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
Related
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)))
I have a list of numbers and would like to find which is the next highest compared to each number in a data.frame. I have:
list <- c(3,6,9,12)
X <- c(1:10)
df <- data.frame(X)
And I would like to add a variable to df being the next highest number in the list. i.e:
X Y
1 3
2 3
3 3
4 6
5 6
6 6
7 9
8 9
9 9
10 12
I've tried:
df$Y <- which.min(abs(list-df$X))
but that gives an error message and would just get the closest value from the list, not the next above.
Another approach is to use findInterval:
df$Y <- list[findInterval(X, list, left.open=TRUE) + 1]
> df
X Y
1 1 3
2 2 3
3 3 3
4 4 6
5 5 6
6 6 6
7 7 9
8 8 9
9 9 9
10 10 12
You could do this...
df$Y <- sapply(df$X, function(x) min(list[list>=x]))
df
X Y
1 1 3
2 2 3
3 3 3
4 4 6
5 5 6
6 6 6
7 7 9
8 8 9
9 9 9
10 10 12
This question already has answers here:
R: define distinct pattern from values of multiple variables [duplicate]
(3 answers)
Closed 5 years ago.
I have a dataset like this:
case x y
1 4 5
2 4 5
3 8 9
4 7 9
5 6 3
6 6 3
I would like to create a grouping variable.
This variable should have the same values when both x and y are the same.
I do not care what this value is but it is to group them. Because in my dataset if x and y are the same for two cases they are probably part of the same organization. I want to see which organizations there are.
So my preferred dataset would look like this:
case x y org
1 4 5 1
2 4 5 1
3 8 9 2
4 7 9 3
5 6 3 4
6 6 3 4
How would I have to program this in R?
As you said , I do not care what this value is, you can just do following
dt$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)
This question already has answers here:
Replacing NAs with latest non-NA value
(21 answers)
Closed 7 years ago.
I have two data.frame as the following:
> a <- data.frame(x=c(1,2,3,4,5,6,7,8), y=c(1,3,5,7,9,11,13,15))
> a
x y
1 1 1
2 2 3
3 3 5
4 4 7
5 5 9
6 6 11
7 7 13
8 8 15
> b <- data.frame(x=c(1,5,7), z=c(2, 4, 6))
> b
x z
1 1 2
2 5 4
3 7 6
Then I use "join" for two data.frames:
> c <- join(a, b, by="x", type="left")
> c
x y z
1 1 1 2
2 2 3 NA
3 3 5 NA
4 4 7 NA
5 5 9 4
6 6 11 NA
7 7 13 6
8 8 15 NA
My requirement is to replace the NAs in the Z column by the last None-Na value before the current place. I want the result like this:
> c
x y z
1 1 1 2
2 2 3 2
3 3 5 2
4 4 7 2
5 5 9 4
6 6 11 4
7 7 13 6
8 8 15 6
This time (if your data is not too large) a loop is an elegant option:
for(i in which(is.na(c$z))){
c$z[i] = c$z[i-1]
}
gives:
> c
x y z
1 1 1 2
2 2 3 2
3 3 5 2
4 4 7 2
5 5 9 4
6 6 11 4
7 7 13 6
8 8 15 6
data:
library(plyr)
a <- data.frame(x=c(1,2,3,4,5,6,7,8), y=c(1,3,5,7,9,11,13,15))
b <- data.frame(x=c(1,5,7), z=c(2, 4, 6))
c <- join(a, b, by="x", type="left")
You might also want to check na.locf in the zoo package.
Eliminate in an increasing order rows in a data frame
x<-c(4,5,6,23,5,6,7,8,0,3)
y<-c(2,4,5,6,23,5,6,7,8,0)
z<-c(1,2,4,5,6,23,5,6,7,8)
df<-data.frame(x,y,z)
df
x y z
1 4 2 1
2 5 4 2
3 6 5 4
4 23 6 5
5 5 23 6
6 6 5 23
7 7 6 5
8 8 7 6
9 0 8 7
10 3 0 8
I would like to eliminate number 23 in the df from all columns by instructing to sequentially increasingly remove a row per column (not by matching the value 23, but by its initial x location).
df
x y z
1 4 2 1
2 5 4 2
3 6 5 4
4 5 6 5
5 6 5 6
6 7 6 5
7 8 7 6
8 0 8 7
9 3 0 8
Thank you
You can iterate through the columns and remove the element from each, then reassemble as a data frame:
result <- as.data.frame(lapply(1:ncol(df), function(x) df[-(x+3),x]))
names(result) <- names(df)
result
## x y z
## 1 4 2 1
## 2 5 4 2
## 3 6 5 4
## 4 5 6 5
## 5 6 5 6
## 6 7 6 5
## 7 8 7 6
## 8 0 8 7
## 9 3 0 8
df[-(x+3),x] is the column with the value removed, by location. To start with row N in column x you would use df[-(x+N-1),x].
You could also try:
n <- 4
df1 <- df[-n,]
df1[] <- unlist(df,use.names=FALSE)[-seq(n, prod(dim(df)), by=nrow(df)+1)]
df1
# x y z
#1 4 2 1
#2 5 4 2
#3 6 5 4
#5 5 6 5
#6 6 5 6
#7 7 6 5
#8 8 7 6
#9 0 8 7
#10 3 0 8