Add frequency into dataframe for each group and unique element (R) - r

I have a table such as
Group Family Nb
1 A 15
2 B 20
3 A 2
3 B 1
3 C 1
4 D 10
4 A 5
5 B 1
5 D 1
And I would like to transform that dataframe such that I have each unique Family element in columns, and for each Group, the frequency of the Nb element, I should then get :
Group A B C D E F
1 1 0 0 0 0 0
2 0 1 0 0 0 0
3 0.5 0.25 0.25 0 0 0
4 0.33 0 0 0.67 0 0
5 0 0.5 0 0.5 0 0
Here is the dput format of the tabel if it can helps :
Family = c("A", "B", "A", "B", "C", "D", "A", "B", "D"),
Nb = c(15L, 20L, 2L, 1L, 1L, 10L, 5L, 1L, 1L)), class = "data.frame", row.names = c(NA,
-9L))


in Base R:
prop.table(xtabs(Nb ~ ., df), 1)
# Family
#Group A B C D
# 1 1.0000000 0.0000000 0.0000000 0.0000000
# 2 0.0000000 1.0000000 0.0000000 0.0000000
# 3 0.5000000 0.2500000 0.2500000 0.0000000
# 4 0.3333333 0.0000000 0.0000000 0.6666667
# 5 0.0000000 0.5000000 0.0000000 0.5000000
If you need it as a data.frame, just wrap the results in as.data.frame.matrix


You can first group_by the Group column, then calculate the frequency and finally pivot the data into a "wide" format.
library(tidyverse)
df %>%
group_by(Group) %>%
mutate(Nb = Nb/sum(Nb)) %>%
pivot_wider(Group, names_from = "Family", values_from = "Nb", values_fill = 0)
# A tibble: 5 × 5
# Groups: Group [5]
Group A B C D
<int> <dbl> <dbl> <dbl> <dbl>
1 1 1 0 0 0
2 2 0 1 0 0
3 3 0.5 0.25 0.25 0
4 4 0.333 0 0 0.667
5 5 0 0.5 0 0.5


Another possible solution:
library(tidyverse)
df %>%
pivot_wider(names_from = Family, values_from = Nb, values_fill = 0) %>%
mutate(aux = rowSums(.[-1]), across(-Group, ~ .x / aux), aux = NULL)
#> # A tibble: 5 × 5
#> Group A B C D
#> <int> <dbl> <dbl> <dbl> <dbl>
#> 1 1 1 0 0 0
#> 2 2 0 1 0 0
#> 3 3 0.5 0.25 0.25 0
#> 4 4 0.333 0 0 0.667
#> 5 5 0 0.5 0 0.5

Related

Conditional replacement of values in a column

I have the following:
ID Value1 Value2 Code
0001 3.3 432 A
0001 0 654 A
0001 0 63 A
0002 0 78 B
0002 1 98 B
0003 0 22 C
0003 0 65 C
0003 0 91 C
I need the following:
ID Value1 Value2 Code
0001 3.3 432 A
0001 0 0 A
0001 0 0 A
0002 0 0 B
0002 1 98 B
0003 0 22 C
0003 0 65 C
0003 0 91 C
i.e., for the same "Code" if there is at least one row with Value1 !=0 then all the other rows referred to the same Code will be set to 0 (meaning that 654 and 63 for 0001 relative to Value2 will be set to 0). If this is not the case (like for 0003 nothing will be done).
Can anyone help me please?
Thank you in advance

