I have 5 lists of different lengths
a <- c(1) #with length of 1
b <- c(4.4,3.5) #length 2
c <- c(5.6,7.8,6.0) #length 3
d <- c(0.8,6.9,8.8,5.8) #length 4
e <- c(1.8,2.5,2.3,6.5,1.1) #length is 5
I am trying to get the mean of elements in all lists:
#since there are 5 values available for 1st element
a[1] b[1] c[1] d[1] e[1] / 5
#since there are 4 values available for 2nd element
b[2] c[2] d[2] e[2] / 4
#next divide by 3 and 2...1
c[3] d[3] e[3] / 3 and so on...
I need the mean of these values in another array so that I can do further processing of the data.
Can anyone give suggestion on what to do to obtain the mean??
CodePudding user response：
Let
l <- list(a, b, c, d, e)
then do:
tapply(unlist(l), sequence(lengths(l)), mean)
1 2 3 4 5
2.720 5.175 5.700 6.150 1.100
Another approach:
rowMeans(sapply(l, `length<-`, max(lengths(l))), na.rm = TRUE)
[1] 2.720 5.175 5.700 6.150 1.100
colMeans(plyr::rbind.fill.matrix(sapply(l, t)), na.rm = TRUE)
1 2 3 4 5
2.720 5.175 5.700 6.150 1.100
CodePudding user response：
With data.table::transpose:
l <- list(a, b, c, d, e)
sapply(data.table::transpose(l), mean, na.rm = TRUE)
# [1] 2.720 5.175 5.700 6.150 1.100
CodePudding user response：
1) Create a list of the vectors and convert each to a ts object. Then use cbind to create a multivariate ts object with each column corresponding to one of the original vectors. Then use rowMeans on that.
nms <- c("a", "b", "c", "d", "e")
L <- mget(nms)
rowMeans(do.call("cbind", lapply(L, ts)), na.rm = TRUE)
## [1] 2.720 5.175 5.700 6.150 1.100
Check
(a[1] b[1] c[1] d[1] e[1]) / 5
## [1] 2.72
( b[2] c[2] d[2] e[2]) / 4
## [1] 5.175
2) Another approach using L from above is to use sapply:
n <- max(lengths(L))
sapply(1:n, function(i) mean(c(a[i], b[i], c[i], d[i], e[i]), na.rm = TRUE))
## [1] 2.720 5.175 5.700 6.150 1.100
3) or use sapply like this where L and n are from above:
rowMeans(sapply(L, `[`, 1:n), na.rm = TRUE)
CodePudding user response：
Another method that put your vectors into a list, then add NA to the vectors to make them equal length. Finally do a rowMeans on them.
mylist <- list(a, b, c, d ,e)
max_L <- max(lengths(mylist))
rowMeans(sapply(mylist, \(x) c(x, rep(NA, max_L - length(x)))), na.rm = T)
[1] 2.720 5.175 5.700 6.150 1.100
CodePudding user response：
Using tidyverse
library(dplyr)
library(tidyr)
library(data.table)
mget(letters[1:5]) %>%
enframe %>%
unnest(value) %>%
group_by(grp = rowid(name)) %>%
summarise(value = mean(value)) %>%
pull(value)
-output
[1] 2.720 5.175 5.700 6.150 1.100
CodePudding user response：
You can try colMeans with option na.rm = TRUE like below
L <- max(lengths(lst))
m <- matrix(nrow = L, ncol = L)
m[cbind(rep(seq_along(lst), lengths(lst)), sequence(lengths(lst)))] <- unlist(lst)
colMeans(m, na.rm = TRUE)
which gives
[1] 2.720 5.175 5.700 6.150 1.100
CodePudding user response：
Interesting to see the wide range of timings on a larger list:
library(plyr)
library(data.table)
library(dplyr)
library(tidyr)
library(tibble)
set.seed(976933858)
l <- unname(split(runif(1e5), sample(1e4, 1e5, TRUE)))
f0 <- function(l) diff(c(0, cumsum(unlist(l)[order(sequence(lengths(l)))])[cumsum(lens <- rev(cumsum(rev(tabulate(lengths(l))))))]))/lens # jblood94
f1 <- function(l) sapply(transpose(l), mean, na.rm = TRUE) # Mael
f2 <- function(l) as.numeric(tapply(unlist(l), sequence(lengths(l)), mean)) # onyambu
f3 <- function(l) rowMeans(sapply(l, `length<-`, max(lengths(l))), na.rm = TRUE) # onyambu
f4 <- function(L) rowMeans(sapply(L, `[`, 1:max(lengths(L))), na.rm = TRUE) # G. Grothendieck
f5 <- function(mylist) {
# benson23
max_L <- max(lengths(mylist))
rowMeans(sapply(mylist, function(x) c(x, rep(NA, max_L - length(x)))), na.rm = T)
}
f6 <- function(l) l %>% enframe %>% unnest(value) %>% group_by(grp = rowid(name)) %>% summarise(value = mean(value)) %>% pull(value) # akrun
f7 <- function(l) unname(colMeans(rbind.fill.matrix(sapply(l, t)), na.rm = TRUE)) # onyambu
f8 <- function(L) rowMeans(do.call("cbind", lapply(L, ts)), na.rm = TRUE) # G. Grothendieck
microbenchmark::microbenchmark(
jblood94 = f0(l),
Mael = f1(l),
onyambu1 = f2(l),
onyambu2 = f3(l),
G.Grothendieck1 = f4(l),
benson23 = f5(l),
akrun = f6(l),
onyambu3 = f7(l),
G.Grothendieck2 = f8(l),
check = "identical"
)
#> Unit: milliseconds
#> expr min lq mean median uq max neval
#> jblood94 1.0303 1.15555 1.544470 1.22690 1.32600 6.4604 100
#> Mael 1.7522 1.88085 3.214797 1.98215 2.11660 85.0957 100
#> onyambu1 2.4569 2.61115 4.030907 2.77880 2.94575 99.1389 100
#> onyambu2 4.1092 4.64760 6.034596 4.91770 6.23725 21.6162 100
#> G.Grothendieck1 4.7224 5.11220 6.485828 5.63770 7.08075 12.9853 100
#> benson23 10.7711 12.03325 16.352950 16.06155 19.34285 32.2033 100
#> akrun 13.8533 16.11800 18.376538 17.09465 20.45810 39.1432 100
#> onyambu3 342.7180 410.80445 465.461878 451.92355 504.14595 800.1453 100
#> G.Grothendieck2 465.0375 509.92070 565.463577 553.64825 607.95115 777.0444 100
CodePudding user response：
Another possible solution, based on the idea that the matrix resulting from row binding a, b, c, d, e is triangular:
m <- matrix(NA, 5, 5)
m[upper.tri(m, diag = T)] <- c(a, b, c, d, e)
colMeans(t(m), na.rm = T)
#> [1] 2.720 5.175 5.700 6.150 1.100