This question already has answers here:
How to create a consecutive group number
(13 answers)
Closed 1 year ago.
I have these set of variables in the column Num I want to create another column that ranks them with size similar to rankt below but I don't like how this is done.
x <- data.frame("Num" = c(2,5,2,7,7,7,2,5,5))
x$rankt <- rank(x$Num)
Num rankt
1 2 2
2 5 5
3 2 2
4 7 8
5 7 8
6 7 8
7 2 2
8 5 5
9 5 5
Desired Outcome I would like for rankt
Num rankt
1 2 1
2 5 2
3 2 1
4 7 3
5 7 3
6 7 3
7 2 1
8 5 2
9 5 2
Well, a crude approach is to turn them to factors, which are just increasing numbers with labels, and then fetch those numbers:
x <- data.frame("Num" = c(2,5,2,7,7,7,2,5,5))
x$rankt <- as.numeric(as.factor( rank(x$Num) ))
x
It produces:
Num rankt
1 2 1
2 5 2
3 2 1
4 7 3
5 7 3
6 7 3
7 2 1
8 5 2
9 5 2
A solution with dplyr
library(dplyr)
x1 <- x %>%
mutate(rankt=dense_rank(desc(-Num)))
Related
I have a factor variable with 6 levels, which simplified looks like:
1 1 2 2 2 3 3 3 4 4 4 4 5 5 5 6 6 6 1 1 1 2 2 2 2... 1 1 1 2 2... (with n = 78)
Note, that each number is repeated mostly but not always three times.
I need to transform this variable into the following pattern:
1 1 2 2 2 3 3 3 4 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 8...
where each repetition of the 6 levels continuous counting ascending.
Is there any way / any function that lets me do that?
Sorry for my bad description!
Assuming that you have a numerical vector that represents your simplified version you posted. i.e. x = c(1,1,1,2,2,3,3,3,1,1,2,2), you can use this:
library(dplyr)
cumsum(x != lag(x, default = 0))
# [1] 1 1 1 2 2 3 3 3 4 4 5 5
which compares each value to its previous one and if they are different it adds 1 (starting from 1).
Maybe you can try rle, i.e.,
v <- rep(seq_along((v<-rle(x))$values),v$lengths)
Example with dummy data
x = c(1,1,1,2,2,3,3,3,4,4,5,6,1,1,2,2,3,3,3,4,4)
then we can get
> v
[1] 1 1 1 2 2 3 3 3 4 4 5 6 7 7 8 8 9 9
[19] 9 10 10
In base you can use diff and cumsum.
c(1, cumsum(diff(x)!=0)+1)
# [1] 1 1 2 2 2 3 3 3 4 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 8
Data:
x <- c(1,1,2,2,2,3,3,3,4,4,4,4,5,5,5,6,6,6,1,1,1,2,2,2,2)
This question already has answers here:
Replace a value NA with the value from another column in R
(5 answers)
Closed 3 years ago.
I don't have the slightest idea of programming, but I need to solve the following problem in R.
Let's suppose I have this data:
x y
5 8
6 5
2
9 8
4
0
6 6
7 3
3 2
I need to create a third column called "z" containing the data of "y" exccept for the missing values where it should have the values of "x". It would be something like this:
x y z
5 8 8
6 5 5
2 2
9 8 8
4 4
0 0
6 6 6
7 3 3
3 2 2
dat <- data.frame(x=c(5,6,2,9,4,0,6,7,3), y = c(8,5,NA,8,NA,NA,6,3,2))
library(tidyverse)
dat %>% mutate(z = ifelse(is.na(y), x, y))
# x y z
# 1 5 8 8
# 2 6 5 5
# 3 2 NA 2
# 4 9 8 8
# 5 4 NA 4
# 6 0 NA 0
# 7 6 6 6
# 8 7 3 3
# 9 3 2 2
I want to replicate a vector with one value within this vector is missing (sequentially).
For example, my vector is
value <- 1:7
First, the series is without 1, second without 2, and so on. In the end, the series is in one vector.
The intended output looks like
2 3 4 5 6 7 1 3 4 5 6 7 1 2 4 5 6 7 1 2 3 5 6 7 1 2 3 4 6 7 1 2 3 4 5 6
Is there any smart way to do this?
You could use the diagonal matrix to set up a logical vector, using it to remove the appropriate values.
n <- 7
rep(1:n, n)[!diag(n)]
# [1] 2 3 4 5 6 7 1 3 4 5 6 7 1 2 4 5 6 7 1 2 3 5 6 7 1 2 3 4 6 7 1 2 3 4 5
# [36] 7 1 2 3 4 5 6
Well, you can certainly do it as a one-liner but I am not sure it qualifies as smart. For example:
x <- 1:7
do.call("c", lapply(as.list(-1:-length(x)), function(a)x[a]))
This simple uses lapply to create a list of copies of x with each of its entries deleted, and then concatenates them using c. The do.call function applies its first argument (a function) to its second argument (a list of arguments to the function).