dplyr
library(dplyr)
quux %>%
group_by(Code) %>%
mutate(Value2 = if_else(abs(Value1) > 0 | !any(abs(Value1) > 0),
Value2, 0L)) %>%
ungroup()
# # A tibble: 8 x 4
# ID Value1 Value2 Code
# <int> <dbl> <int> <chr>
# 1 1 3.3 432 A
# 2 1 0 0 A
# 3 1 0 0 A
# 4 2 0 0 B
# 5 2 1 98 B
# 6 3 0 22 C
# 7 3 0 65 C
# 8 3 0 91 C
base R
quux |>
transform(Value2 = ifelse(ave(abs(Value1), Code, FUN = function(v) abs(v) > 0 | !any(abs(v) > 0)),
Value2, 0L))
# ID Value1 Value2 Code
# 1 1 3.3 432 A
# 2 1 0.0 0 A
# 3 1 0.0 0 A
# 4 2 0.0 0 B
# 5 2 1.0 98 B
# 6 3 0.0 22 C
# 7 3 0.0 65 C
# 8 3 0.0 91 C
data.table
library(data.table)
as.data.table(quux)[, Value2 := fifelse(abs(Value1) > 0 | !any(abs(Value1) > 0), Value2, 0L), by = Code][]
# ID Value1 Value2 Code
# <int> <num> <int> <char>
# 1: 1 3.3 432 A
# 2: 1 0.0 0 A
# 3: 1 0.0 0 A
# 4: 2 0.0 0 B
# 5: 2 1.0 98 B
# 6: 3 0.0 22 C
# 7: 3 0.0 65 C
# 8: 3 0.0 91 C
Data
quux <- structure(list(ID = c(1L, 1L, 1L, 2L, 2L, 3L, 3L, 3L), Value1 = c(3.3, 0, 0, 0, 1, 0, 0, 0), Value2 = c(432L, 654L, 63L, 78L, 98L, 22L, 65L, 91L), Code = c("A", "A", "A", "B", "B", "C", "C", "C")), class = "data.frame", row.names = c(NA, -8L))

This should do it:
df %>% group_by(Code) %>%
mutate(Value2 = if_else(row_number() == 1 & any(Value1 != 0), Value2, 0))
# A tibble: 8 × 4
# Groups: Code [3]
# ID Value1 Value2 Code
# <int> <dbl> <dbl> <fct>
# 1 1 3.3 432 A
# 2 1 0 0 A
# 3 1 0 0 A
# 4 2 0 78 B
# 5 2 1 0 B
# 6 3 0 0 C
# 7 3 0 0 C
# 8 3 0 0 C

We can use an if_else here. For example
library(dplyr)
dd %>%
group_by(ID) %>%
mutate(Value2=if_else(any(Value1!=0) & Value1==0, 0L, Value2))
Basically we use any() to check for non-zero values and then replace with 0s if one is found.

R: Recursive Averages

I am working with the R programming language. I have the following data:
library(dplyr)
my_data = data.frame(id = c(1,1,1,1,2,2,2,3,4,4,5,5,5,5,5), var_1 = sample(c(0,1), 15, replace = TRUE) , var_2 =sample(c(0,1), 15 , replace = TRUE) )
my_data = data.frame(my_data %>% group_by(id) %>% mutate(index = row_number(id)))
my_data = my_data[,c(1,4,2,3)]
The data looks something like this:
id index var_1 var_2
1 1 1 0 1
2 1 2 0 0
3 1 3 1 1
4 1 4 0 1
5 2 1 1 0
6 2 2 1 1
7 2 3 0 1
8 3 1 1 0
9 4 1 0 0
10 4 2 0 0
11 5 1 0 0
12 5 2 1 0
13 5 3 0 1
14 5 4 0 0
15 5 5 0 1
I want to create two new variables (v_1, v_2). For each unique "id":
v_1: I want v_1 to be the average value of the current, previous and previous-to-previous values of var_1 (i.e. index = n, index = n-1 and index = n-2). When this is not possible (e.g. for index = 2 and index = 1), I want this average to be for as back as you can go.
v_2: I want v_2 to be the average value of the current, previous and previous-to-previous values of var_2 (i.e. index = n, index = n-1 and index = n-2). When this is not possible (e.g. for index = 2 and index = 1), I want this average to be for as back as you can go.
This would be something like this:
row 1 (id = 1, index = 1) : v_1 = var_1 (index 1)
row 2 (id = 1, index = 1 ; id = 1 index = 2) : v_1 = (var_1 (index 1) + var_1 (index 2))/2
row 3 (id = 1, index = 1 ; id = 1 index = 2; id = 1, index = 3) : v_1 = (var_1 (index 1) + var_1 (index 2) + var_1 (index 3)) /3
row 4 (id = 1, index = 2 ; id = 1 index = 3; id = 1, index = 4) : v_1 = (var_1 (index 2) + var_1 (index 3) + var_1 (index 4)) /3
etc.
I tried to do this with the following code:
average_data = my_data %>%
group_by(id) %>%
summarise(v_1 = mean(tail(var_1, 3)),
v_2 = mean(tail(var_2, 3)))
# final_result
final_data = merge(x = my_data, y = average_data, by = "id", all.x = TRUE)
But I am not sure if this is correct.
Can someone please show me how to do this?
Thanks!

