I have matrix like
> brdrs <- matrix(c(-1,-0.2,0.2,3),ncol=2,byrow=TRUE)
> brdrs
[,1] [,2]
[1,] -1.0 -0.2
[2,] 0.2 3.0
I want to make sequence, based on this matrix. First column is the start of interval, second- the end. Each row is interval of one sequence.
For example it would be: from -1.0 to -0.2 AND from 0.2 to 3.0 by 0.1.
Is it possible without loops?
Thanks
You could use this :
unlist(sapply(1:nrow(brdrs),function(x){seq(brdrs[x,1],brdrs[x,2],0.1)}))
[1] -1.0 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
[18] 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6
[35] 2.7 2.8 2.9 3.0
Like Ananda Mahto said in the comment, you should specify the increment of the sequence. Here, I used 0.1
Related
I have a network that is connected like the following:
library(igraph)
network <- graph_from_literal(1--2,2--3,3--4,4--5,3--6,6--7,3--8,8--9)
and have the distance between each segment
> data=data.frame(Origin=c(1,2,3,4,3,6,3,8), Destination=c(2,3,4,5,6,7,8,9), km=c(0.3,0.5,0.2,0.1,1,2,0.6,0.4))
> data
Origin Destination km
1 2 0.3
2 3 0.5
3 4 0.2
4 5 0.1
3 6 1.0
6 7 2.0
3 8 0.6
8 9 0.4
I would like to find the distance matrix between all points and I tried using
distMatrix <- shortest.paths(data, v=Origin(data), to=Destination(data))
but cannot make it work.
Thanks!
The distances() function will give you a distance matrix. By setting the edge weights, the distance matrix will include these weights in the calculation. In the code below, I create the graph from your dataset data, such that the edges are correctly ordered:
library("igraph")
data = data.frame(Origin=c(1,2,3,4,3,6,3,8), Destination=c(2,3,4,5,6,7,8,9),
km=c(0.3,0.5,0.2,0.1,1,2,0.6,0.4))
g = graph_from_edgelist(cbind(data$Origin, data$Destination))
E(g)
#> + 8/8 edges from 7ad6d28:
#> [1] 1->2 2->3 3->4 4->5 3->6 6->7 3->8 8->9
E(g)$weight = data$km
distances(g)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
#> [1,] 0.0 0.3 0.8 1.0 1.1 1.8 3.8 1.4 1.8
#> [2,] 0.3 0.0 0.5 0.7 0.8 1.5 3.5 1.1 1.5
#> [3,] 0.8 0.5 0.0 0.2 0.3 1.0 3.0 0.6 1.0
#> [4,] 1.0 0.7 0.2 0.0 0.1 1.2 3.2 0.8 1.2
#> [5,] 1.1 0.8 0.3 0.1 0.0 1.3 3.3 0.9 1.3
#> [6,] 1.8 1.5 1.0 1.2 1.3 0.0 2.0 1.6 2.0
#> [7,] 3.8 3.5 3.0 3.2 3.3 2.0 0.0 3.6 4.0
#> [8,] 1.4 1.1 0.6 0.8 0.9 1.6 3.6 0.0 0.4
#> [9,] 1.8 1.5 1.0 1.2 1.3 2.0 4.0 0.4 0.0
Created on 2021-09-06 by the reprex package (v2.0.1)
Perhaps we can try distances like below
> distances(
+ graph_from_data_frame(data, directed = FALSE),
+ weights = data$km
+ )
1 2 3 4 6 8 5 7 9
1 0.0 0.3 0.8 1.0 1.8 1.4 1.1 3.8 1.8
2 0.3 0.0 0.5 0.7 1.5 1.1 0.8 3.5 1.5
3 0.8 0.5 0.0 0.2 1.0 0.6 0.3 3.0 1.0
4 1.0 0.7 0.2 0.0 1.2 0.8 0.1 3.2 1.2
6 1.8 1.5 1.0 1.2 0.0 1.6 1.3 2.0 2.0
8 1.4 1.1 0.6 0.8 1.6 0.0 0.9 3.6 0.4
5 1.1 0.8 0.3 0.1 1.3 0.9 0.0 3.3 1.3
7 3.8 3.5 3.0 3.2 2.0 3.6 3.3 0.0 4.0
9 1.8 1.5 1.0 1.2 2.0 0.4 1.3 4.0 0.0
I have searched all the lapply questions and solutions, and none of those solutions seems to address and/or work for the following...
