I want to return the number of Transmitter codes that have been seen in both Season Winter1 AND Winter2. The answer should be 6 (6 different codes that were seen in Winter1 and Winter2). But the following command returns 0:
length(unique(Dispersion[(Dispersion$Season == "Winter1") & (Dispersion$Season == "Winter2"),]$Transmitter))
What command is appropriate for my problem?
structure(list(Transmitter = c("A69-1602-59814", "A69-1602-59814",
"A69-1602-59815", "A69-1602-59815", "A69-1602-59819", "A69-1602-59820",
"A69-1602-59821", "A69-1602-59822", "A69-1602-59823", "A69-1602-59824",
"A69-1602-59825", "A69-1602-59826", "A69-1602-59826", "A69-1602-59827",
"A69-1602-59828", "A69-1602-59828", "A69-1602-59830", "A69-1602-59831",
"A69-1602-59831", "A69-1602-59832", "A69-1602-59833", "A69-1602-59834",
"A69-1602-59835", "A69-1602-59835", "A69-1602-59836"), Batch.location = c("Lemmer",
"Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer",
"Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer",
"Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer",
"Lemmer", "Lemmer", "Lemmer"), Location.Dispersion = c("Lemmer",
"Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer",
"Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer",
"Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer",
"Lemmer", "Lemmer", "Lemmer"), Season = c("Winter1", "Winter2",
"Winter1", "Winter2", "Winter1", "Winter1", "Winter1", "Winter1",
"Winter1", "Winter1", "Winter1", "Winter1", "Winter2", "Winter1",
"Winter1", "Winter2", "Winter1", "Winter1", "Winter2", "Winter1",
"Winter1", "Winter1", "Winter1", "Winter2", "Winter1"), Freq = c(1961L,
2075L, 310L, 1L, 2880L, 305L, 366L, 834L, 19L, 2580L, 564L, 997L,
3475L, 6447L, 988L, 2991L, 355L, 3147L, 6155L, 903L, 484L, 321L,
76L, 1921L, 3329L)), row.names = c(NA, -25L), groups = structure(list(
Transmitter = c("A69-1602-59814", "A69-1602-59815", "A69-1602-59819",
"A69-1602-59820", "A69-1602-59821", "A69-1602-59822", "A69-1602-59823",
"A69-1602-59824", "A69-1602-59825", "A69-1602-59826", "A69-1602-59827",
"A69-1602-59828", "A69-1602-59830", "A69-1602-59831", "A69-1602-59832",
"A69-1602-59833", "A69-1602-59834", "A69-1602-59835", "A69-1602-59836"
), Batch.location = c("Lemmer", "Lemmer", "Lemmer", "Lemmer",
"Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer",
"Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer",
"Lemmer", "Lemmer", "Lemmer"), Location.Dispersion = c("Lemmer",
"Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer",
"Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer",
"Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer", "Lemmer"
), .rows = structure(list(1:2, 3:4, 5L, 6L, 7L, 8L, 9L, 10L,
11L, 12:13, 14L, 15:16, 17L, 18:19, 20L, 21L, 22L, 23:24,
25L), ptype = integer(0), class = c("vctrs_list_of",
"vctrs_vctr", "list"))), row.names = c(NA, -19L), class = c("tbl_df",
"tbl", "data.frame"), .drop = TRUE), class = c("grouped_df",
"tbl_df", "tbl", "data.frame"))
CodePudding user response：
You need to group by Transmitter (missing from your attempt) and ensure that both values are in each group of Season.
dplyr
library(dplyr)
out <- dat %>%
group_by(Transmitter) %>%
filter(all(c("Winter1", "Winter2") %in% Season)) %>%
ungroup()
out
# # A tibble: 12 x 5
# Transmitter Batch.location Location.Dispersion Season Freq
# <chr> <chr> <chr> <chr> <int>
# 1 A69-1602-59814 Lemmer Lemmer Winter1 1961
# 2 A69-1602-59814 Lemmer Lemmer Winter2 2075
# 3 A69-1602-59815 Lemmer Lemmer Winter1 310
# 4 A69-1602-59815 Lemmer Lemmer Winter2 1
# 5 A69-1602-59826 Lemmer Lemmer Winter1 997
# 6 A69-1602-59826 Lemmer Lemmer Winter2 3475
# 7 A69-1602-59828 Lemmer Lemmer Winter1 988
# 8 A69-1602-59828 Lemmer Lemmer Winter2 2991
# 9 A69-1602-59831 Lemmer Lemmer Winter1 3147
# 10 A69-1602-59831 Lemmer Lemmer Winter2 6155
# 11 A69-1602-59835 Lemmer Lemmer Winter1 76
# 12 A69-1602-59835 Lemmer Lemmer Winter2 1921
And from here you can use n_distinct or something else to count the unique Transmitter values you need.
summarize(out, n = n_distinct(Transmitter))
# # A tibble: 1 x 1
# n
# <int>
# 1 6
or just
length(unique(out$Transmitter))
# [1] 6
base R, option 1
ind <- ave(dat$Season, dat$Transmitter,
FUN = function(z) all(c("Winter1", "Winter2") %in% z)) == "TRUE"
ind
# [1] TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE TRUE FALSE TRUE TRUE FALSE TRUE TRUE FALSE
# [21] FALSE FALSE TRUE TRUE FALSE
dat[ind,]
# ...
length(unique(dat[ind, "Transmitter"]))
# [1] 6
The == "TRUE" use of a character "TRUE" is because ave forces the return value to be the same class as its first argument, which is dat$Season. Internally it calculates logical but is coerced to string afterwards. (Just run the ave(..) without ==... to see this in action.)
base R, option 2
sum(aggregate(Season ~ Transmitter, data = dat,
FUN = function(z) all(c("Winter1", "Winter2") %in% z))$Season)
# [1] 6
CodePudding user response：
split by season, then using intersect and length.
with(dat,
do.call(\(...) intersect(...), unname(as.list(split(Transmitter, Season))))
) |> length()
# [1] 6
Or using table and count rows with rowSums equal to two.
with(dat, table(Transmitter, Season)) |>
(\(x) x[rowSums(x) == length(unique(dat$Season)), ])() |>
nrow()
# [1] 6
CodePudding user response：
(Dispersion$Season == "Winter1") & (Dispersion$Season == "Winter2") is looking for rows where Season is both "Winter1" and "Winter2" in the same row (at the same time), which is why that doesn't work. Since you're using dplyr, I would do it this way:
Dispersion %>%
group_by(Transmitter) %>%
filter(all(c("Winter1", "Winter2") %in% Season)) %>%
ungroup() %>%
summarize(n_trans = n_distinct(Transmitter))
# # A tibble: 1 × 1
# n_trans
# <int>
# 1 6
CodePudding user response：
Another base solution:
sum(by(dat$Season, dat$Transmitter, FUN = \(x) { all(unique(dat$Season) %in% x) }))
# [1] 6