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Populating a list with lists (replicate, lapply)


Populating a list with lists (replicate, lapply)

By : user2948578
Date : November 14 2020, 04:51 PM
it should still fix some issue How should I generate a list of lists of known size? , You can try
code :
replicate(5, list(vector=rnorm(2)), simplify=FALSE)
# [[1]]
#[[1]]$vector
#[1] -1.5239454 -0.1326934


#[[2]]
#[[2]]$vector
#[1] -1.4369404  0.3701259


#[[3]]
#[[3]]$vector
#[1]  0.3251298 -1.4289498


#[[4]]
#[[4]]$vector
#[1]  0.8346002 -0.2974959


#[[5]]
#[[5]]$vector
#[1]  0.4581858 -0.8066517


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Changing a list of lists with lapply

Changing a list of lists with lapply


By : Alexis
Date : March 29 2020, 07:55 AM
Any of those help I have list of lists that are all the same structure and I want to alter one element of each of the lists in the second tier. In other words, I have a single list with objects x, y & z. This list is repeated multiple times in another list. I need to change x in each of these lists. What is the best way to do this? , Simply return the modified obj instead of re-creating a list.
code :
standardize = function(obj){
  obj$x <- obj$x / max(obj$x)
  obj
}
lapply on each element of a list of lists

lapply on each element of a list of lists


By : iTzKillaPlaysMC Gami
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further I found a new solution that is even simpler than R. Schifini's elegant solution, so now I answer my own question :)
The apply family also provides a recursive version called rapply, and it works perfectly on nested lists, see these examples.
code :
images.list <- rapply(designs.path.list, load.image)
How to create a sequentially ordered list in R with lapply and replicate?

How to create a sequentially ordered list in R with lapply and replicate?


By : Martin
Date : March 29 2020, 07:55 AM
it helps some times I have been trying to avoid for loops in R. However, I have had issues with using lapply with replicate in that the lists I obtain have ordering that are not sequential. For example, the following code: , You can just call unlist() on the result, e.g.:
code :
unlist(lapply(1:5, function(x) replicate(100, rnorm(x))))
unlist(lapply(1:5, function(x) replicate(100, rnorm(1, mean=x))))
values <- unlist(lapply(1:5, function(x) replicate(100, rnorm(1, mean=x))))
plot(seq_along(values), values)
Split, lapply, rbind paradigm. lapply returning lists of numerics instead of date index

Split, lapply, rbind paradigm. lapply returning lists of numerics instead of date index


By : VaughanAshe
Date : March 29 2020, 07:55 AM
seems to work fine Assuming x shown in the Note at the end we can calculate the cummean by year using ave:
code :
transform(x, cummean = ave(win, format(time(x), "%Y"), FUN = cummean))
##            win   cummean
## 2010-04-04   1 1.0000000
## 2010-04-06   0 0.5000000
## 2010-04-07   0 0.3333333
## 2010-04-09   0 0.2500000
## 2010-04-10   1 0.4000000
## 2010-04-11   1 0.5000000
do.call("rbind", lapply(split(x, "years"), transform, cummean = cummean(win)))
Lines <- "date win
2010-04-04   1
2010-04-06   0
2010-04-07   0
2010-04-09   0
2010-04-10   1
2010-04-11   1"
library(xts)
x <- as.xts(read.zoo(text = Lines, header = TRUE, drop = FALSE))
How to names lists under list when using lapply?

How to names lists under list when using lapply?


By : Cyrus
Date : March 29 2020, 07:55 AM
may help you . You should note that lapply() itself is just a wrapper for a well constructed for() loop, so you're not gaining any efficiency, just perhaps readability. That aside, the easiest approach is to add names to the lists going into your nested lapply() calls:
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