I have a list "temp" that contains the names of 100 data frames: "sim_rep1.dat" through "sim_rep100.dat".
Each data frame has 2000 observations and the same 11 variables: ARAND and w1-w10, all of which are numeric.
For all 100 data frames, I am trying to create a new variable called "ps_true" that incorporates certain of the "w" variables, each with a unique coefficient.
The only use of lapply that is working for me is the following:
lapply(mget(paste0("sim_rep", 1:100,".dat")), transform,
ps_true = (1 + exp(-(0.8*w1 - 0.25*w2 + 0.6*w3 -
0.4*w4 - 0.8*w5 - 0.5*w6 + 0.7*w7)))^-1)
When I run the code above, R loops through all 100 data frames and shows newly calculated values for ps_true in the console. Unfortunately, the new column is not getting added to the data frames.
When I try to create a function, the wheels come completely off.
I have tried different variations of the following:
lapply(temp, function(x){
ps_true = (1 + exp(-(0.8*w1 - 0.25*w2 + 0.6*w3 -
0.4*w4 - 0.8*w5 - 0.5*w6 + 0.7*w7)))^-1
cbind(x, ps_true)
return(x)
})
Error in FUN(X[[i]], ...) : object 'w1' not found results from the function shown above
Error in x$w1 : $ operator is invalid for atomic vectors results if I try to reference x$w1 instead
Error in FUN(X[[i]], ...) : object 'w1' not found results if I try to reference x[[w1]] instead
Error in x[["w1"]] : subscript out of bounds results if I try to reference x[["w1"]] instead
I am hoping there is something obvious that I am missing. I'd appreciate your insights and suggestions to solve this frustrating problem.
In response to Uwe's addendum:
The code I had used to read all the files was the following:
temp = list.files(pattern='*.dat')
for (i in 1:length(temp)) {
assign(temp[i], read.csv(temp[i], header=F,sep="",
col.names = c("ARAND", "w1", "w2", "w3", "w4", "w5", "w6", "w7", "w8", "w9", "w10")))
}
According to the OP, there are 100 data.frames with identical columns names. The OP wants to create a new column in all of the data.frames using exactly the same formula.
This indicates a fundamental flaw in the design of the data structure. I guess, no data base admin would create 100 identical tables where only the data contents differs. Instead, he would create one table with an additional column identifying the origin of each row. Then, all subsequent operations would be applied on one table instead to be repeated for each of many.
In R, the data.table package has the convenient rbindlist() function which can be used for this purpose:
library(data.table) # CRAN version 1.10.4 used
# get list of data.frames from the given names and
# combine the rows of all data sets into one large data.table
DT <- rbindlist(mget(temp), idcol = "origin")
# now create new column for all rows across all data sets
DT[, ps_true := (1 + exp(-(0.8*w1 - 0.25*w2 + 0.6*w3 -
0.4*w4 - 0.8*w5 - 0.5*w6 + 0.7*w7)))^-1]
DT
origin ARAND w1 w2 w3 w4 w5 w6 w7 w8 w9 w10 ps_true
1: sim_rep1.dat -0.6 -0.5 0.2 -0.7 0.5 2.4 -0.2 -0.9 -1.1 0.3 -0.8 0.0287485
2: sim_rep1.dat -0.2 0.2 0.7 1.0 1.8 -0.2 0.8 0.3 -1.3 -1.6 -0.2 0.4588433
3: sim_rep1.dat 1.6 -0.5 0.7 -0.7 -1.7 0.9 -1.2 -1.0 1.1 -0.3 -2.1 0.2432395
4: sim_rep1.dat 0.1 1.2 -1.3 -0.1 0.3 -0.6 0.4 0.3 0.8 -1.2 -1.7 0.8313184
5: sim_rep1.dat 0.1 0.2 -2.0 0.6 -0.3 0.2 0.2 0.5 -0.9 -0.8 -1.1 0.7738186
---
199996: sim_rep100.dat 0.1 -1.4 1.6 -0.7 -1.0 -0.6 0.8 -0.6 -0.5 -0.4 -0.8 0.1323889
199997: sim_rep100.dat 0.3 1.3 -2.4 -0.7 -0.4 0.0 1.0 -0.2 1.0 -0.1 0.3 0.6769959
199998: sim_rep100.dat 0.3 1.2 0.0 -1.3 -0.8 -0.7 -0.3 0.1 0.9 0.9 -1.3 0.7824498
199999: sim_rep100.dat 0.5 -0.7 0.2 0.5 1.1 -0.3 0.3 -0.5 -0.8 1.9 -0.7 0.2669799
200000: sim_rep100.dat -0.5 1.1 0.8 0.2 -0.6 -0.5 -0.4 1.1 -1.8 0.9 -1.3 0.9175867
DT consists now of 200 K rows. Performance is no reason to worry as data.tablewas built to deal with large (even larger) data efficiently.