data
df <- data.frame(
id = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 4L, 4L, 5L, 5L, 5L, 5L, 5L),
index = c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 1L, 1L, 2L, 1L, 2L, 3L, 4L, 5L),
var_1 = c(0L, 0L, 1L, 0L, 1L, 1L, 0L, 1L, 0L, 0L, 0L, 1L, 0L, 0L, 0L),
var_2 = c(1L, 0L, 1L, 1L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 1L)
)
tidyverse
library(tidyverse)
df %>%
group_by(id) %>%
mutate(across(starts_with("var_"),
.fns = ~zoo::rollapply(data = .x, width = 3, FUN = mean, partial = TRUE, align = "right"),
.names = "new_{.col}")) %>%
ungroup()
#> # A tibble: 15 × 6
#> id index var_1 var_2 new_var_1 new_var_2
#> <int> <int> <int> <int> <dbl> <dbl>
#> 1 1 1 0 1 0 1
#> 2 1 2 0 0 0 0.5
#> 3 1 3 1 1 0.333 0.667
#> 4 1 4 0 1 0.333 0.667
#> 5 2 1 1 0 1 0
#> 6 2 2 1 1 1 0.5
#> 7 2 3 0 1 0.667 0.667
#> 8 3 1 1 0 1 0
#> 9 4 1 0 0 0 0
#> 10 4 2 0 0 0 0
#> 11 5 1 0 0 0 0
#> 12 5 2 1 0 0.5 0
#> 13 5 3 0 1 0.333 0.333
#> 14 5 4 0 0 0.333 0.333
#> 15 5 5 0 1 0 0.667
Created on 2022-06-06 by the reprex package (v2.0.1)
data.table
library(data.table)
COLS <- gsub("ar", "", grep("var_", names(df), value = TRUE))
setDT(df)[,
(COLS) := lapply(.SD, function(x) zoo::rollapply(data = x, width = 3, FUN = mean, partial = TRUE, align = "right")),
by = id,
.SDcols = patterns("var_")][]
#> id index var_1 var_2 v_1 v_2
#> 1: 1 1 0 1 0.0000000 1.0000000
#> 2: 1 2 0 0 0.0000000 0.5000000
#> 3: 1 3 1 1 0.3333333 0.6666667
#> 4: 1 4 0 1 0.3333333 0.6666667
#> 5: 2 1 1 0 1.0000000 0.0000000
#> 6: 2 2 1 1 1.0000000 0.5000000
#> 7: 2 3 0 1 0.6666667 0.6666667
#> 8: 3 1 1 0 1.0000000 0.0000000
#> 9: 4 1 0 0 0.0000000 0.0000000
#> 10: 4 2 0 0 0.0000000 0.0000000
#> 11: 5 1 0 0 0.0000000 0.0000000
#> 12: 5 2 1 0 0.5000000 0.0000000
#> 13: 5 3 0 1 0.3333333 0.3333333
#> 14: 5 4 0 0 0.3333333 0.3333333
#> 15: 5 5 0 1 0.0000000 0.6666667
Created on 2022-06-06 by the reprex package (v2.0.1)

I would say this is moving average, and it can be impemented by a function f like below, using embed (preferrable) or sapply (less efficient, not recommanded), and run it group-wisely using ave:
f <- function(v, n = 3) {
rowMeans(embed(c(rep(NA, n-1), v), n), na.rm = TRUE)
}
or
f <- function(v, n = 3) {
sapply(
seq_along(v),
function(k) sum(v[pmax(k - n + 1, 1):k]) / pmin(k, n)
)
}
and then we run
transform(
df,
v1 = ave(var_1, id, FUN = f),
v2 = ave(var_2, id, FUN = f)
)
such that
id index var_1 var_2 v1 v2
1 1 1 0 1 0.0000000 1.0000000
2 1 2 0 0 0.0000000 0.5000000
3 1 3 1 1 0.3333333 0.6666667
4 1 4 0 1 0.3333333 0.6666667
5 2 1 1 0 1.0000000 0.0000000
6 2 2 1 1 1.0000000 0.5000000
7 2 3 0 1 0.6666667 0.6666667
8 3 1 1 0 1.0000000 0.0000000
9 4 1 0 0 0.0000000 0.0000000
10 4 2 0 0 0.0000000 0.0000000
11 5 1 0 0 0.0000000 0.0000000
12 5 2 1 0 0.5000000 0.0000000
13 5 3 0 1 0.3333333 0.3333333
14 5 4 0 0 0.3333333 0.3333333
15 5 5 0 1 0.0000000 0.6666667

