I have 100+ files and have starting and ending coordinates for each file. So based on starting and ending coordinates, I want to extract the regions from all data sets and want to store in file. I have used following approach but its not giving me the expected out put.
startco have the starting indices of 1st 2nd 3rd file respectively and endco have ending indices of 1st 2nd 3rd file respectively. And if the indices is going beyond the files indices want to put NA
Example:
startco<-c(3,4,1)
endco<-c(5,6,2)
ctc<-c(1,2,3)
for (hm0 in 1:length(ctc)) {
for (hm1 in 1:length(startco)) {
for (hm2 in 1:length(endco)) {
methd1<-read.table( paste0("path/to folder/","file_",ctc[hm0],".txt"))
methd2<- methd1[,startco[hm1]:endco[hm2]]
}
}
}
File_1.txt
V1 V2 V3 V4 V5
41 42 43 45 46
0.31 0.21 0.87 0.65 0.54
0.32 0.28 0.74 0.87 0.65
0.19 0.12 0.99 0.99 0.89
File_2.txt
V1 V2 V3 V4 V5
12 24 13 14 16
0.89 0.78 0.50 0.22 0.34
0.54 0.78 0.50 0.34 0.41
0.78 0.54 0.66 0.26 0.14
File_3.txt
V1 V2 V3 V4 V5
1 2 3 5 6
0.20 0.40 0.50 0.49 0.52
Expected output :
43 45 46
0.87 0.65 0.54
0.74 0.87 0.65
0.99 0.99 0.89
0.22 0.34 NA
0.34 0.41 NA
0.99 0.89 NA
1 2
0.20 0.40
in Base R you could do:
fun <- function(path, start, end){
id <- basename(path)
dat <- read.table(path, header = TRUE)
p <- ncol(dat)
n <- nrow(dat)
neg <- if(start<0) -start else 0
add <- matrix(nrow = n, ncol = neg)
if (start < 1) start <- 1
if (end > p) end <- p
d <- cbind(add, dat[, start:end])
names(d) <- paste0('X', seq(ncol(d)))
cbind(id,r = seq(nrow(d)), d)
}
startco<-c(3,4,-2) # TAKES NEGATIVE INDICES
endco<-c(5,6,2)
ctc<-c(1,2,3)
files <- file.path('path/to/folder', ctc)
A <- Map(fun, files, startco, endco)
Reduce(function(x, y)merge(x,y, all =TRUE), A)[, -(1:2)]
X1 X2 X3 X4
1 43.00 45.00 46.00 NA
2 0.87 0.65 0.54 NA
3 0.74 0.87 0.65 NA
4 0.99 0.99 0.89 NA
5 14.00 16.00 NA NA
6 0.22 0.34 NA NA
7 0.34 0.41 NA NA
8 0.26 0.14 NA NA
9 NA NA 1.00 2.0
10 NA NA 0.20 0.4
The one with no negatives
startco<-c(3,4,1)
B <- Map(fun, files, startco, endco)
Reduce(function(x, y)merge(x,y, all =TRUE), B)[, -(1:2)]
X1 X2 X3
1 43.00 45.00 46.00
2 0.87 0.65 0.54
3 0.74 0.87 0.65
4 0.99 0.99 0.89
5 14.00 16.00 NA
6 0.22 0.34 NA
7 0.34 0.41 NA
8 0.26 0.14 NA
9 1.00 2.00 NA
10 0.20 0.40 NA
I would use a readfun,
readfun <- \(i, s, e) {
stopifnot(s != 0)
r <- read.table(paste0("foo1/", "file_", i, ".txt"), header=TRUE)
if (e > ncol(r)) { ## inserts cols to the right if e > ncol
e1 <- e - ncol(r)
nm <- paste0('V', as.numeric(substring(colnames(r), 2)[ncol(r)]) + seq_len(e1))
m <- matrix(NA_real_, nrow(r), e1, dimnames=list(NULL, nm))
r <- cbind(r, m)
}
if (s < 0) { ## inserts cols to the left if s < 0
m <- matrix(NA_real_, nrow(r), -s)
r <- cbind(m, r)
e <- e + -s
s <- 1
}
out <- r[, s:e]
unname(as.matrix(out))
}
in Map.
