I created a vector list, aa, with 50 elements. And I need to split aa into two vector lists called bb and cc. bb has the first 20 elements of aa while cc has the last 30 elements of aa. How do I do it?
Creation of original vector list
aa <- list (sample (1:50))
aa
#[[1]]
# [1] 29 30 39 45 17 11 43 14 24 34 3 1 28 2 21 23 6 31 5 27 44 7 4 46 49 22 33 38 50 36 15 48 8 16 25 42 13 41 47
#[40] 37 26 32 35 9 18 10 20 40 19 12
Sorry all, I know my question is really basic. Maybe it is because the question is too simple and the solution is thus not easily found from the internet.
Since I couldn't a direct question answering this adding an answer. We can first subset the list using [[ and then select individual elements in it with [.
bb <- aa[[1]][1:20]
cc <- aa[[1]][21:50]
We can also use head and tail to select first 20 and last 30 elements respectively.
bb <- head(aa[[1]], 20)
cc <- tail(aa[[1]], 30)
We can use split to create a list of vectors
lst1 <- split(aa[[1]], rep(1:2, c(20, 30)))
and extract the vector with [[
lst[[1]]
lst1[[2]]
It can be extended to any number of splits (i.e. generalized version) where we just need to change the rep
Related
How can I create a matrix , with random number on row and not replace.
like this
5 29 24 20 31 33
2 18 35 4 11 21
30 40 22 14 2 28
33 14 4 18 5 10
10 33 15 2 28 18
7 22 9 25 31 20
12 29 31 22 37 26
7 31 34 28 19 23
7 34 11 6 31 28
my code :
matrix(sample(1:42, 60, replace = FALSE), ncol = 6)
But I receive this error message:
Error in sample.int(length(x), size, replace, prob) : cannot take a
sample larger than the population when 'replace = FALSE'
but it's wrong because only 1~42, it can't create a 60 matrix.
You can not generate all 60 of the numbers with one sample function as you want to allow replacement of numbers in a different row. Therefore you have to do one sample per row. #Jav provided very neat code to accomplish this in the comment to the question:
t(sapply(1:10, function(x) sample(1:42, 6, replace = FALSE)))
if you want to have a different sample in each row, then replicate can help you -- but replicate (as pretty much everything else in R) works naturally columnwise, so you have to transpose the result:
t(replicate(10, sample(1:42, 6)))
replace = FALSE is the default, so I didn't include it
after transposing, 10 becomes the number of rows and 6 becomes the number of columns
I have a data frame like this.
> abc
ID 1.x 2.x 1.y 2.y
1 4 10 20 30 40
2 16 5 10 5 10
3 42 16 17 18 19
4 91 20 20 20 20
5 103 103 42 56 84
How do I create two additional columns '1' and '2' by multiplying 1.x * 1.y and 2.x * 2.y in a generalized way?
I am trying to get a generalized solution where number of columns can be too many. So I want to multiply all x with all y. While x and y are fixed, n has to be figured out from data frame.
For simplicity lets assume n is also fixed however it is a large number.
One thing i can try is :-
abc[,c(6,7)]=abc[,c(2,3)]*abc[,c(4,5)]
It will work only if col positions are contiguous. This is good enough for me. If anyone can have more generalized solution, it will benefit us all.
If there are only couple of variables to multiply, we can do this with transform by multiplying the columns of interest
transform(abc, new1 = `1.x`*`1.y`, new2 = `2.x`*`2.y`, check.names = FALSE)
# ID 1.x 2.x 1.y 2.y new1 new2
#1 4 10 20 30 40 300 800
#2 16 5 10 5 10 25 100
#3 42 16 17 18 19 288 323
#4 91 20 20 20 20 400 400
#5 103 103 42 56 84 5768 3528
If we have lots of columns, then one approach is to split the dataset into a list of data.frames by taking the substring of names and then loop through the list and multiply the rows with do.call
abc[paste0("new", 1:2)] <- lapply(split.default(abc[-1],
sub("\\.[a-z]+$", "", names(abc)[-1])), function(x) do.call(`*`, x))
Or another option is (based on the pairwise column multiplication)
apply(aperm(array(unlist(abc[-1]), c(5, 2, 2)),
c(3, 1, 2)), 3, matrixStats::colProds)
Mutate will preserve the original variables. Mutate_all will allow you to multiply all columns in your dataframe.
abc %>%
mutate(new_vary1 = `1.x`* `2.x`,
new_vary2 = `1.y`* `2.y`) %>%
mutate_all(funs(.*`1.x`))
I have created the following simple function in R:
fun <- function(a,b,c,d,e){b+(c-a)*((e-b)/(d-a))}
That I want to apply this function to a data.frame that looks something like:
> data.frame("x1"=seq(55,75,5),"x2"=round(rnorm(5,50,10),0),"x3"=seq(30,10,-5))
x1 x2 x3
1 55 51 30
2 60 45 25
3 65 43 20
4 70 57 15
5 75 58 10
I want to apply fun to each separate row to create a new variable x4, but now comes the difficult part (to me at least..): for the arguments d and e I want to use the values x2 and x3 from the next row. So for the first row of the example that would mean: fun(a=55,b=51,c=30,d=45,e=25). I know that I can use mapply() to apply a function to each row, but I have no clue on how to tell mapply that it should use some values from the next row, or whether I should be looking for a different approach than mapply()?