For fun, it's also possible to just use rep:
> n <- 7
> rep(1:n, n)[rep(c(FALSE, rep(TRUE, n)), length.out=n^2)]
[1] 2 3 4 5 6 7 1 3 4 5 6 7 1 2 4 5 6 7 1 2 3 5 6 7 1 2 3 4 6 7 1 2 3 4 5 7 1 2
[39] 3 4 5 6
But lapply is cleaner, I think.
You could also do:
n <- 7
rep(seq(n), n)[-seq(1,n*n,n+1)]
#[1] 2 3 4 5 6 7 1 3 4 5 6 7 1 2 4 5 6 7 1 2 3 5 6 7 1 2 3 4 6 7 1 2 3 4 5 7 1 2 3 4 5 6
This question already has answers here:
R: define distinct pattern from values of multiple variables [duplicate]
(3 answers)
Closed 5 years ago.
I have a dataset like this:
case x y
1 4 5
2 4 5
3 8 9
4 7 9
5 6 3
6 6 3
I would like to create a grouping variable.
This variable should have the same values when both x and y are the same.
I do not care what this value is but it is to group them. Because in my dataset if x and y are the same for two cases they are probably part of the same organization. I want to see which organizations there are.
So my preferred dataset would look like this:
case x y org
1 4 5 1
2 4 5 1
3 8 9 2
4 7 9 3
5 6 3 4
6 6 3 4
How would I have to program this in R?
As you said , I do not care what this value is, you can just do following
dt$new=as.numeric(as.factor(paste(dt$x,dt$y)))
dt
case x y new
1 1 4 5 1
2 2 4 5 1
3 3 8 9 4
4 4 7 9 3
5 5 6 3 2
6 6 6 3 2
A solution from dplyr using the group_indices.
library(dplyr)
dt2 <- dt %>%
mutate(org = group_indices(., x, y))
dt2
case x y org
1 1 4 5 1
2 2 4 5 1
3 3 8 9 4
4 4 7 9 3
5 5 6 3 2
6 6 6 3 2
If the group numbers need to be in order, we can use the rleid from the data.table package after we create the org column as follows.
library(dplyr)
library(data.table)
dt2 <- dt %>%
mutate(org = group_indices(., x, y)) %>%
mutate(org = rleid(org))
dt2
case x y org
1 1 4 5 1
2 2 4 5 1
3 3 8 9 2
4 4 7 9 3
5 5 6 3 4
6 6 6 3 4
Update
Here is how to arrange the columns in dplyr.
library(dplyr)
dt %>%
arrange(x)
case x y
1 1 4 5
2 2 4 5
3 5 6 3
4 6 6 3
5 4 7 9
6 3 8 9
We can also do this for more than one column, such as arrange(x, y) or use desc to reverse the oder, like arrange(desc(x)).
DATA
dt <- read.table(text = " case x y
1 4 5
2 4 5
3 8 9
4 7 9
5 6 3
6 6 3",
header = TRUE)
I would like to refer to values in a data frame column with the row index being dependent on the value of another column.
Example:
value lag laggedValue
1 1 2
2 2 4
3 3 6
4 2 6
5 1 6
6 3 9
7 3 10
8 1 9
9 1 10
10 2
In Excel I use this formula in column "laggedValue":
=INDIRECT("B"&(ROW(B2)+C2))
How can I do this in an R data frame?
Thanks!
For row r with associated lag value lag[r] it looks like you're trying to create a new column that is the (r+lag[r])th element of value (or a missing value if this is out of bounds). You can do this with:
dat$laggedValue <- dat$value[seq(nrow(dat)) + dat$lag]
dat
value lag laggedValue
1 1 1 2
2 2 2 4
3 3 3 6
4 4 2 6
5 5 1 6
6 6 3 9
7 7 3 10
8 8 1 9
9 9 1 10
10 10 2 NA
Other commenters are mentioning that it looks like you're just adding the value and lag columns because your value column has the elements 1 through 10, but this solution will work even when your value column has other data stored in it.
Assuming the same thing as #rawr here:
dat <- data.frame(value=c(1:10),
lag=c(1,2,3,2,1,3,3,1,1,2))
dat$laggedValue <- dat$value + dat$lag
dat
value lag laggedValue
1 1 1 2
2 2 2 4
3 3 3 6
4 4 2 6
5 5 1 6
6 6 3 9
7 7 3 10
8 8 1 9
9 9 1 10
10 10 2 12