The origin of each row can be identified in case the data of the individual data sets need to be treated separately. E.g.,
DT[origin == "sim_rep47.dat"]
origin ARAND w1 w2 w3 w4 w5 w6 w7 w8 w9 w10 ps_true
1: sim_rep47.dat -0.6 -0.5 0.2 -0.7 0.5 2.4 -0.2 -0.9 -1.1 0.3 -0.8 0.0287485
2: sim_rep47.dat -0.2 0.2 0.7 1.0 1.8 -0.2 0.8 0.3 -1.3 -1.6 -0.2 0.4588433
3: sim_rep47.dat 1.6 -0.5 0.7 -0.7 -1.7 0.9 -1.2 -1.0 1.1 -0.3 -2.1 0.2432395
4: sim_rep47.dat 0.1 1.2 -1.3 -0.1 0.3 -0.6 0.4 0.3 0.8 -1.2 -1.7 0.8313184
5: sim_rep47.dat 0.1 0.2 -2.0 0.6 -0.3 0.2 0.2 0.5 -0.9 -0.8 -1.1 0.7738186
---
1996: sim_rep47.dat 0.1 -1.4 1.6 -0.7 -1.0 -0.6 0.8 -0.6 -0.5 -0.4 -0.8 0.1323889
1997: sim_rep47.dat 0.3 1.3 -2.4 -0.7 -0.4 0.0 1.0 -0.2 1.0 -0.1 0.3 0.6769959
1998: sim_rep47.dat 0.3 1.2 0.0 -1.3 -0.8 -0.7 -0.3 0.1 0.9 0.9 -1.3 0.7824498
1999: sim_rep47.dat 0.5 -0.7 0.2 0.5 1.1 -0.3 0.3 -0.5 -0.8 1.9 -0.7 0.2669799
2000: sim_rep47.dat -0.5 1.1 0.8 0.2 -0.6 -0.5 -0.4 1.1 -1.8 0.9 -1.3 0.9175867
extracts all row belonging to data set sim_rep47.dat.
Data
For test and demonstration, I've created 100 sample data.frames using the code below:
# create vector of file names
temp <- paste0("sim_rep", 1:100, ".dat")
# create one sample data.frame
nr <- 2000L
nc <- 11L
set.seed(123L)
foo <- as.data.frame(matrix(round(rnorm(nr * nc), 1), nrow = nr))
names(foo) <- c("ARAND", paste0("w", 1:10))
str(foo)
# create 100 individually named data.frames by "copying" foo
for (t in temp) assign(t, foo)
# print warning message on using assign
fortunes::fortune(236)
# verify objects have been created
ls()
Addendum: Reading all files at once
The OP has named the single data.frames sim_rep1.dat, sim_rep2.dat, etc. which resemble typical file names. Just in case the OP indeed has 100 files on disk I would like to suggest a way to read all files at once. Let's suppose all files are stored in one directory.