Many thanks in advance!
Use mapply, but shift the fourth and fifth columns by one row. You can do it manually, or use taRifx::shift.
> dat
x1 x2 x3
1 55 25 30
2 60 58 25
3 65 59 20
4 70 68 15
5 75 43 10
library(taRifx)
> shift(dat$x2)
[1] 58 59 68 43 25
> mapply( dat$x1, dat$x2, dat$x3, shift(dat$x2), shift(dat$x3) , FUN=fun )
[1] 25.00000 -1272.00000 719.00000 -50.14815 26.10000
If you want the last row to be NA rather than wrapping, use wrap=FALSE,pad=TRUE:
> shift(dat$x2,wrap=FALSE,pad=TRUE)
[1] 58 59 68 43 NA
At some point in my code, I get a list of tables that looks much like this:
[[1]]
cluster_size start end number p_value
13 2 12 13 131 4.209645e-233
12 1 12 12 100 6.166824e-185
22 11 12 22 132 6.916323e-143
23 12 12 23 133 1.176194e-139
13 1 13 13 31 3.464284e-38
13 68 13 117 34 3.275941e-37
23 78 23 117 2 4.503111e-32
....
[[2]]
cluster_size start end number p_value
13 2 12 13 131 4.209645e-233
12 1 12 12 100 6.166824e-185
22 11 12 22 132 6.916323e-143
23 12 12 23 133 1.176194e-139
13 1 13 13 31 3.464284e-38
....
While I don't show the full table here I know they are all the same size. What I want to do is make one table where I add up the p-values. Problem is that the $cluster_size, start, $end and $number columns don't necessarily correspond to the same row when I look at the table in different list elements so I can't just do a simple sum.
The brute force way to do this is to: 1) make a blank table 2) copy in the appropriate $cluster_size, $start, $end, $number columns from the first table and pull the correct p-values using a which() statement from all the tables. Is there a more clever way of doing this? Or is this pretty much it?
Edit: I was asked for a dput file of the data. It's located here:
http://alrig.com/code/
In the sample case, the order of the rows happen to match. That will not always be the case.
Seems like you can do this in two steps
Convert your list to a data.frame
Use any of the split-apply-combine approaches to summarize.
Assuming your data was named X, here's what you could do:
library(plyr)
#need to convert to data.frame since all of your list objects are of class matrix
XDF <- as.data.frame(do.call("rbind", X))
ddply(XDF, .(cluster_size, start, end, number), summarize, sump = sum(p_value))
#-----
cluster_size start end number sump
1 1 12 12 100 5.550142e-184
2 1 13 13 31 3.117856e-37
3 1 22 22 1 9.000000e+00
...
29 105 23 117 2 6.271469e-16
30 106 22 146 13 7.266746e-25
31 107 23 146 12 1.382328e-25
Lots of other aggregation techniques are covered here. I'd look at data.table package if your data is large.
I am trying to do unions on several lists (these are actually GRanges objects not integer lists but the priciple is the same), basically one big union.
x<-sort(sample(1:20, 9))
y<-sort(sample(10:30, 9))
z<-sort(sample(20:40, 9))
mylists<-c("x","y","z")
emptyList<-list()
sapply(mylists,FUN=function(x){emptyList<-union(emptyList,get(x))})
That is just returning the list contents.
I need the equivalent of
union(x,union(y,z))
[1] 2 3 5 6 7 10 13 15 20 14 19 21 24 27 28 29 26 31 36 39
but written in an extensible and non-"variable explicit" form
A not necessarily memory efficient paradigm that will work with GRanges is
Reduce(union, list(x, y, z))
The argument might also be a GRangesList(x, y, z) for appropriate values of x etc.
x<-sort(sample(1:20, 9))
y<-sort(sample(10:30, 9))
z<-sort(sample(20:40, 9))
Both of the below produce the same output
unique(c(x,y,z))
[1] 1 2 4 6 7 8 11 15 17 14 16 18 21 23 26 28 29 20 22 25 31 32 35
union(x,union(y,z))
[1] 1 2 4 6 7 8 11 15 17 14 16 18 21 23 26 28 29 20 22 25 31 32 35
unique(unlist(mget(mylists, globalenv())))
will do the trick. (Possibly changing the environment given in the call to mget, as required.)
I think it would be cleaner to separate the "dereference" part from the n-ary union part, e.g.
dereflist <- function(l) lapply(a,get)
nunion <- function(l) Reduce(union,l)
But if you look at how union works, you'll see that you could also do
nunion <- function(l) unique(do.call(c,l))
which is faster in all the cases I've tested (much faster for long lists).
-s
This can be done by using the reduce function in the purrr package.
purrr::reduce(list(x, y, z),union)
ok this works but I am curious why sapply seems to have its own scope
x<-sort(sample(1:20, 9))
y<-sort(sample(10:30, 9))
z<-sort(sample(20:40, 9))
mylists<-c("x","y","z")
emptyList<-vector()
for(f in mylists){emptyList<-union(emptyList,get(f))}