Consider the following sample dataset. Id is an individual identifier.
rm(list=ls()); set.seed(1)
n<-100
X<-rbinom(n, 1, 0.5) #binary covariate
j<-rep (1:n)
dat<-data.frame(id=1:n, X)
ntp<- rep(4, n)
mat<-matrix(ncol=3,nrow=1)
m=0; w <- mat
for(l in ntp)
{
m=m+1
ft<- seq(from = 2, to = 8, length.out = l)
# ft<- seq(from = 1, to = 9, length.out = l)
ft<-sort(ft)
seq<-rep(ft,each=2)
seq<-c(0,seq,10)
matid<-cbind( matrix(seq,ncol=2,nrow=l+1,byrow=T ) ,m)
w<-rbind(w,matid)
}
d<-data.frame(w[-1,])
colnames(d)<-c("time1","time2","id")
D <- round( merge(d,dat,by="id") ,2) #merging dataset
nr<-nrow(D)
D$Survival_time<-round(rexp(nr, 0.1)+1,3)
head(D,15)
id time1 time2 X Survival_time
1 1 0 2 0 21.341
2 1 2 4 0 18.987
3 1 4 6 0 4.740
4 1 6 8 0 13.296
5 1 8 10 0 6.397
6 2 0 2 0 10.566
7 2 2 4 0 2.470
8 2 4 6 0 14.907
9 2 6 8 0 8.620
10 2 8 10 0 13.376
11 3 0 2 1 45.239
12 3 2 4 1 11.545
13 3 4 6 1 11.352
14 3 6 8 1 19.760
15 3 8 10 1 7.547
How can I obtain the value at which Survival_time is less that time2 for the very first time per individual. I should end up with the following values
id Survival_time
1 4.740
2 2.470
3 7.547
Also, how can I subset the data to stop individualwise when this condition occurs. i.e obtain
id time1 time2 X Survival_time
1 1 0 2 0 21.341
2 1 2 4 0 18.987
3 1 4 6 0 4.740
6 2 0 2 0 10.566
7 2 2 4 0 2.470
11 3 0 2 1 45.239
12 3 2 4 1 11.545
13 3 4 6 1 11.352
14 3 6 8 1 19.760
15 3 8 10 1 7.547
Using data.table
library(data.table)
setDT(D)[, .SD[seq_len(.N) <= which(Survival_time < time2)[1]], id]
-output
id time1 time2 X Survival_time
1: 1 0 2 0 21.341
2: 1 2 4 0 18.987
3: 1 4 6 0 4.740
4: 2 0 2 0 10.566
5: 2 2 4 0 2.470
6: 3 0 2 1 45.239
7: 3 2 4 1 11.545
8: 3 4 6 1 11.352
9: 3 6 8 1 19.760
10: 3 8 10 1 7.547
Slight variation:
library(dplyr)
D %>% # Take D, and then
group_by(id) %>% # group by id, and then
filter(Survival_time < time2) %>% # keep Survival times < time2, and then
slice(1) %>% # keep the first row per id, and then
ungroup() # ungroup
You can use -
library(dplyr)
D %>%
group_by(id) %>%
summarise(Survival_time = Survival_time[match(TRUE, Survival_time < time2)])
#Also using which.max
#summarise(Survival_time = Survival_time[which.max(Survival_time < time2)])