# path to data directory
data_dir <- file.path("path", "to", "data", "directory")
# create vector of file paths
files <- dir(data_dir, pattern = "sim_rep\\d+\\.dat", full.names = TRUE)
# read all files and create one large data.table
# NB: it might be necessary to add parameters to fread()
# or to use another file reader depending on the file type
DT <- rbindlist(lapply(files, fread), idcol = "origin")
# rename origin to contain the file names without path
DT[, origin := factor(origin, labels = basename(files))]
DT
origin ARAND w1 w2 w3 w4 w5 w6 w7 w8 w9 w10 ps_true
1: sim_rep1.dat -0.6 -0.5 0.2 -0.7 0.5 2.4 -0.2 -0.9 -1.1 0.3 -0.8 0.0287485
2: sim_rep1.dat -0.2 0.2 0.7 1.0 1.8 -0.2 0.8 0.3 -1.3 -1.6 -0.2 0.4588433
3: sim_rep1.dat 1.6 -0.5 0.7 -0.7 -1.7 0.9 -1.2 -1.0 1.1 -0.3 -2.1 0.2432395
4: sim_rep1.dat 0.1 1.2 -1.3 -0.1 0.3 -0.6 0.4 0.3 0.8 -1.2 -1.7 0.8313184
5: sim_rep1.dat 0.1 0.2 -2.0 0.6 -0.3 0.2 0.2 0.5 -0.9 -0.8 -1.1 0.7738186
---
199996: sim_rep99.dat 0.1 -1.4 1.6 -0.7 -1.0 -0.6 0.8 -0.6 -0.5 -0.4 -0.8 0.1323889
199997: sim_rep99.dat 0.3 1.3 -2.4 -0.7 -0.4 0.0 1.0 -0.2 1.0 -0.1 0.3 0.6769959
199998: sim_rep99.dat 0.3 1.2 0.0 -1.3 -0.8 -0.7 -0.3 0.1 0.9 0.9 -1.3 0.7824498
199999: sim_rep99.dat 0.5 -0.7 0.2 0.5 1.1 -0.3 0.3 -0.5 -0.8 1.9 -0.7 0.2669799
200000: sim_rep99.dat -0.5 1.1 0.8 0.2 -0.6 -0.5 -0.4 1.1 -1.8 0.9 -1.3 0.9175867
All data sets are now stored in one large data.table DT consisting of 200 k rows. However, the order of data sets is different as files is sorted alphabetically, i.e.,
head(files)
[1] "./data/sim_rep1.dat" "./data/sim_rep10.dat" "./data/sim_rep100.dat"
[4] "./data/sim_rep11.dat" "./data/sim_rep12.dat" "./data/sim_rep13.dat"
probably just need single brackets.
test = data.frame('w1' = c(1,2,3),'w2' = c(2,3,4))
temp = list(test,test,test)
temp2 = lapply(temp,function(x){cbind(x,setNames(x['w1'] + x['w2'],'ps_true'))})
temp2
[[1]]
w1 w2 ps_true
1 1 2 3
2 2 3 5
3 3 4 7
[[2]]
w1 w2 ps_true
1 1 2 3
2 2 3 5
3 3 4 7
[[3]]
w1 w2 ps_true
1 1 2 3
2 2 3 5
3 3 4 7
I have a data table similar to this except much larger:
set.seed(1)
dt <- data.table(t1=round(rnorm(5),1), t2=round(rnorm(5),1), t3=round(rnorm(5),1),
t4=round(rnorm(5),1), t5=round(rnorm(5),1), t6=round(rnorm(5),1),
t7=round(rnorm(5),1),t8=round(rnorm(5),1))
Which outputs:
t1 t2 t3 t4 t5 t6 t7 t8
1: -0.6 -0.8 1.5 0.0 0.9 -0.1 1.4 -0.4
2: 0.2 0.5 0.4 0.0 0.8 -0.2 -0.1 -0.4
3: -0.8 0.7 -0.6 0.9 0.1 -1.5 0.4 -0.1
4: 1.6 0.6 -2.2 0.8 -2.0 -0.5 -0.1 1.1
5: 0.3 -0.3 1.1 0.6 0.6 0.4 -1.4 0.8
I would like to rename columns t3:t8 as hour_t3:hour_t8, to output like this:
t1 t2 hour_t3 hour_t4 hour_t5 hour_t6 hour_t7 hour_t8
1: -0.6 -0.8 1.5 0.0 0.9 -0.1 1.4 -0.4
2: 0.2 0.5 0.4 0.0 0.8 -0.2 -0.1 -0.4
3: -0.8 0.7 -0.6 0.9 0.1 -1.5 0.4 -0.1
4: 1.6 0.6 -2.2 0.8 -2.0 -0.5 -0.1 1.1
5: 0.3 -0.3 1.1 0.6 0.6 0.4 -1.4 0.8
These two methods work:
names(dt)[3:8] <- c(paste0("hour_t", 3:8))
and
setnames(dt, 3:8, c(paste0("hour_t", 3:8)))
but, I would like to be able to subset by reference using something like this:
setnames(dt, "t3":"t8", c(paste0("hour_t", 3:8)))
When I use such syntax or subset with c("t3":"t8"), I get the following error:
Error in "t3":"t8" : NA/NaN argument
In addition: Warning messages:
1: In setnames(dt, c("t3":"t8"), c(paste0("hour_t", 3:8))) :
NAs introduced by coercion
2: In setnames(dt, c("t3":"t8"), c(paste0("hour_t", 3:8))) :
NAs introduced by coercion
Any thoughts on how to subset the columns to rename by reference/column name instead of by position would be greatly appreciated. Thanks.