You could create a function that acomplishes this:
library(tidyverse)
fun <- function(x, k){
y <- cummean(first(x, k-1))
if(k > length(x)) y else c(y, zoo::rollmean(x, k))
}
df %>%
group_by(id) %>%
mutate(v_1 = fun(var_1, 3), v_2 = fun(var_2, 3))
# Groups: id [5]
id index var_1 var_2 v_1 v_2
<int> <int> <int> <int> <dbl> <dbl>
1 1 1 0 1 0 1
2 1 2 0 0 0 0.5
3 1 3 1 1 0.333 0.667
4 1 4 0 1 0.333 0.667
5 2 1 1 0 1 0
6 2 2 1 1 1 0.5
7 2 3 0 1 0.667 0.667
8 3 1 1 0 1 0
9 4 1 0 0 0 0
10 4 2 0 0 0 0
11 5 1 0 0 0 0
12 5 2 1 0 0.5 0
13 5 3 0 1 0.333 0.333
14 5 4 0 0 0.333 0.333
15 5 5 0 1 0 0.667

Here is a solution using only built in functions and dplyr:
my_data %>%
mutate(
row = seq_along(id),
v_1 = (var_1 + lag(var_1, default = 0) + lag(var_1, 2, default = 0))/pmin(row, 3),
v_2 = (var_2 + lag(var_2, default = 0) + lag(var_2, 2, default = 0))/pmin(row, 3)
)
#> id index var_1 var_2 row v_1 v_2
#> 1 1 1 0 1 1 0.0000000 1.0000000
#> 2 1 2 1 0 2 0.5000000 0.5000000
#> 3 1 3 1 1 3 0.6666667 0.6666667
#> 4 1 4 1 0 4 1.0000000 0.3333333
#> 5 2 1 0 1 5 0.6666667 0.6666667
#> 6 2 2 0 1 6 0.3333333 0.6666667
#> 7 2 3 1 1 7 0.3333333 1.0000000
#> 8 3 1 1 1 8 0.6666667 1.0000000
#> 9 4 1 1 1 9 1.0000000 1.0000000
#> 10 4 2 1 1 10 1.0000000 1.0000000
#> 11 5 1 0 1 11 0.6666667 1.0000000
#> 12 5 2 0 1 12 0.3333333 1.0000000
#> 13 5 3 0 0 13 0.0000000 0.6666667
#> 14 5 4 0 0 14 0.0000000 0.3333333
#> 15 5 5 1 0 15 0.3333333 0.0000000
Created on 2022-06-09 by the reprex package (v2.0.1)
dplyr::lag() gives you the previuous values of your variable. If they don't exist, we swap them for 0, basically ignoring them. To get the average we divide on pmax(seq_along(<any variable>, 3)), which will be 1 for the first row, 2 for the second, and 3 for all other rows.
This will also work on a grouped dataframe.

You can use filter (which is hidden when loading dplyr) or convolve and ave for grouping.
fun <- function(x) {
. <- if(length(x) > 2) stats::filter(x, c(1,1,1)/3, side=1)[-2:-1] else NULL
#. <- if(length(x) > 2) convolve(x, c(1,1,1)/3, , type = "filter") else NULL #Alternative
c(cummean(x[1:min(2, length(x))]), .)
}
my_data$v_1 <- ave(my_data$var_1, my_data$id, FUN=fun)
my_data$v_2 <- ave(my_data$var_2, my_data$id, FUN=fun)
my_data
# id index var_1 var_2 v_1 v_2
#1 1 1 1 1 1.0000000 1.0000000
#2 1 2 1 1 1.0000000 1.0000000
#3 1 3 0 1 0.6666667 1.0000000
#4 1 4 1 1 0.6666667 1.0000000
#5 2 1 0 1 0.0000000 1.0000000
#6 2 2 0 0 0.0000000 0.5000000
#7 2 3 1 0 0.3333333 0.3333333
#8 3 1 0 0 0.0000000 0.0000000
#9 4 1 0 1 0.0000000 1.0000000
#10 4 2 0 0 0.0000000 0.5000000
#11 5 1 1 0 1.0000000 0.0000000
#12 5 2 0 1 0.5000000 0.5000000
#13 5 3 0 0 0.3333333 0.3333333
#14 5 4 1 0 0.3333333 0.3333333
#15 5 5 0 1 0.3333333 0.3333333
Or using cumsum:
fun2 <- function(x, n=3) {
(cumsum(x) - head(cumsum(c(rep(0, n), x)), -n)) / pmin(n, seq_along(x)) }
my_data$v_1 <- ave(my_data$var_1, my_data$id, FUN=fun2)
my_data$v_2 <- ave(my_data$var_2, my_data$id, FUN=fun2)

