dplyr relative frequency within group - r

Question

(hopefully) simplified
I have asked farmers of a specific farmtype (organic and conventional) that I asked for a report on species (A,B) occur (0/1) on their land.
So, I have
df<-data.frame(id=1:10,
farmtype=c(rep("org",4), rep("conv",6)),
spA=c(0,0,0,1,1,1,1,1,1,1),
spB=c(1,1,1,0,0,0,0,0,0,0)
)
And my question is pretty simple... In what percentage of organic or conventional farms do the species occur?
solution
sp A occurs in 25% of org farms and 100% of conv farms
sp B occurs in 75% of org farms and 0% of conv farms
None of the solutions outlined below achieve that.
**additional question **
All I want is a simple ggplot with the species on the x-axis and the percentage of detection on the y-axis (once for org and once for conv).
ggplot(df.melt)+
geom_bar(aes(x=species, fill=farmtype))
### but, of course the species recognitions not just the farm types

Answer 1

janitor's tabyl is your friend. What you're calculating is "row"-percentages, but what you want is "col"-percentages. E.g.
set.seed(1234)
df <- data.frame(farmtype=sample(c("organic","conventional"),100, replace=T),
species=sample(letters[1:4], 100, replace=T),
occ=sample(c("yes","no"),100, replace=T))
df |>
tabyl(species,farmtype) |>
adorn_percentages("col")
# species conventional organic
# a 0.2553191 0.2641509
# b 0.2765957 0.2452830
# c 0.2553191 0.1886792
# d 0.2127660 0.3018868
But you could also use your own approach. Group by farmtype in the second group_by and remember to save the dataframe. This would be easier to use with ggplot2 as it is already in a long format.
df <-
df %>%
group_by(species, farmtype) %>%
dplyr::summarise(count = n()) %>%
group_by(farmtype) %>%
dplyr::mutate(prop = count/sum(count))
df
# A tibble: 8 × 4
# Groups: farmtype [2]
# species farmtype count prop
# <chr> <chr> <int> <dbl>
# a conventional 12 0.255
# a organic 14 0.264
# b conventional 13 0.277
# b organic 13 0.245
# c conventional 12 0.255
# c organic 10 0.189
# d conventional 10 0.213
# d organic 16 0.302
df %>%
ggplot(aes(x = species, y = prop, fill = farmtype)) +
geom_col()
Update: A variant of second option also suggested by Isaac Bravo.

Answer 2

Here you can have another option using your approach:
df %>%
group_by(farmtype, species) %>%
summarize(n = n()) %>%
mutate(percentage = n/sum(n))
OUTPUT:
farmtype species n percentage
<chr> <chr> <int> <dbl>
1 conventional a 12 0.235
2 conventional b 12 0.235
3 conventional c 12 0.235
4 conventional d 15 0.294
5 organic a 16 0.327
6 organic b 9 0.184
7 organic c 14 0.286
8 organic d 10 0.204

Answer 3

If I understand the poster's first question correctly, the poster seeks the proportion of organic versus conventional farm types among farms that grew a given species. This can also be accomplished using the data.table package as follows.
First, the example data set is recreated by setting the seed.
set.seed(1234) ##setting seed for reproducible example
df<-data.frame(farmtype=sample(c("organic","conventional"),100, replace=T),
species=sample(letters[1:4], 100, replace=T),
occ=sample(c("yes","no"),100, replace=T))
require(data.table)
df = data.table(df)
Next, the "no" answers are filtered out because we are only interested in farms that reported growing the species in the "occur" column. We then count the occurrences of the species for each farm type. The column "N" gives the count.
#Filter out "no" answers because they shouldn't affect the result sought
#and count the number of farmtypes that reported each species
ans = df[occ == "yes",.N,by = .(farmtype,species)]
ans
# farmtype species N
#1: conventional a 8
#2: conventional c 8
#3: organic a 6
#4: conventional d 11
#5: organic d 5
#6: organic c 7
#7: organic b 4
#8: conventional b 6
The total occurrences of each species for either farm type are then counted. As a check for this result, each row for a given species should give the same species total.
#Total number of farms that reported the species
ans[,species_total := sum(N), by = species] #
ans
# farmtype species N species_total
#1: conventional a 8 14
#2: conventional c 8 15
#3: organic a 6 14
#4: conventional d 11 16
#5: organic d 5 16
#6: organic c 7 15
#7: organic b 4 10
#8: conventional b 6 10
Finally, the columns are combined to calculate the proportion of organic or conventional farms for each species that was reported. As a check against the result, the proportion of organic and the proportion of conventional for each species should sum to 1 because there are only two farm types.
##Calculate the proportion of each farm type reported for each species
ans[, proportion := N/species_total]
ans
# farmtype species N species_total proportion
#1: conventional a 8 14 0.5714286
#2: conventional c 8 15 0.5333333
#3: organic a 6 14 0.4285714
#4: conventional d 11 16 0.6875000
#5: organic d 5 16 0.3125000
#6: organic c 7 15 0.4666667
#7: organic b 4 10 0.4000000
#8: conventional b 6 10 0.6000000
##Gives the proportion of organic farms specifically
ans[farmtype == "organic"]
# farmtype species N species_total proportion
#1: organic a 6 14 0.4285714
#2: organic d 5 16 0.3125000
#3: organic c 7 15 0.4666667
#4: organic b 4 10 0.4000000
If, on the other hand, one wanted to calculate the fraction of each species to all species occurrences reported for organic or conventional farms, you could use this code:
ans = df[,.N, by = .(species, farmtype,occ)] ##count by species,farmtype, and occurrence
ans[, spf := sum(N), by = .(occ,farmtype)] ##spf is the total number of times an occurrence was reported for each type
ans[, prop := N/spf]
ans = ans[occ == "yes"] ##proportion of the given species to all species occurrences reported for each farm type
ans
# species farmtype occ N spf prop
#1: a conventional yes 8 33 0.2424242
#2: c conventional yes 8 33 0.2424242
#3: a organic yes 6 22 0.2727273
#4: d conventional yes 11 33 0.3333333
#5: d organic yes 5 22 0.2272727
#6: c organic yes 7 22 0.3181818
#7: b organic yes 4 22 0.1818182
#8: b conventional yes 6 33 0.1818182
This result means that, for example, conventional farmers reported species "a" about 24.2% of the times that they reported any species. The result can be verified by selecting a species and farmtype and calculating manually as a spot check.