Weighted mean by group in long-data - r

Question

My questing is kinda similar to this question, and is building on this answer, only thing is that my data is long format, not wide, and I would like to keep it that way.
Wondered if there's smart way to calculate the weighted.mean() shown in this answer, but with long data.
Say my data lookslike this
library(tidyverse)
dft_w <- structure(list(obs = c(1L, 1L, 2L, 2L, 3L, 3L, 4L, 4L), education = c("A",
"A", "B", "B", "B", "B", "A", "A"), Item = c("income", "weight",
"income", "weight", "income", "weight", "income", "weight"),
Amount = c(1000L, 10L, 2000L, 1L, 1500L, 5L, 2000L, 2L)), row.names = c(NA,
-8L), class = c("tbl_df", "tbl", "data.frame")); dft_w
# A tibble: 8 x 4
obs education Item Amount
<int> <chr> <chr> <int>
1 1 A income 1000
2 1 A weight 10
3 2 B income 2000
4 2 B weight 1
5 3 B income 1500
6 3 B weight 5
7 4 A income 2000
8 4 A weight 2
and I would like to get to something like this
# A tibble: 12 x 4
obs education Item Amount
<int> <chr> <chr> <dbl>
1 1 A income 1000
2 1 A weight 10
3 1 A weighted_income 1167.
4 2 B income 2000
5 2 B weight 1
6 2 B weighted_income 1583.
7 3 B income 1500
8 3 B weight 5
9 3 B weighted_income 1583.
10 4 A income 2000
11 4 A weight 2
12 4 A weighted_income 1167.

Answer 1

dft_w %>%
group_by(education) %>%
summarize(
Amount = rep(weighted.mean(Amount[Item == "income"], Amount[Item == "weight"]), length(unique(obs))),
obs = unique(obs),
Item = "weighted_income"
) %>%
bind_rows(dft_w, .) %>%
arrange(obs, education, Item)
# # A tibble: 12 x 4
# obs education Item Amount
# <int> <chr> <chr> <dbl>
# 1 1 A income 1000
# 2 1 A weight 10
# 3 1 A weighted_income 1167.
# 4 2 B income 2000
# 5 2 B weight 1
# 6 2 B weighted_income 1583.
# 7 3 B income 1500
# 8 3 B weight 5
# 9 3 B weighted_income 1583.
# 10 4 A income 2000
# 11 4 A weight 2
# 12 4 A weighted_income 1167.
Note that this will error if the data does not contain equal numbers of "income" and "weight" (erring with 'x' and 'w' must have the same length).
This can be preempted with sufficient filtering, perhaps this:
dft_w %>%
slice(-1) %>% # just to trigger the fail, test the filter
group_by(obs, education) %>%
filter(all(c("income", "weight") %in% Item)) %>%
group_by(education) %>%
summarize(
Amount = rep(weighted.mean(Amount[Item == "income"], Amount[Item == "weight"]), length(unique(obs))),
obs = unique(obs),
Item = "weighted_income"
) %>%
bind_rows(slice(dft_w, -1), .) %>% # slice() only to keep the output consistent
arrange(obs, education, Item)
# # A tibble: 10 x 4
# obs education Item Amount
# <int> <chr> <chr> <dbl>
# 1 1 A weight 10
# 2 2 B income 2000
# 3 2 B weight 1
# 4 2 B weighted_income 1583.
# 5 3 B income 1500
# 6 3 B weight 5
# 7 3 B weighted_income 1583.
# 8 4 A income 2000
# 9 4 A weight 2
# 10 4 A weighted_income 2000
noting that the obs/education pair without both will not gain the "weighted_income" value.

Answer 2

Another way around is to use tidyr's pivot_wider and pivot_longer in the same pipe chain so you can actually work with wide data before going back to long format. It may not be the most efficient way but it allows to keep "wide-format" tips & tricks.
library(dplyr)
dft_w %>%
tidyr::pivot_wider(names_from = Item, values_from = Amount) %>%
group_by(education) %>%
mutate(weighted_income = weighted.mean(income, weight)) %>%
tidyr::pivot_longer(3:last_col(), names_to = "Item", values_to = "Amount")
Output:
# A tibble: 12 x 4
# Groups: education [2]
obs education Item Amount
<int> <chr> <chr> <dbl>
1 1 A income 1000
2 1 A weight 10
3 1 A weighted_income 1167.
4 2 B income 2000
5 2 B weight 1
6 2 B weighted_income 1583.
7 3 B income 1500
8 3 B weight 5
9 3 B weighted_income 1583.
10 4 A income 2000
11 4 A weight 2
12 4 A weighted_income 1167.

Answer 3

Here is just another way of doing this using tibble::add_row. I just opted for only one summary per grouping variable:
library(dplyr)
library(purrr)
dft_w %>%
group_split(education) %>%
map_dfr(~ .x %>%
add_row(obs = .x$obs[1], education = .x$education[1],
Item = "weighted.mean", Amount = weighted.mean(.x$Amount[.x$Item == "income"],
.x$Amount[.x$Item == "weight"])))
# A tibble: 10 x 4
obs education Item Amount
<int> <chr> <chr> <dbl>
1 1 A income 1000
2 1 A weight 10
3 4 A income 2000
4 4 A weight 2
5 1 A weighted.mean 1167.
6 2 B income 2000
7 2 B weight 1
8 3 B income 1500
9 3 B weight 5
10 2 B weighted.mean 1583.