CODEWITHSUNDEEP
rapplyequalitylapply
dasf
dasf

Reputation: 1045

R: Lapply multiple equality

Given list test, how do I find any numbers that repeat exactly 3 times in test[[2]] and 4 times in test[[3]]? In this particular example, I want an output that gives me 3 and 6 in [[2]] and 10 in [[3]].

test <- list(c(1:10),c(1:10,3,3,6,6),c(1:10,10,10,10))

I am able to generate a table for each vector, test2.

lapply(test,table)

But when I tried to equate the table to 1:3, which I stored as the names of test, I get logical(0) for each vector.

names(test) <- 1:3
lapply(test,function(x) table(x)==names(x))

Any idea what went wrong and what the simplest solution is?

Upvotes: 0

Views: 832

Answers (3)

MichaelChirico
MichaelChirico

Reputation: 34733

I'm not sure what you were after with your approach, but this will return the numbers appearing twice:

lapply(seq_along(test), function(ii) which(tabulate(test[[ii]]) == ii))
# [[1]]
# integer(0)
# 
# [[2]]
# [1] 10
# 
# [[3]]
# integer(0)

Upvotes: 0

Sraffa
Sraffa

Reputation: 1668

If you run lapply(test, names) you would see that the names function applied to each list element just returns NULL. Under the hood I think lapply is just passing the contents of the sublist and not the sublist itself to the function (just like when you use double square brackets to subset - try comparing names(test[[1]]) to names(test[1])).

Using your original idea, if what you want is a vector of logicals, you could try this instead:

lapply(1:3, function(x) table(test[x]) == x)

Upvotes: 1

Adam Quek
Adam Quek

Reputation: 7163

To generate a list of table: 

 tab <- lapply(test, table)

To get the number of times 10 appear in each object in the list:

lapply(tab, '[', "10")

[[1]]
10 
 1 

[[2]]
10 
 2 

[[3]]
10 
 3

Upvotes: 0

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