data.table, this is a one-liner:
require(data.table)
DT <- data.table(x)
DT[x%in%2:3,`:=`(ab=0,cd=1)]
Original answer: You can pass a matrix of row-column pairs:
ijs <- expand.grid(with(x, which(x==2|x==3)),c(2:3))
ijs <- ijs[order(ijs$Var1),]
x[as.matrix(ijs)] <- c(0,1)
which yields
x ab cd
1 0 0 0
2 2 0 1
3 2 0 1
4 2 0 1
5 3 0 1
6 2 0 1
7 0 0 0
8 0 0 0
9 2 0 1
10 2 0 1
My original answer worked on my computer, but not a commenter's.
Generalized for multi-columns and multi-values:
mycol<-as.list(names(x)[-1])
myvalue<-as.list(c(0,1))
kk<-Map(function(y,z) list(x[x[,1] %in% c(2,3),y]<-z,x),mycol, myvalue)
myresult<-data.frame(kk[[2]][[2]])
x ab cd
1 1 0 0
2 1 0 0
3 0 0 0
4 0 0 0
5 0 0 0
6 3 0 1
7 2 0 1
8 3 0 1
9 3 0 1
10 0 0 0
You could use ifelse:
> set.seed(1234)
> dat<-data.frame(x=sample(c(0:3), 10, replace=T))
> dat$ab <- 0
> dat$cd <- ifelse(dat$x==2 | dat$x==3, 1, 0)
x ab cd
1 0 0 0
2 2 0 1
3 2 0 1
4 2 0 1
5 3 0 1
6 2 0 1
7 0 0 0
8 0 0 0
9 2 0 1
10 2 0 1
What about that?
x[x$x%in%c(2,3),c(2,3)]=matrix(rep(c(0,1),sum(x$x%in%c(2,3))),ncol=2,byrow=TRUE)
x$ab[x$x==2 | x$x==3] <- 0
x$cd[x$x==2 | x$x==3] <- 1
EDIT
Here is a general approach that would work with lots of columns. You simply create a vector of the replacement values you wish to use for each column.
set.seed(1234)
y<-data.frame(x=sample(c(0:3), 10, replace=T))
y$ab<-4 #column that will be replaced
y$cd<-2 #column that will be replaced
y$ef<-0 #column that will be replaced
y
# x ab cd ef
#1 0 4 2 0
#2 2 4 2 0
#3 2 4 2 0
#4 2 4 2 0
#5 3 4 2 0
#6 2 4 2 0
#7 0 4 2 0
#8 0 4 2 0
#9 2 4 2 0
#10 2 4 2 0
replacement.values <- c(10,20,30)
y2 <- y
y2[,2:ncol(y)] <- sapply(2:ncol(y), function(j) {
apply(y, 1, function(i) {
ifelse((i[1] %in% c(2,3)), replacement.values[j-1], i[j])
})
})
y2
# x ab cd ef
#1 0 4 2 0
#2 2 10 20 30
#3 2 10 20 30
#4 2 10 20 30
#5 3 10 20 30
#6 2 10 20 30
#7 0 4 2 0
#8 0 4 2 0
#9 2 10 20 30
#10 2 10 20 30
I have got a vector which is as under
a<- c(1,1,1,2,3,2,2,2,2,1,0,0,0,0,2,3,4,4,1,1)
Here we can see that there are lot of duplicate elements, ie. they are repeated ones.
I want a code which can replace all the elements which are consecutive and duplicate by 0 except for the first element. The result which i require is
a<- c(1,0,0,2,3,2,0,0,0,1,0,0,0,0,2,3,4,0,1,0)
I've tried
unique(a)
#which gives
[1] 1 2 3 0 4
You can created a lagged series and compare
> a
[1] 1 1 1 2 3 2 2 2 2 1 0 0 0 0 2 3 4 4 1 1
> ifelse(a == c(a[1]-1,a[(1:length(a)-1)]) , 0 , a)
[1] 1 0 0 2 3 2 0 0 0 1 0 0 0 0 2 3 4 0 1 0
replace(a, duplicated(c(0, cumsum(abs(diff(a))))), 0)
# [1] 1 0 0 2 3 2 0 0 0 1 0 0 0 0 2 3 4 0 1 0