Here is a try with a simple function avg to return this type of average
library(dplyr , warn.conflicts = FALSE)
set.seed(1978)
my_data = data.frame(id = c(1,1,1,1,2,2,2,3,4,4,5,5,5,5,5), var_1 = sample(c(0,1), 15, replace = TRUE) , var_2 =sample(c(0,1), 15 , replace = TRUE) )
my_data = data.frame(my_data %>% group_by(id) %>% mutate(index = row_number(id)))
my_data = my_data[,c(1,4,2,3)]
#===================================
avg <- function(x){
t <- rep(c(T,NA) , c(3 , length(x) - 1))
m <- numeric(length(x))
for(i in 1:length(x)){
m [i]<- mean(x[t[3:length(t)]] , na.rm = TRUE)
t <- lag(t)
}
m
}
#===================================
library(tidyverse)
my_data %>%
group_by(id) %>%
mutate(v_1 = avg(var_1), v_2 = avg(var_2))
#> # A tibble: 15 × 6
#> # Groups: id [5]
#> id index var_1 var_2 v_1 v_2
#> <dbl> <int> <dbl> <dbl> <dbl> <dbl>
#> 1 1 1 0 0 0 0
#> 2 1 2 1 0 0.5 0
#> 3 1 3 1 0 0.667 0
#> 4 1 4 1 1 1 0.333
#> 5 2 1 0 1 0 1
#> 6 2 2 1 1 0.5 1
#> 7 2 3 0 0 0.333 0.667
#> 8 3 1 1 0 1 0
#> 9 4 1 1 1 1 1
#> 10 4 2 0 1 0.5 1
#> 11 5 1 1 1 1 1
#> 12 5 2 1 0 1 0.5
#> 13 5 3 0 1 0.667 0.667
#> 14 5 4 1 0 0.667 0.333
#> 15 5 5 1 0 0.667 0.333
Created on 2022-06-09 by the reprex package (v2.0.1)

This uses dplyr's across with slider's slide_dbl; both from the tidyverse. Slider handles partial windows, so is well-suited to this problem.
(%>% may be used instead of the native pipe |>.)
library(dplyr)
library(slider)
# Sample Data
df <- data.frame(
id = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 4L, 4L, 5L, 5L, 5L, 5L, 5L),
index = c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 1L, 1L, 2L, 1L, 2L, 3L, 4L, 5L),
var_1 = c(0L, 0L, 1L, 0L, 1L, 1L, 0L, 1L, 0L, 0L, 0L, 1L, 0L, 0L, 0L),
var_2 = c(1L, 0L, 1L, 1L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 1L)
)
# Possible answer
df |>
group_by(id) |>
mutate(across(starts_with("var_"), ~ slide_dbl(., mean, .before = 2), .names = "{.col}_mean")) |>
ungroup()
#> # A tibble: 15 × 6
#> id index var_1 var_2 var_1_mean var_2_mean
#> <int> <int> <int> <int> <dbl> <dbl>
#> 1 1 1 0 1 0 1
#> 2 1 2 0 0 0 0.5
#> 3 1 3 1 1 0.333 0.667
#> 4 1 4 0 1 0.333 0.667
#> 5 2 1 1 0 1 0
#> 6 2 2 1 1 1 0.5
#> 7 2 3 0 1 0.667 0.667
#> 8 3 1 1 0 1 0
#> 9 4 1 0 0 0 0
#> 10 4 2 0 0 0 0
#> 11 5 1 0 0 0 0
#> 12 5 2 1 0 0.5 0
#> 13 5 3 0 1 0.333 0.333
#> 14 5 4 0 0 0.333 0.333
#> 15 5 5 0 1 0 0.667
Created on 2022-06-12 by the reprex package (v2.0.1)