ctc <- c(1, 2, 3); startco <- c(3, 4, -2); endco <- c(5, 6, 2)
Map(readfun, ctc, startco, endco)
# [[1]]
# [,1] [,2] [,3]
# [1,] 43.00 45.00 46.00
# [2,] 0.87 0.65 0.54
# [3,] 0.74 0.87 0.65
# [4,] 0.99 0.99 0.89
#
# [[2]]
# [,1] [,2] [,3]
# [1,] 14.00 16.00 NA
# [2,] 0.22 0.34 NA
# [3,] 0.34 0.41 NA
# [4,] 0.26 0.14 NA
#
# [[3]]
# [,1] [,2] [,3] [,4]
# [1,] NA NA 1.0 2.0
# [2,] NA NA 0.2 0.4
Data:
dir.create('foo1')
write.table(read.table(header=TRUE, text='
V1 V2 V3 V4 V5
41 42 43 45 46
0.31 0.21 0.87 0.65 0.54
0.32 0.28 0.74 0.87 0.65
0.19 0.12 0.99 0.99 0.89'), './foo1/file_1.txt', row.names=F)
write.table(read.table(header=TRUE, text='
V1 V2 V3 V4 V5
12 24 13 14 16
0.89 0.78 0.50 0.22 0.34
0.54 0.78 0.50 0.34 0.41
0.78 0.54 0.66 0.26 0.14'), './foo1/file_2.txt', row.names=F)
write.table(read.table(header=TRUE, text='
V1 V2 V3 V4 V5
1 2 3 5 6
0.20 0.40 0.50 0.49 0.52 '), './foo1/file_3.txt', row.names=F)
Related
# Create a data frame
> df <- data.frame(a = rnorm(7), b = rnorm(7), c = rnorm(7), threshold = rnorm(7))
> df <- round(abs(df), 2)
>
> df
a b c threshold
1 1.17 0.27 1.26 0.19
2 1.41 1.57 1.23 0.97
3 0.16 0.11 0.35 1.34
4 0.03 0.04 0.10 1.50
5 0.23 1.10 2.68 0.45
6 0.99 1.36 0.17 0.30
7 0.28 0.68 1.22 0.56
>
>
# Replace values in columns a, b, and c with NA if > value in threshold
> df[1:3][df[1:3] > df[4]] <- "NA"
Error in Ops.data.frame(df[1:3], df[4]) :
‘>’ only defined for equally-sized data frames
There could be some obvious solutions that I am incapable of producing. The intent is to replace values in columns "a", "b", and "c" with NA if the value is larger than that in "threshold". And I need to do that row-by-row.
If I had done it right, the df would look like this:
a b c threshold
1 NA NA NA 0.19
2 NA NA NA 0.97
3 0.16 0.11 0.35 1.34
4 0.03 0.04 0.10 1.50
5 0.23 NA NA 0.45
6 NA NA 0.17 0.30
7 0.28 NA NA 0.56
I had also tried the apply() approach but to no avail. Can you help, please??
You should use dplyr for most of such use cases.
One way below:
> set.seed(10)
> df <- data.frame(a = rnorm(7), b = rnorm(7), c = rnorm(7), threshold = rnorm(7))
> df <- round(abs(df), 2)
> df
a b c threshold
1 0.02 0.36 0.74 2.19
2 0.18 1.63 0.09 0.67
3 1.37 0.26 0.95 2.12
4 0.60 1.10 0.20 1.27
5 0.29 0.76 0.93 0.37
6 0.39 0.24 0.48 0.69
7 1.21 0.99 0.60 0.87
>
> df %>%
+ mutate_at(vars(a:c), ~ifelse(.x > df$threshold, NA, .x))
a b c threshold
1 0.02 0.36 0.74 2.19
2 0.18 NA 0.09 0.67
3 1.37 0.26 0.95 2.12
4 0.60 1.10 0.20 1.27
5 0.29 NA NA 0.37
6 0.39 0.24 0.48 0.69
7 NA NA 0.60 0.87
You can use apply function across dataframe
df[,c(1:3)]<- apply(df[,c(1:3),drop=F], 2, function(x){ ifelse(x>df[,4],NA,x)})
The problem with your code was the usage of df[4] instead of df[, 4]. The difference is that df[4] returns a data.frame with one column and df[, 4] returns a vector.
