CODEWITHSUNDEEP
rfunctionlapply
SavedByJESUS
SavedByJESUS

Reputation: 3314

Nested lapply() in a list?

I have a list l, which has the following features:

l = list(a = c(2, 3, 1, 5, 1), b = c(4, 3, 3, 5, 2), c = c(5, 1, 3, 2, 4))

I want to do two things:

First

I want to know how many times each number occurs in the entire list and I want each result in a vector (or any form that can allow me to perform computations with the results later):

Code 1:

> a <- table(sapply(l, "["))
> x <- as.data.frame(a)
> x

  Var1   Freq
1    1   3
2    2   3
3    3   4
4    4   2
5    5   3

Is there anyway to do it without using the table() function. I would like to do it "manually". I try to do it right below.

Code 2: (I know this is not very efficient!)

x <- data.frame(
"1" <- sum(sapply(l, "[")) == 1
"2" <- sum(sapply(l, "[")) == 2
"3" <- sum(sapply(l, "[")) == 3
"4" <- sum(sapply(l, "[")) == 4
"5" <- sum(sapply(l, "[")) == 5)

I tried the following, but I did not work. I actually did not understand the result.

> sapply(l, "[") == 1:5        

     a     b     c
[1,] FALSE FALSE FALSE
[2,] FALSE FALSE FALSE
[3,] FALSE  TRUE  TRUE
[4,] FALSE FALSE FALSE
[5,] FALSE FALSE FALSE

> sum(sapply(l, "[") == 1:5)

[1] 2

Second

Now, I would like to get the number of times each number appears in the list, but now in each element $a, $b and $c. I thought about using the lapply() but I don't know how exactly. Following is what I tried, but it is inefficient just like Code 2:

lapply(l, function(x) sum(x == 1))
lapply(l, function(x) sum(x == 2))
lapply(l, function(x) sum(x == 3))
lapply(l, function(x) sum(x == 4))
lapply(l, function(x) sum(x == 5))

What I get with these 5 lines of code are 5 lists of 3 elements each containing a single numeric value. For example, the second line of code tells me how many times number 2 appears in each element of l.

Code 3:

> lapply(l, function(x) sum(x == 2))

$a
[1] 1

$b
[1] 1

$c
[1] 1

What I would like to obtain is a list with three elements containing all the information I am looking for.

Please, use the references "Code 1", "Code 2" and "Code 3" in your answers. Thank you very much.

Upvotes: 5

Views: 15483

Answers (3)

and-bri
and-bri

Reputation: 1664

here you have another example with nested lapply().

created data:

list = NULL  
list[[1]] = c(1:5)  
list[[2]] = c(1:5)+3  
list[[2]] = c(1:5)+4  
list[[3]] = c(1:5)-1  
list[[4]] = c(1:5)*3  

list2 = NULL  
list2[[1]] = rep(1,5)  
list2[[2]] = rep(2,5)  
list2[[3]] = rep(0,5)

The result is this; it serve to subtract each element of one list with all elements of the other list.

lapply(list, function(d){ lapply(list2, function(a,b) {a-b}, b=d)})

Upvotes: 2

IRTFM
IRTFM

Reputation: 263301

Just use as.data.frame(l) for the second part and table(unlist(l)) for the first.

> table(unlist(l))

1 2 3 4 5 
3 3 4 2 3 

> data.frame(lapply(l, tabulate))
  a b c
1 2 0 1
2 1 1 1
3 1 2 1
4 0 1 1
5 1 1 1`

Upvotes: 9

josliber
josliber

Reputation: 44299

For code 1/2, you could use sapply to obtain the counts for whichever values you wanted:

l = list(a = c(2, 3, 1, 5, 1), b = c(4, 3, 3, 5, 2), c = c(5, 1, 3, 2, 4))
data.frame(number = 1:5,
           freq = sapply(1:5, function(x) sum(unlist(l) == x)))
#   number freq
# 1      1    3
# 2      2    3
# 3      3    4
# 4      4    2
# 5      5    3

For code 3, if you wanted to get the counts for lists a, b, and c, you could just apply your frequency function to each element of the list with the lapply function:

freqs = lapply(l, function(y) sapply(1:5, function(x) sum(unlist(y) == x)))
data.frame(number = 1:5, a=freqs$a, b=freqs$b, c=freqs$c)
#   number a b c
# 1      1 2 0 1
# 2      2 1 1 1
# 3      3 1 2 1
# 4      4 0 1 1
# 5      5 1 1 1

Upvotes: 5

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