I am still quite new to data.table and am using data.table version 1.9.6.
I have a data table similar to this (except it has 150 columns and about 5 million rows):
set.seed(1)
dt <- data.table(ID=1:10, Status=c(rep("OUT",2),rep("IN",2),"ON",rep("OUT",2),rep("IN",2),"ON"),
t1=round(rnorm(10),1), t2=round(rnorm(10),1), t3=round(rnorm(10),1),
t4=round(rnorm(10),1), t5=round(rnorm(10),1), t6=round(rnorm(10),1),
t7=round(rnorm(10),1),t8=round(rnorm(10),1))
which outputs:
ID Status t1 t2 t3 t4 t5 t6 t7 t8
1: 1 OUT -0.6 1.5 0.9 1.4 -0.2 0.4 2.4 0.5
2: 2 OUT 0.2 0.4 0.8 -0.1 -0.3 -0.6 0.0 -0.7
3: 3 IN -0.8 -0.6 0.1 0.4 0.7 0.3 0.7 0.6
4: 4 IN 1.6 -2.2 -2.0 -0.1 0.6 -1.1 0.0 -0.9
5: 5 ON 0.3 1.1 0.6 -1.4 -0.7 1.4 -0.7 -1.3
6: 6 OUT -0.8 0.0 -0.1 -0.4 -0.7 2.0 0.2 0.3
7: 7 OUT 0.5 0.0 -0.2 -0.4 0.4 -0.4 -1.8 -0.4
8: 8 IN 0.7 0.9 -1.5 -0.1 0.8 -1.0 1.5 0.0
9: 9 IN 0.6 0.8 -0.5 1.1 -0.1 0.6 0.2 0.1
10: 10 ON -0.3 0.6 0.4 0.8 0.9 -0.1 2.2 -0.6
Using data.table, I would like to add a new column (using :=) called Total that would contain the following:
For each row,
if Status=OUT, sum columns t1:t4 and t8
if Status=IN, sum columns t5,t6,t8
if Status=ON, sum columns t1:t3 and t6:t8
The final output should look like this:
ID Status t1 t2 t3 t4 t5 t6 t7 t8 Total
1: 1 OUT -0.6 1.5 0.9 1.4 -0.2 0.4 2.4 0.5 3.7
2: 2 OUT 0.2 0.4 0.8 -0.1 -0.3 -0.6 0.0 -0.7 0.6
3: 3 IN -0.8 -0.6 0.1 0.4 0.7 0.3 0.7 0.6 1.6
4: 4 IN 1.6 -2.2 -2.0 -0.1 0.6 -1.1 0.0 -0.9 -1.4
5: 5 ON 0.3 1.1 0.6 -1.4 -0.7 1.4 -0.7 -1.3 1.4
6: 6 OUT -0.8 0.0 -0.1 -0.4 -0.7 2.0 0.2 0.3 -1.0
7: 7 OUT 0.5 0.0 -0.2 -0.4 0.4 -0.4 -1.8 -0.4 -0.5
8: 8 IN 0.7 0.9 -1.5 -0.1 0.8 -1.0 1.5 0.0 -0.2
9: 9 IN 0.6 0.8 -0.5 1.1 -0.1 0.6 0.2 0.1 0.6
10: 10 ON -0.3 0.6 0.4 0.8 0.9 -0.1 2.2 -0.6 2.2
I am fairly new to data.table (currently using version 1.9.6) and would like to try for a solution using efficient data.table syntax.