# id Survival_time
# <int> <dbl>
#1 1 4.74
#2 2 2.47
#3 3 7.55
To select the rows you may till that point you may use -
D %>%
group_by(id) %>%
filter(row_number() <= match(TRUE, Survival_time < time2)) %>%
ungroup
# id time1 time2 X Survival_time
# <int> <int> <int> <int> <dbl>
# 1 1 0 2 0 21.3
# 2 1 2 4 0 19.0
# 3 1 4 6 0 4.74
# 4 2 0 2 0 10.6
# 5 2 2 4 0 2.47
# 6 3 0 2 1 45.2
# 7 3 2 4 1 11.5
# 8 3 4 6 1 11.4
# 9 3 6 8 1 19.8
#10 3 8 10 1 7.55
Related
I have a data frame like so:
ID <- c('A','A','A','A','A','A','A','A','A','A','A','B','B','B','B','B')
val1 <- c(0,1,2,3,4,5,6,7,8,9,10,11,0,1,2,3)
val2 <- c(0,1,2,3,4,5,0,1,0,1,2,0,1,0,1,2)
df <- data.frame(ID, val1, val2)
Output:
ID val1 val2
1 A 0 0
2 A 1 1
3 A 2 2
4 A 3 3
5 A 4 4
6 A 5 5
7 A 6 0
8 A 7 1
9 A 8 0
10 A 9 1
11 A 10 2
12 B 11 0
13 B 0 1
14 B 1 0
15 B 2 1
16 B 3 2
I am trying to create a third column (val 3) which is like an index. When val1 = 0 and val 2 = 0 it should be 1 (this is also grouped by ID). It should stay as one and then increment by 1 until val2 = 0 again, like so showing desired output:
ID val1 val2 val3
1 A 0 0 1
2 A 1 1 1
3 A 2 2 1
4 A 3 3 1
5 A 4 4 1
6 A 5 5 1
7 A 6 0 2
8 A 7 1 2
9 A 8 0 3
10 A 9 1 3
11 A 10 2 3
12 B 11 0 1
13 B 0 1 1
14 B 1 0 2
15 B 2 1 2
16 B 3 2 2
How can this be achieved? I tried:
df <- df %>%
group_by(ID, val2) %>%
mutate(val3 = row_number())
And:
df$val3 <- cumsum(c(1,diff(df$val2)==0))
But neither provide the desired outcome.
Inside cumsum use the logical comparison val2==0
df %>%
group_by(ID) %>%
mutate(val3 = cumsum(val2==0))
# A tibble: 16 × 4
# Groups: ID [2]
ID val1 val2 val3
<chr> <dbl> <dbl> <int>
1 A 0 0 1
2 A 1 1 1
3 A 2 2 1
4 A 3 3 1
5 A 4 4 1
6 A 5 5 1
7 A 6 0 2
8 A 7 1 2
9 A 8 0 3
10 A 9 1 3
11 A 10 2 3
12 B 11 0 1
13 B 0 1 1
14 B 1 0 2
15 B 2 1 2
16 B 3 2 2
This question already has answers here:
Numbering rows within groups in a data frame
(10 answers)
Closed 1 year ago.
I am trying to create an additional variable (new variable-> flag) that will number the repetition of observation in my variable starting from 0.
dataset <- data.frame(id = c(1,1,1,2,2,4,6,6,6,7,7,7,7,8))
intended results will look like:
id flag
1 0
1 1
1 2
2 0
2 1
4 0
6 0
6 1
6 2
7 0
7 1
7 2
7 3
8 0
Thank You!
