Manipulating all split data sets - r

Question

I'm drawing a blank-- I have 51 sets of split data from a data frame that I had, and I want to take the mean of the height of each set.
print(dataset)
$`1`
ID Species Plant Height
1 A 1 42.7
2 A 1 32.5
$`2`
ID Species Plant Height
3 A 2 43.5
4 A 2 54.3
5 A 2 45.7
...
...
...
$`51`
ID Species Plant Height
134 A 51 52.5
135 A 51 61.2
I know how to run each individually, but with 51 split sections, it would take me ages.
I thought that
mean(dataset[,4])
might work, but it says that I have the wrong number of dimensions. I get now why that is incorrect, but I am no closer to figuring out how to average all of the heights.

Answer 1

The dataset is a list. We could use lapply/sapply/vapply etc to loop through the list elements and get the mean of the 'Height' column. Using vapply, we can specify the class and length of the output (numeric(1)). This will be useful for debugging.
vapply(dataset, function(x) mean(x[,4], na.rm=TRUE), numeric(1))
# 1 2 51
#37.60000 47.83333 56.85000
Or another option (if we have the same columnames/number of columns for the data.frames in the list), would be to use rbindlist from data.table with the optionidcol=TRUEto generate a singledata.table. The '.id' column shows the name of thelistelements. We group by '.id' and get themeanof theHeight`.
library(data.table)
rbindlist(dataset, idcol=TRUE)[, list(Mean=mean(Height, na.rm=TRUE)), by = .id]
# .id Mean
#1: 1 37.60000
#2: 2 47.83333
#3: 51 56.85000
Or a similar option as above is unnest from library(tidyr) to return a single dataset with the '.id' column, grouped by '.id', we summarise to get the mean of 'Height'.
library(tidyr)
library(dplyr)
unnest(dataset, .id) %>%
group_by(.id) %>%
summarise(Mean= mean(Height, na.rm=TRUE))
# .id Mean
#1 1 37.60000
#2 2 47.83333
#3 51 56.85000
The syntax for plyr is
df1 <- unnest(dataset, .id)
ddply(df1, .(.id), summarise, Mean=mean(Height, na.rm=TRUE))
# .id Mean
#1 1 37.60000
#2 2 47.83333
#3 51 56.85000
data
dataset <- structure(list(`1` = structure(list(ID = 1:2, Species = c("A",
"A"), Plant = c(1L, 1L), Height = c(42.7, 32.5)), .Names = c("ID",
"Species", "Plant", "Height"), class = "data.frame", row.names = c(NA,
-2L)), `2` = structure(list(ID = 3:5, Species = c("A", "A", "A"
), Plant = c(2L, 2L, 2L), Height = c(43.5, 54.3, 45.7)), .Names = c("ID",
"Species", "Plant", "Height"), class = "data.frame", row.names = c(NA,
-3L)), `51` = structure(list(ID = 134:135, Species = c("A", "A"
), Plant = c(51L, 51L), Height = c(52.5, 61.2)), .Names = c("ID",
"Species", "Plant", "Height"), class = "data.frame", row.names = c(NA,
-2L))), .Names = c("1", "2", "51"))

Answer 2

This also works, though it uses dplyr.
library(dplyr)
1:length(dataset) %>%
lapply(function(i)
test[[i]] %>%
mutate(section = i ) ) %>%
bind_rows %>%
group_by(section) %>%
summarize(mean_height = mean(height) )