This problem is very similar to Consecutive value after column value change in R
So for
SOG <- c(4,4,0,0,0,3,4,5,0,0,1,2,0,0,0)
the difference is that now I'd like to count how many groups of SOG there are. For example:
SOG Trips
--- -----
4 1
4 1
0 0
0 0
0 0
3 2
4 2
5 2
0 0
0 0
1 3
2 3
0 0
0 0
0 0
Anyone?
Assuming you mean a "group of SOG" is a set of consecutive non-zero SOG values, i.e. starts with a non-zero SOG value and ends with a non-zero SOG value (not necessarily the same value):
Trips <- ifelse(SOG>0, cumsum(c(SOG[1]>0, diff(SOG>0)) == 1), 0)
# [1] 1 1 0 0 0 2 2 2 0 0 3 3 0 0 0
This is one option:
replace(cumsum(c(SOG[1], abs(diff(SOG))) == SOG & SOG != 0), SOG == 0, 0)
# [1] 1 1 0 0 0 2 2 2 0 0 3 3 0 0 0
You can try my TrueSeq function from my GitHub-only "SOfun" package.
Usage would be:
library(SOfun)
TrueSeq(as.logical(SOG))
# [1] 1 1 0 0 0 2 2 2 0 0 3 3 0 0 0
To get the inverse, just negate the as.logical step:
TrueSeq(!as.logical(SOG))
# [1] 0 0 1 1 1 0 0 0 2 2 0 0 3 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$w <- apply(a, 1, max)
I'm looking for a function in R which can do the permutation. For example, I have a vector with five 1 and ten 0 like this:
> status=c(rep(1,5),rep(0,10))
> status
[1] 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0
Now I'd like to randomly permute the position of these numbers but keep the same number of 0 and 1 in vector and to get new series of number, for example to get something like this:
1 1 0 1 0 1 0 0 0 0 0 1 0 0 0
or
1 0 0 0 0 0 0 1 1 0 0 1 0 1 0
I found the function sample() can help us to sample, but the number of 1 and 0 is not the same each time. Do you know how can I do this with R? Thanks in advance.
We can use sample
sample(status)
#[1] 1 0 0 1 0 0 1 0 0 0 0 1 0 1 0
sample(status)
#[1] 0 0 0 0 1 1 0 0 1 1 0 0 0 1 0
If we use sample to return the entire vector, it will do the permutation and give the frequency count same for each of the unique elements
colSums(replicate(5, sample(status)))
#[1] 5 5 5 5 5
i.e. we get 5 one's in each of the sampling. So, the remaining 0's would be 10.
This question already has answers here:
Cumulative sum for positive numbers only [duplicate]
(9 answers)
Closed 6 years ago.
If I have the following vector:
x = c(1,1,1,0,0,0,0,1,1,0,0,1,1,1,0,0,1,1,1,1,0,0,0,0,1,1,1)
how can I calculate the cumulative sum for all of the consecutive 1's, resetting each time I hit a 0?
So, the desired output would look like this:
> y
[1] 1 2 3 0 0 0 0 1 2 0 0 1 2 3 0 0 1 2 3 4 0 0 0 0 1 2 3
This works:
unlist(lapply(rle(x)$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 have newly started to learn R, so my question may be utterly ridiculous. I have a data frame
data<- data.frame('number'=1:11, 'col1'=sample(10:20),'col2'=sample(10:20),'col3'=sample(10:20),'col4'=sample(10:20),'col5'=sample(10:20), 'date'= c('12-12-2014','12-11-2014','12-10-2014','12-09-2014', '12-08-2014','12-07-2014','12-06-2014','12-05-2014','12-04-2014', '12-04-2014', '12-03-2014') )
The number column is an 'id' column and the last column is a date.
I want to count the number of times that each number occurs across (not per column, but the whole data frame containing data) the columns 2:6 and when they occurred.
I am stuck on the first part having tried the following using data.table:
count <- function(){
i = 1
DT <-data.table(data[2:6])
for (i in 10:20){
DT[, .N, by =i]
i = i + 1
}
}
which gives an error that I don't begin to understand
Error in `[.data.table`(DT, , .N, by = i) :
The items in the 'by' or 'keyby' list are length (1). Each must be same length as rows in x or number of rows returned by i (11)
Can someone help, please. Also with the second part that I have not even attempted yet i.e. associating a date or a row number with each occurrence of a number
Perhaps you may want this
library(reshape2)
table(melt(data[,-1], id.var='date')[,-2])
# value
#date 10 11 12 13 14 15 16 17 18 19 20
# 12-03-2014 0 0 1 0 0 1 0 0 1 2 0
# 12-04-2014 2 0 0 2 2 0 1 0 1 1 1
# 12-05-2014 0 0 0 0 0 0 1 1 2 0 1
# 12-06-2014 1 1 0 0 0 1 0 1 0 0 1
# 12-07-2014 0 1 0 1 0 1 1 1 0 0 0
# 12-08-2014 1 1 0 0 1 0 0 1 1 0 0
# 12-09-2014 0 0 2 0 1 2 0 0 0 0 0
# 12-10-2014 0 0 1 1 0 0 1 0 0 1 1
# 12-11-2014 0 1 1 0 0 0 1 0 0 1 1
# 12-12-2014 1 1 0 1 1 0 0 1 0 0 0
Or if you need a data.table solution (from #Arun's comments)
library(data.table)
dcast.data.table(melt(setDT(data),
id="date", measure=2:6), date ~ value)
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