You may try
dataset$flag <- unlist(sapply(rle(dataset$id)$length, function(x) seq(1,x)-1))
id flag
1 1 0
2 1 1
3 1 2
4 2 0
5 2 1
6 4 0
7 6 0
8 6 1
9 6 2
10 7 0
11 7 1
12 7 2
13 7 3
14 8 0
data.table:
library(data.table)
setDT(dataset)[, flag := rowid(id) - 1]
dataset
id flag
1: 1 0
2: 1 1
3: 1 2
4: 2 0
5: 2 1
6: 4 0
7: 6 0
8: 6 1
9: 6 2
10: 7 0
11: 7 1
12: 7 2
13: 7 3
14: 8 0
Base R:
dataset$flag = sequence(rle(dataset$id)$lengths) - 1
dataset
id flag
1 1 0
2 1 1
3 1 2
4 2 0
5 2 1
6 4 0
7 6 0
8 6 1
9 6 2
10 7 0
11 7 1
12 7 2
13 7 3
14 8 0
Another base option:
transform(dataset,
flag = Reduce(function(x, y) y * x + y, duplicated(id), accumulate = TRUE))
id flag
1 1 0
2 1 1
3 1 2
4 2 0
5 2 1
6 4 0
7 6 0
8 6 1
9 6 2
10 7 0
11 7 1
12 7 2
13 7 3
14 8 0
dplyr -
library(dplyr)
dataset %>% group_by(id) %>% mutate(flag = row_number() - 1)
# id flag
# <dbl> <dbl>
# 1 1 0
# 2 1 1
# 3 1 2
# 4 2 0
# 5 2 1
# 6 4 0
# 7 6 0
# 8 6 1
# 9 6 2
#10 7 0
#11 7 1
#12 7 2
#13 7 3
#14 8 0
Base R with similar logic
transform(dataset, flag = ave(id, id, FUN = seq_along) - 1)
another way to reach what you expect but writing a little more
x <- dataset %>%
group_by(id) %>%
summarise(nreg=n())
df <- data.frame()
for(i in 1:nrow(x)){
flag <- data.frame(id = rep( x$id[i], x$nreg[i] ),
flag = seq(0, x$nreg [i] -1 )
)
df <- rbind(df, flag)
}
Consider the following data simulation mechanism:
set.seed(1)
simulW <- function(G)
{
# Let G be the number of groups
n<-2*G #Assume 2 individuals per group
i<-rep(1:G, rep(2,G)) # Group index
j<-rep (1:n)
Y<-rbinom(n, 1, 0.5) # binary
data.frame(id=1:n, i,Y)
}
r<-5 #5 replicates
dat1 <- replicate(r, simulW(G = 10 ), simplify=FALSE)
#For example the first data replicate will be
> dat1[[1]]
id i Y
1 1 1 0
2 2 1 1
3 3 2 0
4 4 2 0
5 5 3 0
6 6 3 0
7 7 4 0
8 8 4 1
9 9 5 1
10 10 5 0
The code below can perform group wise (i is the group) sum of Y but by default considers only the first replicate i.e dat1[[1]].
Di<-aggregate( Y, by=list ( i ),FUN=sum) #Sum per group for the first dataset
e<-colSums(Di [ 2 ] ) #Total sum of Y for all groups for dataset 1
> e
x
8
di<-Di [ 2 ] # Groupwise sum for replicate 1
> di
x
1 2
2 2
3 2
4 0
5 2
How can I use the same function to perform the group wise sum for the other replicates.
Maybe something like:
for (m in 1:r )
{
Di[m]<-
e[m]<-
di[m]<-
}
You may use aggregate in lapply -
result <- lapply(dat1, function(x) aggregate(Y~i, x, sum))
result
#[[1]]
# i Y
#1 1 1
#2 2 1
#3 3 0
#4 4 0
#5 5 1
#6 6 1
#7 7 0
#8 8 2
#9 9 1
#10 10 1
#[[2]]
# i Y
#1 1 2
#2 2 2
#3 3 2
#4 4 0
#5 5 2
#6 6 1
#7 7 0
#8 8 0
#9 9 1
#10 10 1
#...
#...
