3

I have a list with the following structure: 

> str(test)
List of 5
$ :List of 2
..$ : num [1:4] 0.0544 0.0839 0.0486 0.043
..$ : num [1:4] 0.0799 0.2434 0.0373 0.2166
$ :List of 2
..$ : num [1:6] 0.938 1.047 1.022 0.689 0.39 ...
..$ : num [1:6] 0.871 0.25 0.824 0.664 0.481 ...
$ :List of 2
..$ : num [1:4] 0.000598 0.000923 0.000535 0.000473
..$ : num [1:4] 0.001039 0.003164 0.000485 0.002816
$ :List of 2
..$ : num [1:6] 0.01032 0.01152 0.01124 0.00758 0.00429 ...
..$ : num [1:6] 0.01133 0.00325 0.01071 0.00864 0.00625 ...
$ :List of 2
..$ : num -0.659
..$ : num -0.962

I want to compute the mean of each entry, i.e. for the first entry I want:

x <- mean(c(0.0544, 0.0839, 0.0486, 0.043))
y <- mean(c(0.0799, 0.2434, 0.0373, 0.2166))

And finally the mean of those two results = mean(c(x,y).

What is the most efficient way to do this?

João Carvalho
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5 Answers5

4

Sample data - one (parent) list of (child) lists.

set.seed(1)
foo <- list(
            list(runif(4),runif(4)),
            list(runif(6),runif(6))
           )
#> [[1]]
#> [[1]][[1]]
#> [1] 0.2655087 0.3721239 0.5728534 0.9082078
#> 
#> [[1]][[2]]
#> [1] 0.2016819 0.8983897 0.9446753 0.6607978
#> 
#> 
#> [[2]]
#> [[2]][[1]]
#> [1] 0.62911404 0.06178627 0.20597457 0.17655675 0.68702285 0.38410372
#> 
#> [[2]][[2]]
#> [1] 0.7698414 0.4976992 0.7176185 0.9919061 0.3800352 0.7774452

Get the means of each child list:

foo_means <- lapply(foo, lapply, mean)
#> [[1]]
#> [[1]][[1]]
#> [1] 0.5296734
#> 
#> [[1]][[2]]
#> [1] 0.6763862
#> 
#> 
#> [[2]]
#> [[2]][[1]]
#> [1] 0.3574264
#> 
#> [[2]][[2]]
#> [1] 0.6890909

To get the average of (child) averages for each (parent) list element:

lapply(foo_means, function(x) mean(unlist(x)))
#> [[1]]
#> [1] 0.6030298
#> 
#> [[2]]
#> [1] 0.5232587
ngm
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3

The first question appears to be a perfect use for rapply. Using the same sample data as provided above in @ngm's answer, you can do:

foo2 <- rapply(foo, mean, how = "replace")

foo2
#[[1]]
#[[1]][[1]]
#[1] 0.5296734
#
#[[1]][[2]]
#[1] 0.6763862
#
#
#[[2]]
#[[2]][[1]]
#[1] 0.3574264
#
#[[2]][[2]]
#[1] 0.6890909
Mike H.
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1

This has been answered many times. It is as simple as 

mean(unlist(lapply(mylist,sapply,mean)))

Follow this link.

Grouping functions (tapply, by, aggregate) and the *apply family

user5249203
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  • Yeah, I knew that lapply would be the correct way to do it, but I was having some trouble with the lists inside a list issue... However, I found that lapply(test, FUN = function(x) {mean(unlist(x))}) also seems to work – João Carvalho Apr 20 '18 at 14:32
1

Since the elements of your list are also lists, then use some nested lapply's to get the means of each list.

lapply(test, function(x) lapply(x, mean))

If you like a matrix as an output, you can do

sapply(test, function(x) sapply(x, mean))

With the sapply, you can get the means of each pair with

colMeans(sapply(test, function(x) sapply(x, mean, na.rm = TRUE)))
hpesoj626
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1

You can also use modify_depth function from the purrr package. Take example posted by @ngm

library(purrr)

set.seed(1)
foo <- list(list(runif(4), runif(4)),
            list(runif(6), runif(6)))

modify_depth(foo, 2, mean)

#> [[1]]
#> [[1]][[1]]
#> [1] 0.5296734
#> 
#> [[1]][[2]]
#> [1] 0.6763862
#> 
#> 
#> [[2]]
#> [[2]][[1]]
#> [1] 0.3574264
#> 
#> [[2]][[2]]
#> [1] 0.6890909

Edit: to calculate the mean of all elements in the 1st level of that list

modify_depth(foo, 2, mean) %>% 
  map(flatten_dbl) %>% 
  map_dbl(mean)

modify_depth(foo, 2, mean) %>% 
  simplify_all() %>% 
  map_dbl(mean)

[1] 0.60303 0.52326

Created on 2018-04-20 by the reprex package (v0.2.0).

Tung
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