I think doing it one by one, as suggested in comments, is perfectly fine, but you can also create a lookup table:
cond = data.table(Status = c("OUT", "IN", "ON"),
cols = Map(paste0, 't', list(c(1:4, 8), c(5,6,8), c(1:3, 6:8))))
# Status cols
#1: OUT t1,t2,t3,t4,t8
#2: IN t5,t6,t8
#3: ON t1,t2,t3,t6,t7,t8
dt[cond, Total := Reduce(`+`, .SD[, cols[[1]], with = F]), on = 'Status', by = .EACHI]
# ID Status t1 t2 t3 t4 t5 t6 t7 t8 Total
# 1: 1 OUT -0.6 1.5 0.9 1.4 -0.2 0.4 2.4 0.5 3.7
# 2: 2 OUT 0.2 0.4 0.8 -0.1 -0.3 -0.6 0.0 -0.7 0.6
# 3: 3 IN -0.8 -0.6 0.1 0.4 0.7 0.3 0.7 0.6 1.6
# 4: 4 IN 1.6 -2.2 -2.0 -0.1 0.6 -1.1 0.0 -0.9 -1.4
# 5: 5 ON 0.3 1.1 0.6 -1.4 -0.7 1.4 -0.7 -1.3 1.4
# 6: 6 OUT -0.8 0.0 -0.1 -0.4 -0.7 2.0 0.2 0.3 -1.0
# 7: 7 OUT 0.5 0.0 -0.2 -0.4 0.4 -0.4 -1.8 -0.4 -0.5
# 8: 8 IN 0.7 0.9 -1.5 -0.1 0.8 -1.0 1.5 0.0 -0.2
# 9: 9 IN 0.6 0.8 -0.5 1.1 -0.1 0.6 0.2 0.1 0.6
#10: 10 ON -0.3 0.6 0.4 0.8 0.9 -0.1 2.2 -0.6 2.2
I need to apply function to vector b3, but insert 0 (and do not apply function) when b3 < 0.
> b3
[1] 1.5 0.5 5.5 0.5 9.5 8.5 4.5 3.5 3.5 3.5 1.5 -0.5 1.5 5.5 9.5
[16] 5.5 2.5 1.5 2.5 3.5 9.5 -0.5 4.5 2.5 1.5 4.5 -0.5 -0.5 1.5 0.5
> unlist(lapply(b3, function(x) {seq(from = 0.5, to = x)}))
[1] 0.5 1.5 0.5 0.5 1.5 2.5 3.5 4.5 5.5 0.5 0.5 1.5 2.5 3.5 4.5
[16] 5.5 6.5 7.5 8.5 9.5 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 0.5
[31] 1.5 2.5 3.5 4.5 0.5 1.5 2.5 3.5 0.5 1.5 2.5 3.5 0.5 1.5 2.5
[46] 3.5 0.5 1.5 0.5 -0.5 0.5 1.5 0.5 1.5 2.5 3.5 4.5 5.5 0.5 1.5
[61] 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5 0.5 1.5 2.5 3.5 4.5 5.5 0.5
[76] 1.5 2.5 0.5 1.5 0.5 1.5 2.5 0.5 1.5 2.5 3.5 0.5 1.5 2.5 3.5
[91] 4.5 5.5 6.5 7.5 8.5 9.5 0.5 -0.5 0.5 1.5 2.5 3.5 4.5 0.5 1.5
[106] 2.5 0.5 1.5 0.5 1.5 2.5 3.5 4.5 0.5 -0.5 0.5 -0.5 0.5 1.5 0.5
gives me close to what I need, but I have problems with negatives values of b3. So how to introduce 0 each times function get to negative value in b3 ? Here's what I came with :
> unlist(lapply(b3, function(x) {ifelse(x>0,seq(from = 0.5, to = x),0)}))
[1] 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.0 0.5 0.5 0.5 0.5 0.5 0.5 0.5
[20] 0.5 0.5 0.0 0.5 0.5 0.5 0.5 0.0 0.0 0.5 0.5
works for negative values, but my function doesn't behave as expected...
Sorry for the previous answer, I read too quickly. You probably just want to use if rather than ifelse:
#Small, similar example
b3 <- c(1.5,0.5,5.5,-2.5,1.5)
unlist(lapply(b3, function(x) {if (x < 0) 0 else seq(from = 0.5, to = x)}))
[1] 0.5 1.5 0.5 0.5 1.5 2.5 3.5 4.5 5.5 0.0 0.5 1.5