We may use tidyverse
library(purrr)
library(dplyr)
map(dat1, ~ .x %>%
group_by(i) %>%
summarise(Y = sum(Y)))
-output
[[1]]
# A tibble: 10 × 2
i Y
<int> <int>
1 1 0
2 2 2
3 3 1
4 4 2
5 5 1
6 6 0
7 7 1
8 8 1
9 9 2
10 10 1
[[2]]
# A tibble: 10 × 2
i Y
<int> <int>
1 1 1
2 2 1
3 3 0
4 4 0
5 5 1
6 6 1
7 7 0
8 8 2
9 9 1
10 10 1
[[3]]
# A tibble: 10 × 2
i Y
<int> <int>
1 1 2
2 2 2
3 3 2
4 4 0
5 5 2
6 6 1
7 7 0
8 8 0
9 9 1
10 10 1
[[4]]
# A tibble: 10 × 2
i Y
<int> <int>
1 1 1
2 2 0
3 3 1
4 4 1
5 5 1
6 6 1
7 7 0
8 8 1
9 9 1
10 10 2
[[5]]
# A tibble: 10 × 2
i Y
<int> <int>
1 1 1
2 2 0
3 3 1
4 4 1
5 5 0
6 6 0
7 7 2
8 8 2
9 9 0
10 10 2
I'd like to count the rows in the column input if the values are smaller than the current row (Please see the results wanted below). The issue to me is that the condition is based on current row value, so it is very different from general case where the condition is a fixed number.
data <- data.frame(input = c(1,1,1,1,2,2,3,5,5,5,5,6))
input
1 1
2 1
3 1
4 1
5 2
6 2
7 3
8 5
9 5
10 5
11 5
12 6
The results I expect to get are like this. For example, for observations 5 and 6 (with value 2), there are 4 observations with value 1 less than their value 2. Hence count is given value 4.
input count
1 1 0
2 1 0
3 1 0
4 1 0
5 2 4
6 2 4
7 3 6
8 5 7
9 5 7
10 5 7
11 5 7
12 6 11
Edit: as I am dealing with grouped data with dplyr, the ultimate results I wish to get is like below, that is, I am wishing the conditions could be dynamic within each group.
data <- data.frame(id = c(1,1,2,2,2,3,3,4,4,4,4,4),
input = c(1,1,1,1,2,2,3,5,5,5,5,6),
count=c(0,0,0,0,2,0,1,0,0,0,0,4))
id input count
1 1 1 0
2 1 1 0
3 2 1 0
4 2 1 0
5 2 2 2
6 3 2 0
7 3 3 1
8 4 5 0
9 4 5 0
10 4 5 0
11 4 5 0
12 4 6 4
Here is an option with tidyverse
library(tidyverse)
data %>%
mutate(count = map_int(input, ~ sum(.x > input)))
# input count
#1 1 0
#2 1 0
#3 1 0
#4 1 0
#5 2 4
#6 2 4
#7 3 6
#8 5 7
#9 5 7
#10 5 7
#11 5 7
#12 6 11
Update
With the updated data, add the group by 'id' in the above code
data %>%
group_by(id) %>%
mutate(count1 = map_int(input, ~ sum(.x > input)))
# A tibble: 12 x 4
# Groups: id [4]
# id input count count1
# <dbl> <dbl> <dbl> <int>
# 1 1 1 0 0
# 2 1 1 0 0
# 3 2 1 0 0
# 4 2 1 0 0
# 5 2 2 2 2
# 6 3 2 0 0
# 7 3 3 1 1
# 8 4 5 0 0
# 9 4 5 0 0
#10 4 5 0 0
#11 4 5 0 0
#12 4 6 4 4
In base R, we can use sapply and for each input count how many values are greater than itself.
data$count <- sapply(data$input, function(x) sum(x > data$input))
data
# input count
#1 1 0
#2 1 0
#3 1 0
#4 1 0
#5 2 4
#6 2 4
#7 3 6
#8 5 7
#9 5 7
#10 5 7
#11 5 7
#12 6 11
With dplyr one way would be using rowwise function and following the same logic.
library(dplyr)
data %>%
rowwise() %>%
mutate(count = sum(input > data$input))
1. outer and rowSums
data$count <- with(data, rowSums(outer(input, input, `>`)))
2. table and cumsum
tt <- cumsum(table(data$input))
v <- setNames(c(0, head(tt, -1)), c(head(names(tt), -1), tail(names(tt), 1)))
data$count <- v[match(data$input, names(v))]
3. data.table non-equi join
Perhaps more efficient with a non-equi join in data.table. Count number of rows (.N) for each match (by = .EACHI).