That's why
df[1:3] > df[4]
returns
error in Ops.data.frame(df[1:3], df[4]) :
‘>’ only defined for equally-sized data frames
While this works as expected
df[1:3][df[1:3] > df[, 4]] <- NA
df
# a b c threshold
#1 0.63 0.74 NA 0.78
#2 NA NA 0.04 0.07
#3 0.84 0.31 0.02 1.99
#4 NA NA NA 0.62
#5 NA NA NA 0.06
#6 NA NA NA 0.16
#7 0.49 NA 0.92 1.47
data
set.seed(1)
df <- data.frame(a = rnorm(7), b = rnorm(7), c = rnorm(7), threshold = rnorm(7))
df <- round(abs(df), 2)
You can use a for-loop like this:
for(i in 1:(ncol(df)-1)){
df[, i] <- ifelse(df[, i] > df[, 4], NA, df[, i])
}
I have the following dataframe (DF_A):
PARTY_ID PROBS_3001 PROBS_3002 PROBS_3003 PROBS_3004 PROBS_3005 PROBS_3006 PROBS_3007 PROBS_3008
1: 1000000 0.03 0.58 0.01 0.42 0.69 0.98 0.55 0.96
2: 1000001 0.80 0.37 0.10 0.95 0.77 0.69 0.23 0.07
3: 1000002 0.25 0.73 0.79 0.83 0.24 0.82 0.81 0.01
4: 1000003 0.10 0.96 0.53 0.59 0.96 0.10 0.98 0.76
5: 1000004 0.36 0.87 0.76 0.03 0.95 0.40 0.53 0.89
6: 1000005 0.15 0.78 0.24 0.21 0.03 0.87 0.67 0.64
And I have this other dataframe (DF_B):
V1 V2 V3 V4 PARTY_ID
1 0.58 0.69 0.96 0.98 1000000
2 0.69 0.77 0.80 0.95 1000001
3 0.79 0.81 0.82 0.83 1000002
4 0.76 0.96 0.96 0.98 1000003
5 0.76 0.87 0.89 0.95 1000004
6 0.64 0.67 0.78 0.87 1000005
I need to find the position of the elements of the DF_A in the DF_B to have something like this:
PARTY_ID P1 P2 P3 P4
1 1000000 3 6 9 7
...
Currently I'm working with match function but it takes a lot of time (I have 400K rows). I'm doing this:
i <- 1
while(i < nrow(DF_A)){
position <- match(DF_B[i,],DF_A[i,])
i <- i + 1
}
Although it works, it's very slow and I know that it's not the best answer to my problem. Can anyone help me please??
You can merge and then Map with a by group operation:
df_a2 <- df_a[setDT(df_b), on = "PARTY_ID"]
df_a3 <- df_a2[, c(PARTY_ID,
Map(f = function(x,y) which(x==y),
x = list(.SD[,names(df_a), with = FALSE]),
y = .SD[, paste0("V",1:4), with = FALSE])), by = 1:nrow(df_a2)]
setnames(df_a3, paste0("V",1:5), c("PARTY_ID", paste0("P", 1:4)))[,nrow:=NULL]
df_a3
# PARTY_ID P1 P2 P3 P4
#1: 1000000 3 6 9 7
#2: 1000001 7 6 2 5
#3: 1000002 4 8 7 5
#4: 1000003 9 3 3 8
#5: 1000003 9 6 6 8
#6: 1000004 4 3 9 6
#7: 1000005 9 8 3 7
Here is an example on 1 milion rows with two columns. It takes 14 ms on my computer.
# create data tables with matching ids but on different positions
x <- as.data.table(data.frame(id=sample(c(1:1000000), 1000000, replace=FALSE), y=sample(LETTERS, 1000000, replace=TRUE)))
y <- as.data.table(data.frame(id=sample(c(1:1000000), 1000000, replace=FALSE), z=sample(LETTERS, 1000000, replace=TRUE)))
# add column to both data tables which will store the position in x and y
x$x_row_nr <- 1:nrow(x)
y$y_row_nr <- 1:nrow(y)
# set key in both data frames using matching columns name
setkey(x, "id")
setkey(y, "id")
# merge data tables into one
z <- merge(x,y)
# now you just use this to extract what is the position
# of 100 hundreth record in x data table in y data table
z[x_row_nr==100, y_row_nr]
z will contain matching row records from both datasets with there columns attached.