Creating multiple proportion table in R dataframe

I've the following table
Result_Group
Review
A
1
B
4
A
1
C
1
D
5
D
4
E
5
C
1
C
2
A
2
B
3
E
2
df = structure(list(Result_Group = structure(c(1L, 2L, 1L, 3L, 4L, 4L, 5L, 3L, 3L, 1L, 2L, 5L), .Label = c("A", "B", "C", "D", "E"
), class = "factor"), Review = c(1L, 4L, 1L, 1L, 5L, 4L, 5L, 1L, 2L, 2L, 3L, 2L)),
class = "data.frame", row.names = c(NA, -12L))
Does anyone know how can create a table for the proportion of the review for each group? Currently I'm doing it group by group and it's taking quite a while just to subset the data.
i.e. the table as follows:
Review
A
B
C
D
E
1
0.67
0
0.67
0
0
2
0.33
0
0.33
0
0.50
3
0
0.50
0
0
0
4
0
0.50
0
0.5
0
5
0
0
0
0.5
0.50
Thanks!

You can do:
library(tidyverse)
df |>
group_by(Result_Group) |>
count(Review) |>
mutate(prop = n/sum(n)) |>
ungroup() |>
select(-n) |>
pivot_wider(names_from = Result_Group,
values_from = prop,
values_fill = 0)
# A tibble: 5 x 6
Review A B C D E
<int> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 0.667 0 0.667 0 0
2 2 0.333 0 0.333 0 0.5
3 3 0 0.5 0 0 0
4 4 0 0.5 0 0.5 0
5 5 0 0 0 0.5 0.5

Here is a tidy approach using dplyr and tidyr
library(dplyr)
df %>%
# Add count values (all equal to 1)
mutate(count = 1) %>%
# Pivot wider to get A, B, C.. as column names, and sum of count as values
tidyr::pivot_wider(
id_cols = Review,
names_from = Result_Group,
values_from = count,
values_fn = sum,
values_fill = 0 # NAs are turned into 0
) %>%
# Mutate to get fractions instead of count
mutate(
across(
-Review,
~ .x / sum(.x)
)
) %>%
# Sort by review
arrange(Review)
#> # A tibble: 5 × 6
#> Review A B C D E
#> <int> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 1 0.667 0 0.667 0 0
#> 2 2 0.333 0 0.333 0 0.5
#> 3 3 0 0.5 0 0 0
#> 4 4 0 0.5 0 0.5 0
#> 5 5 0 0 0 0.5 0.5
Created on 2022-03-22 by the reprex package (v2.0.1)

r convert summary data to presence/absence data

I conducted 5 presence/absence measures at multiple sites and summed them together and ended up with a dataframe that looked something like this:
df <- data.frame("site" = c("a", "b", "c"),
"species1" = c(0, 2, 1),
"species2" = c(5, 2, 4))
ie. at site "a" species1 was recorded 0/5 times and species2 was recorded 5/5 times.
What I would like to do is convert this back into presence/absence data. Something like this:
data.frame("site" = ("a", "b", "c"),
"species1" = c(0,0,0,0,0, 1,1,0,0,0, 1,0,0,0,0),
"species2" = c(1,1,1,1,1, 1,1,0,0,0, 1,1,1,1,0))
I can duplicate each row 5 times with:
df %>% slice(rep(1:n(), each = 5))
but I can't figure out how to change "2" into "1,1,0,0,0". Ideally the order of the 1s and 0s (within each site) would also be randomised (ie. "0,0,1,0,1"), but that might be too difficult.
Any help would be appreciated.

