CODEWITHSUNDEEP
r
ayshelina
ayshelina

Reputation: 137

Shuffle elements in list with same names

I have a large list of lists of matrices. The format can be seen here: Input Data

These are 50 matrices but divided into sublists and ordered. What I need is one list of 50 matrices in random order while keeping the names which are characters 0-10. Example:

0: [1:28, 1:28] 0 0 0 ...
5: [1:28, 1:28] 0 0 0 ...
3: [1:28, 1:28] 0 0 0 ...
0: [1:28, 1:28] 0 0 0 ...
8: [1:28, 1:28] 0 0 0 ...
8: [1:28, 1:28] 0 0 0 ...

Upvotes: 0

Views: 107

Answers (3)

CPak
CPak

Reputation: 13581

You can try this

Data

data <- lapply(1:10, function(i) lapply(1:5, function(j) matrix(1:(28*28), nrow=28)))
names(data) <- 1:10
str(data)
# List of 10
 # $ 1 :List of 5
  # ..$ : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
  # ..$ : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
  # ..$ : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
  # ..$ : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
  # ..$ : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 2 :List of 5
  # ..$ : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
  # ..$ : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
  # ..$ : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
  # ..$ : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
  # ..$ : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # etc

purrr::flatten your list

EDIT If each list has a different length, you can use names(result) <- rep(1:length(data), times=(lengths(data))) instead

library(purrr)
result <- flatten(data)
names(result) <- rep(1:length(data), times=(lengths(data)))
str(result)
# List of 50
 # $ 1 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 1 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 1 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 1 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 1 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 2 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 2 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 2 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # etc

Shuffle

set.seed(1)
ans <- result[sample(1:50)]
str(ans)
# List of 50
 # $ 3 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 4 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 6 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 9 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 2 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 9 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 9 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # $ 6 : int [1:28, 1:28] 1 2 3 4 5 6 7 8 9 10 ...
 # etc

Upvotes: 1

Rui Barradas
Rui Barradas

Reputation: 76460

Without a dataset example I don't know if this runs but here it goes.

set.seed(3694)
inx <- sample(names(trainData), 50, TRUE)

tmp <- lapply(inx, function(i){
    lst <- trainData[[i]]
    mat <- lst[[sample(seq_along(lst), 1)]]
    list(i, mat)
})

result <- lapply(tmp, `[[`, 2)
names(result) <- sapply(tmp, `[[`, 1)
rm(tmp)

EDIT.
Tested with the data by CPak, a bug in the names of the result was corrected. It now seems to do what the OP is asking for.

Upvotes: 2

Heikki
Heikki

Reputation: 2254

What about using sample(...)? Running it second time gives you new permutation:

> a = list('0'=list(2,3),'1'=2,'2'=3)
> a
$`0`
$`0`[[1]]
[1] 2

$`0`[[2]]
[1] 3


$`1`
[1] 2

$`2`
[1] 3

> sample(a,length(a))
$`1`
[1] 2

$`2`
[1] 3

$`0`
$`0`[[1]]
[1] 2

$`0`[[2]]
[1] 3


> sample(a,length(a))
$`2`
[1] 3

$`1`
[1] 2

$`0`
$`0`[[1]]
[1] 2

$`0`[[2]]
[1] 3

Upvotes: 1

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