How to select specific element from nested dataframes - r

Question

I have a list of nested data frames and I want to extract the observations of the earliest year, my problem is the first year change with the data frames. the year is either 1992 or 2005.
I want to create a list to stock them, I tried with which, but since there is the same year, observations are repeated, and I want them apart
new_df<- which(df[[i]]==1992 | df[[i]]==2005)
I've tried with ifelse() but I have to do an lm operation after, and it doesn't work. And I can't take only the first rows, because the year are repeated
my code looks like this:
df<- list(a<-data.frame(a_1<-(1992:2015),
a_2<-sample(1:24)),
b<-data.frame(b_1<-(1992:2015),
b_2<-sample(1:24)),
c<-data.frame(c_1<-(2005:2015),
c_2<-sample(1:11)),
d<-data.frame(d_1<-(2005:2015),
d_2<-sample(1:11)))

Answer 1

You can define a function to get the data on one data.frame and loop on the list to extract values.
Below I use map from the purrr package but you can also use lapply and for loops
Please do not use <- when assigning values in a function call (here data.frame() ) because it will mess colnames. = is used in function calls for arguments variables and it's okay to use it. You can read this ;)
df<- list(a<-data.frame(a_1 = (1992:2015),
a_2 = sample(1:24)),
b<-data.frame(b_1 = (1992:2015),
b_2 = sample(1:24)),
c<-data.frame(c_1 = (2005:2015),
c_2 = sample(1:11)),
d<-data.frame(d_1 = (2005:2015),
d_2 = sample(1:11)))
extract_miny <- function(df){
miny <- min(df[,1])
res <- df[df[,1] == miny, 2]
names(res) <- miny
return(res)
}
map(df, extract_miny)

Answer 2

If the data is sorted as the example, you can slice() the first row for the information. Notice the use of = rather than <- in creating a nested dataframe.
library(tidyverse)
df <- list(
a = data.frame(a_1 = (1992:2015),
a_2 = sample(1:24)),
b = data.frame(b_1 = (1992:2015),
b_2 = sample(1:24)),
c = data.frame(c_1 = (2005:2015),
c_2 = sample(1:11)),
d = data.frame(d_1 = (2005:2015),
d_2 = sample(1:11))
)
df %>%
imap_dfr( ~ slice(.x, 1) %>%
set_names(c("year", "value")) %>%
mutate(dataframe = .y) %>%
as_tibble())
# A tibble: 4 x 3
year value dataframe
<int> <int> <chr>
1 1992 19 a
2 1992 2 b
3 2005 1 c
4 2005 5 d

Answer 3

You may subset anonymeously.
lapply(df, \(x) setNames(x[x[[1]] == min(x[[1]]), ], c('year', 'value'))) |> do.call(what=rbind)
# year value
# 1 1992 6
# 2 1992 9
# 3 2005 11
# 4 2005 11
Or maybe better by creating a variable from which sample the value stems from.
Map(`[<-`, df, 'sample', value=letters[seq_along(df)]) |>
lapply(\(x) setNames(x[x[[1]] == min(x[[1]]), ], c('year', 'value', 'sample'))) |>
do.call(what=rbind)
# year value sample
# 1 1992 6 a
# 2 1992 9 b
# 3 2005 11 c
# 4 2005 11 d
Data:
df <- list(structure(list(a_1.....1992.2015. = 1992:2015, a_2....sample.1.24. = c(6L,
18L, 23L, 5L, 7L, 14L, 4L, 10L, 19L, 17L, 15L, 1L, 11L, 22L,
13L, 8L, 20L, 16L, 2L, 3L, 24L, 21L, 9L, 12L)), class = "data.frame", row.names = c(NA,
-24L)), structure(list(b_1.....1992.2015. = 1992:2015, b_2....sample.1.24. = c(9L,
24L, 18L, 8L, 16L, 11L, 13L, 23L, 15L, 20L, 19L, 21L, 12L, 22L,
7L, 3L, 6L, 17L, 2L, 5L, 4L, 10L, 1L, 14L)), class = "data.frame", row.names = c(NA,
-24L)), structure(list(c_1.....2005.2015. = 2005:2015, c_2....sample.1.11. = c(11L,
2L, 5L, 10L, 9L, 6L, 1L, 7L, 3L, 8L, 4L)), class = "data.frame", row.names = c(NA,
-11L)), structure(list(d_1.....2005.2015. = 2005:2015, d_2....sample.1.11. = c(11L,
2L, 5L, 1L, 6L, 9L, 3L, 7L, 10L, 4L, 8L)), class = "data.frame", row.names = c(NA,
-11L)))