I would like to sort values in columns of the xy1 dataframe, based on the increasing order of values in columns of the xy dataframe.
x <- c(3,1,7,45,22,2)
y <- c(23,65,1,23,2,11)
xy <- data.frame(x,y)
x1 <- c(0.34,0.3,0.7,0.22,0.67,0.87)
y1 <- c(0.4,0.13,0.17,0.72,0.61,0.7)
xy1 <- data.frame(x1,y1)
> xy
x y
1 3 23
2 1 65
3 7 1
4 45 23
5 22 2
6 2 11
> xy1
x1 y1
1 0.34 0.40
2 0.30 0.13
3 0.70 0.17
4 0.22 0.72
5 0.67 0.61
6 0.87 0.70
The following is a new data.frame result that I desire - note it deals with repeated observations (two the same values in y). x1 and y1 are now sorted according to the order of values in each column of xy dataframe.
x1 y1
1 0.30 0.17
2 0.87 0.61
3 0.34 0.70
4 0.70 0.40
5 0.67 0.72
6 0.22 0.13
You can use the order function to get the sorting order of a vector.
x <- c(3,1,7,45,22,2)
y <- c(23,65,1,23,2,11)
xy <- data.frame(x,y)
x1 <- c(0.34,0.3,0.7,0.22,0.67,0.87)
y1 <- c(0.4,0.13,0.17,0.72,0.61,0.7)
xy1 <- data.frame(x1,y1)
result <- data.frame(x1[order(x)], y1[order(y)])
result
This produces
x1.order.x.. y1.order.y..
1 0.30 0.17
2 0.87 0.61
3 0.34 0.70
4 0.70 0.40
5 0.67 0.72
6 0.22 0.13
You can beautify the output by setting the column names in the result:
data.frame(x1=x1[order(x)], y1=y1[order(y)])
Now if you don't want to manually type in everything but have two data frames with the same dimensions that you can use this one-liner
sapply(1:ncol(xy1), function(i) {xy1[order(xy[,i]), i]})
which produces
[,1] [,2]
[1,] 0.30 0.17
[2,] 0.87 0.61
[3,] 0.34 0.70
[4,] 0.70 0.40
[5,] 0.67 0.72
[6,] 0.22 0.13
As this is based on ordering corresponding columns on both datasets, Map can be used
xy1[] <- Map(function(x,y) x[order(y)], xy1, xy)
xy1
# x1 y1
#1 0.30 0.17
#2 0.87 0.61
#3 0.34 0.70
#4 0.70 0.40
#5 0.67 0.72
#6 0.22 0.13
Or another option is to order based on the col of 'xy', 'xy'
xy1[] <- as.matrix(xy1)[order(col(xy), xy)]
xy1
# x1 y1
#1 0.30 0.17
#2 0.87 0.61
#3 0.34 0.70
#4 0.70 0.40
#5 0.67 0.72
#6 0.22 0.13
You could try this:
library(tidyverse)
df_1 <- xy %>%
bind_cols(xy1) %>%
arrange(x) %>%
select(x1)
df_2 <- xy %>%
bind_cols(xy1) %>%
arrange(y) %>%
select(y1)
df <- bind_cols(df_1, df_2)
Which returns:
# A tibble: 6 x 2
x1 y1
<dbl> <dbl>
1 0.30 0.17
2 0.87 0.61
3 0.34 0.70
4 0.70 0.40
5 0.67 0.72
6 0.22 0.13
Basically just arrange x1 and y1 by x and y separately, then combine x1 and y1.
I am having some problems sorting my dataset into bins, that based on the numeric value of the data value. I tried doing it with the function shingle from the lattice which seem to split it accurately.
I can't seem to extract the desired output which is the knowledge how the data is divided into the predefined bins. I seem only able to print it.
bin_interval = matrix(c(0.38,0.42,0.46,0.50,0.54,0.58,0.62,0.66,0.70,0.74,0.78,0.82,0.86,0.90,0.94,0.98,
0.40,0.44,0.48,0.52,0.56,0.60,0.64,0.68,0.72,0.76,0.80,0.84,0.88,0.92,0.96,1.0),
ncol = 2, nrow = 16)
bin_1 = shingle(data_1,intervals = bin_interval)
How do i extract the intervals which is outputted by the shingle function, and not only print it...
the intervals being the output:
Intervals:
min max count
1 0.38 0.40 0
2 0.42 0.44 6
3 0.46 0.48 46
4 0.50 0.52 251
5 0.54 0.56 697
6 0.58 0.60 1062
7 0.62 0.64 1215
8 0.66 0.68 1227
9 0.70 0.72 1231
10 0.74 0.76 1293
11 0.78 0.80 1330
12 0.82 0.84 1739
13 0.86 0.88 2454
14 0.90 0.92 3048
15 0.94 0.96 8936
16 0.98 1.00 71446
As an variable, that can be fed to another function.
The shingle() function returns the values using attributes().
The levels are specifically given by attr(bin_1,"levels").