We can also use uncount
library(dplyr)
library(tidyr)
df %>%
uncount(max(species2), .remove = FALSE) %>%
group_by(site) %>%
mutate(across(starts_with('species'), ~ as.integer(row_number() <= first(.))))
# A tibble: 15 x 3
# Groups: site [3]
# site species1 species2
# <chr> <int> <int>
# 1 a 0 1
# 2 a 0 1
# 3 a 0 1
# 4 a 0 1
# 5 a 0 1
# 6 b 1 1
# 7 b 1 1
# 8 b 0 0
# 9 b 0 0
#10 b 0 0
#11 c 1 1
#12 c 0 1
#13 c 0 1
#14 c 0 1
#15 c 0 0

After repeating the rows you can compare the row number with any value of the respective column and assign 1 if the current row number is less than the value.
library(dplyr)
df %>%
slice(rep(seq_len(n()), each = 5)) %>%
group_by(site) %>%
mutate(across(starts_with('species'), ~+(row_number() <= first(.))))
#Use mutate_at with old dplyr
#mutate_at(vars(starts_with('species')), ~+(row_number() <= first(.)))
# site species1 species2
# <chr> <int> <int>
# 1 a 0 1
# 2 a 0 1
# 3 a 0 1
# 4 a 0 1
# 5 a 0 1
# 6 b 1 1
# 7 b 1 1
# 8 b 0 0
# 9 b 0 0
#10 b 0 0
#11 c 1 1
#12 c 0 1
#13 c 0 1
#14 c 0 1
#15 c 0 0

Finding values in consecutive rows

An example of the dataframe I have is given below.
ID X
1 1
2 2
3 1
4 0
5 0
6 1
7 4
8 5
9 6
10 7
11 0
12 0
I want to apply logic to it that looks to see whether 3 or more consecutive rows have a value >0 in it. If they do I want to flag them in another column. Hence the output will look as follows.
ID X Y
1 1 1
2 2 1
3 1 1
4 0 0
5 0 0
6 1 1
7 4 1
8 5 1
9 6 1
10 7 1
11 0 0
12 0 0
EXTENSION -
How would I get the following output, givibng a different Y value for each group?
ID X Y
1 1 1
2 2 1
3 1 1
4 0 0
5 0 0
6 1 2
7 4 2
8 5 2
9 6 2
10 7 2
11 0 0
12 0 0

One option with base R. Using rle to find the adjacent values in 'X' that are greater than 0, then do the replication based on the lengths
df1$Y <- with(rle(df1$X > 0), as.integer(rep(values & lengths > 2, lengths)))
df1$Y
#[1] 1 1 1 0 0 1 1 1 1 1 0 0
For the updated case in the OP's post
df1$Y <- inverse.rle(within.list(rle(df1$X > 0), {
i1 <- values & (lengths > 2)
values[i1] <- seq_along(values[i1])}))
df1$Y
#[1] 1 1 1 0 0 2 2 2 2 2 0 0
Or using rleid from data.table
library(data.table)
setDT(df1)[, Y := as.integer((.N > 2) * (X > 0)),rleid(X > 0)]
data
df1 <- structure(list(ID = 1:12, X = c(1L, 2L, 1L, 0L, 0L, 1L, 4L, 5L,
6L, 7L, 0L, 0L)), class = "data.frame", row.names = c(NA, -12L
))

We can use rleid from data.table to create groups and use it in ave and get length of each group and assign 1 to groups which has length greater than equal to 3.
library(data.table)
df$Y <- as.integer(ave(df$X, rleid(df$X > 0), FUN = length) >= 3)
df
# ID X Y
#1 1 1 1
#2 2 2 1
#3 3 1 1
#4 4 0 0
#5 5 0 0
#6 6 1 1
#7 7 4 1
#8 8 5 1
#9 9 6 1
#10 10 7 1
#11 11 0 0
#12 12 0 0
EDIT
For updated post we could include the above data.table part with dplyr by doing
library(dplyr)
library(data.table)
df %>%
group_by(group = rleid(X > 0)) %>%
mutate(Y = ifelse(n() >= 3 & row_number() == 1, 1, 0)) %>%
ungroup() %>%
mutate(Y = cumsum(Y) * Y) %>%
group_by(group) %>%
mutate(Y = first(Y)) %>%
ungroup() %>%
select(-group)
# ID X Y
# <int> <int> <dbl>
# 1 1 1 1
# 2 2 2 1
# 3 3 1 1
# 4 4 0 0
# 5 5 0 0
# 6 6 1 2
# 7 7 4 2
# 8 8 5 2
# 9 9 6 2
#10 10 7 2
#11 11 0 0
#12 12 0 0

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