library(data.table)
setDT(data)
data[data, on = .(input < input), .N, by = .EACHI]
If your data is grouped by 'id', as in your update, join on that variable as well:
data[data, on = .(id, input < input), .N, by = .EACHI]
# id input N
# 1: 1 1 0
# 2: 1 1 0
# 3: 2 1 0
# 4: 2 1 0
# 5: 2 2 2
# 6: 3 2 0
# 7: 3 3 1
# 8: 4 5 0
# 9: 4 5 0
# 10: 4 5 0
# 11: 4 5 0
# 12: 4 6 4
Hi I have a data frame like this
df <-data.frame(x=rep(rep(seq(0,3),each=2),2 ),gr=gl(2,8))
x gr
1 0 1
2 0 1
3 1 1
4 1 1
5 2 1
6 2 1
7 3 1
8 3 1
9 0 2
10 0 2
11 1 2
12 1 2
13 2 2
14 2 2
15 3 2
16 3 2
I want to add a new column numbering sequence of numbers when the x value ==0
I tried
library(dplyr)
df%>%
group_by(gr)%>%
mutate(numbering=seq(2,8,2))
Error in mutate_impl(.data, dots) :
Column `numbering` must be length 8 (the group size) or one, not 4
?
Just for side note mutate(numbering=rep(seq(2,8,2),each=2)) would work for this minimal example but for the general case its better to look x value change from 0!
the expected output
x gr numbering
1 0 1 2
2 0 1 2
3 1 1 4
4 1 1 4
5 2 1 6
6 2 1 6
7 3 1 8
8 3 1 8
9 0 2 2
10 0 2 2
11 1 2 4
12 1 2 4
13 2 2 6
14 2 2 6
15 3 2 8
16 3 2 8
Do you mean something like this?
library(tidyverse);
df %>%
group_by(gr) %>%
mutate(numbering = cumsum(c(1, diff(x) != 0)))
## A tibble: 16 x 3
## Groups: gr [2]
# x gr numbering
# <int> <fct> <dbl>
# 1 0 1 1.
# 2 0 1 1.
# 3 1 1 2.
# 4 1 1 2.
# 5 2 1 3.
# 6 2 1 3.
# 7 3 1 4.
# 8 3 1 4.
# 9 0 2 1.
#10 0 2 1.
#11 1 2 2.
#12 1 2 2.
#13 2 2 3.
#14 2 2 3.
#15 3 2 4.
#16 3 2 4.
Or if you must have a numbering sequence 2,4,6,... instead of 1,2,3,... you can do
df %>%
group_by(gr) %>%
mutate(numering = 2 * cumsum(c(1, diff(x) != 0)));
## A tibble: 16 x 3
## Groups: gr [2]
# x gr numering
# <int> <fct> <dbl>
# 1 0 1 2.
# 2 0 1 2.
# 3 1 1 4.
# 4 1 1 4.
# 5 2 1 6.
# 6 2 1 6.
# 7 3 1 8.
# 8 3 1 8.
# 9 0 2 2.
#10 0 2 2.
#11 1 2 4.
#12 1 2 4.
#13 2 2 6.
#14 2 2 6.
#15 3 2 8.
#16 3 2 8.
Here is an option using match to get the index and then pass on the seq values to fill
df %>%
group_by(gr) %>%
mutate(numbering = seq(2, length.out = n()/2, by = 2)[match(x, unique(x))])
# A tibble: 16 x 3
# Groups: gr [2]
# x gr numbering
# <int> <fct> <dbl>
# 1 0 1 2
# 2 0 1 2
# 3 1 1 4
# 4 1 1 4
# 5 2 1 6
# 6 2 1 6
# 7 3 1 8
# 8 3 1 8
# 9 0 2 2
#10 0 2 2
#11 1 2 4
#12 1 2 4
#13 2 2 6
#14 2 2 6
#15 3 2 8
#16 3 2 8