So:
set.seed(1337)
data_1 = runif(100)
bin_interval = matrix(c(0.38,0.42,0.46,0.50,0.54,0.58,0.62,0.66,0.70,0.74,0.78,0.82,0.86,0.90,0.94,0.98,
0.40,0.44,0.48,0.52,0.56,0.60,0.64,0.68,0.72,0.76,0.80,0.84,0.88,0.92,0.96,1.0),
ncol = 2, nrow = 16)
bin_1 = shingle(data_1,intervals = bin_interval)
attr(bin_1,"levels")
This gives:
[,1] [,2]
[1,] 0.38 0.40
[2,] 0.42 0.44
[3,] 0.46 0.48
[4,] 0.50 0.52
[5,] 0.54 0.56
[6,] 0.58 0.60
[7,] 0.62 0.64
[8,] 0.66 0.68
[9,] 0.70 0.72
[10,] 0.74 0.76
[11,] 0.78 0.80
[12,] 0.82 0.84
[13,] 0.86 0.88
[14,] 0.90 0.92
[15,] 0.94 0.96
[16,] 0.98 1.00
Edit
The count information for each interval is only computed within the print.shingle method. Thus, you would need to run the following code:
count.shingle = function(x){
l <- levels(x)
n <- nlevels(x)
int <- data.frame(min = numeric(n), max = numeric(n),
count = numeric(n))
for (i in 1:n) {
int$min[i] <- l[[i]][1]
int$max[i] <- l[[i]][2]
int$count[i] <- length(x[x >= l[[i]][1] & x <= l[[i]][2]])
}
int
}
a = count.shingle(bin_1)
This gives:
> a
min max count
1 0.38 0.40 0
2 0.42 0.44 1
3 0.46 0.48 3
4 0.50 0.52 1
5 0.54 0.56 2
6 0.58 0.60 2
7 0.62 0.64 2
8 0.66 0.68 4
9 0.70 0.72 1
10 0.74 0.76 3
11 0.78 0.80 2
12 0.82 0.84 2
13 0.86 0.88 5
14 0.90 0.92 1
15 0.94 0.96 1
16 0.98 1.00 2
where a$min is lower range, a$max is upper range, and a$count is the number within the bins.
I have generated a file that looks like this (here is an extract, roughly each one of two files with names having columns 1 and 2 in common are multiplied by a different parameter and the best fit, i.e. the lowest chi2, is returned in column 3 with the appropriate parameters in columns 4 and 5):
10 05 0.42 0.13 0.01
10 10 0.30 0.12 0.01
10 15 0.25 0.11 0.07
15 05 0.29 0.12 0.01
15 10 0.25 0.11 0.06
15 15 0.23 0.10 0.02
20 05 0.25 0.11 0.03
20 10 0.23 0.12 0.04
20 15 0.23 0.13 0.05
25 05 0.23 0.10 0.03
25 10 0.23 0.10 0.08
25 15 0.24 0.09 0.05
I am starting/learning to use lists as my codes are really to slow using for loops (currently I am using 4 for loops so it's insanely long), and I don't know enough to re-write my optimisation code so it does not take 8 hours to work. So instead, to reorganise the output I was wondering if it would be possible to create a list, say mytemplist, that reads:
> mytemplist
$5
[1] 10 15 20 25
[2] 0.42 0.29 0.25 0.23
[3] 0.13 0.12 0.11 0.10
[4] 0.01 0.01 0.03 0.03
$10
[1] 10 15 20 25
[2] 0.30 0.25 0.23 0.23
[3] 0.12 0.11 0.12 0.10
[4] 0.01 0.06 0.04 0.08
$15
[1] 10 15 20 25
[2] 0.25 0.23 0.23 0.24
[3] 0.11 0.10 0.13 0.09
[4] 0.07 0.02 0.05 0.05
I have looked at questions about lists and I could only sort that out by creating lists within lists which is not helping here.
EDIT:
the accepted answer replies to the specific question above, to answer #rawr post I join how the file is generated (it's not pretty, I am not using opt so far as I will evolve the code to optimise with bigger freedom around the data points):
note: typical file to read are 2-column file (just lists of numbers) and named a10b05s and a10b05t
dataname is also a 2-column file
in those 3 files the first column is the same and represents the pivots
need to find par[1] and par[2] such that par[1]*a10b05s + par[2]*a10b05t best fit data
par <- rep(NA, 2)
pivot <- read.table(dataname)[[1]]
data2fit <- read.table(dataname)[[2]]
for (i in 1:10){
vala <- 10+5*(i-1)
namei <- paste("a", vala, sep="")
for (j in 1:10){
#creates a coordinates for storage
cglobal <- (i-1) * 10 + j
valb <- 5+5*(j-1)
namej1 <- paste(namei, "b", valb, "s", sep="")
namej2 <- paste(namei, "b", valb, "t", sep="")
infile1 <- read.table(namej1)
infile2 <- read.table(namej2)
# infile1 prominent wrt infile2 so first quick determination of par1
tempspace1 <- seq(0.001, 0.009, 0.001)
par1_s1 <- c(tempspace1, tempspace1*10, tempspace1*100)
opt1_par1 <- rep(NA, length(par1))
# set a pivot for comparison at position named temppivot find par1 wrt temppivot
for(k in 1:length(par1){
opt1_par1[k] <- abs(par1_s1[k]*infile1[[1]][temppivot] - data2fit[temppivot])
}
par[1] <- par1_s1[match(min(opt1_par1)), opt1_par1]
# set a space for a finer fit for par[1]
par1_s2 <- seq(par[1]-5*par[1]/10, par[1]+5*par[1]/10, par[1]/100)
# set a space to fit par[2] note that there is an option in the code to choose btw 0.001-0.01, 0.01-0.1 etc.
tempspace2 <- seq(0.001, 0.009, 0.0001)
par2 <- c(tempspace2, tempspace2*10, tempspace2*100)
chi2 <- rep(NA, length(par1_s2)*length(par2))
#data2fit
for(z in 1:length(par1_s2)){
for(w in 1:length(par2)){
par[1] <- par1_s2[z]
par[2] <- par2[w]
thesum <- rep(NA, length(pivot))
for(h in 1:length(pivot)){
c1 <- pivot[h]
thesum[h] <- par[1] * infile[[1]][c1] + par[2] * infile2[[1]][c1]
}
c2 <- (z-1) * length(par2) + w
chi2[c2] <- sum((thesum-data2fit)^2/thesum)
}
}
whichbestfit <- match(min(chi2), chi2)
chi2min <- min(chi2)
localparfinder <- function(x){
temp1 <- trunc(x/length(par2)) + 1
temp2 <- x - (temp1 -1) * length(par2)
y <- c(par1_s2[temp1], par2[temp2])
}
par <- localparfinder(whichbestfit)
# creates the table of the original post
storage[cglobal,] <- c(vala, valb, chi2min, par[1], par[2])
}
}
write.table(storage, file=paste("storage_", format(Sys.time(), "%d%b_%H%M"), ".dat", sep="")
You can use by and transpose, like this:
by(mydf[-2], mydf[[2]], t)
# mydf[[2]]: 5
# 1 4 7 10
# V1 10.00 15.00 20.00 25.00
# V3 0.42 0.29 0.25 0.23
# V4 0.13 0.12 0.11 0.10
# V5 0.01 0.01 0.03 0.03
# -----------------------------------------------------------
# mydf[[2]]: 10
# 2 5 8 11
# V1 10.00 15.00 20.00 25.00
# V3 0.30 0.25 0.23 0.23
# V4 0.12 0.11 0.12 0.10
# V5 0.01 0.06 0.04 0.08
# -----------------------------------------------------------
# mydf[[2]]: 15
# 3 6 9 12
# V1 10.00 15.00 20.00 25.00
# V3 0.25 0.23 0.23 0.24
# V4 0.11 0.10 0.13 0.09
# V5 0.07 0.02 0.05 0.05
The result of the above is a list with a class of by. If you used unclass on it, it would be similar to the split + lapply approach.
Here's one possibility:
lst <- split(df[-2], df$V2) # split according to colum 2 and drop column 2 in the output
lst <- lapply(lst, t) # transpose each list element
lst
# $`5`
# 1 4 7 10
#V1 10.00 15.00 20.00 25.00
#V3 0.42 0.29 0.25 0.23
#V4 0.13 0.12 0.11 0.10
#V5 0.01 0.01 0.03 0.03
#
#$`10`
# 2 5 8 11
#V1 10.00 15.00 20.00 25.00
#V3 0.30 0.25 0.23 0.23
#V4 0.12 0.11 0.12 0.10
#V5 0.01 0.06 0.04 0.08
#
#$`15`
# 3 6 9 12
#V1 10.00 15.00 20.00 25.00
#V3 0.25 0.23 0.23 0.24
#V4 0.11 0.10 0.13 0.09
#V5 0.07 0.02 0.05 0.05
If you like it more compact, you could also nest the split and lapply like this:
lst <- lapply(split(df[-2], df$V2), t)