I have a dataframe of points in different groups. My actual dataframe is over a thousand lines long. For every combination of groups, I need to find the distance between each point in the combination with every other point. I sum the distances of each point. I have a solution, but it is slow, when I am dealing with say 63 combinations.
To illustrate my current solution, consider the example where I have only three groups. I sort them into all possible combinations i.e. Combination 1 only contains group 1, combination 4 contains group 1 and 2.... (reproducible data below)
I then transform my dataframe into a shapefile of points:
points <- points_csv %>%st_as_sf(coords = c('longitude', 'latitude'))
I then make a vector of the distinct combinations:
Combination_list = points$combination
Combination_list <- unique(Combination_list)
And use the following loop:
Density_total = data.frame()
for (b in Combination_list){
filtered <- filter(points, combination == b)
x <- filtered$geometry
for (t in filtered$geometry){
test_point <- filtered$geometry[t]
M <- st_distance(test_point,x)
M <- unclass(M)
D <- sum(M)
df1 <- data.frame(D)
Density_total <- rbind(Density_total,df1)
}}
Reproducible data:
structure(list(Name = c("Group1", "Group1", "Group2", "Group3",
"Group1", "Group1", "Group2", "Group1", "Group1", "Group3", "Group2",
"Group3", "Group1", "Group2", "Group3"), combination = c("Combination1",
"Combination1", "Combination2", "Combination3", "Combination4",
"Combination4", "Combination4", "Combination5", "Combination5",
"Combination5", "Combination6", "Combination6", "Combination7",
"Combination7", "Combination7"), latitude = c(0.1989, 0.1989,
0.201, 0.201, 0.1989, 0.1989, 0.201, 0.1989, 0.1989, 0.201, 0.201,
0.201, 0.1989, 0.201, 0.201), longitude = c(-0.001, -0.0015,
-0.0015, -0.001, -0.001, -0.0015, -0.0015, -0.001, -0.0015, -0.001,
-0.0015, -0.001, -0.0015, -0.0015, -0.001)), class = "data.frame", row.names = c(NA,
-15L), spec = structure(list(cols = list(Name = structure(list(), class =
c("collector_character",
"collector")), combination = structure(list(), class = c("collector_character",
"collector")), latitude = structure(list(), class = c("collector_double",
"collector")), longitude = structure(list(), class = c("collector_double",
"collector"))), default = structure(list(), class = c("collector_guess",
"collector")), skip = 1), class = "col_spec"))
Desired output should look something like this:
Distance X Y Combination
0.000500000 0.1989 -0.0010 Combination1
0.000500000 0.1989 -0.0015 Combination1
0.000000000 0.2010 -0.0015 Combination2
0.000000000 0.2010 -0.0010 Combination3
0.002658703 0.1989 -0.0010 Combination4
0.002600000 0.1989 -0.0015 Combination4
0.004258703 0.2010 -0.0015 Combination4
0.002600000 0.1989 -0.0010 Combination5
0.002658703 0.1989 -0.0015 Combination5
0.004258703 0.2010 -0.0010 Combination5
0.000500000 0.2010 -0.0015 Combination6
0.000500000 0.2010 -0.0010 Combination6
0.004758703 0.1989 -0.0010 Combination7
0.004758703 0.1989 -0.0015 Combination7
0.004758703 0.2010 -0.0015 Combination7
0.004758703 0.2010 -0.0010 Combination7
Assigning your data to a data.frame named points. Here is a dplyr way to do it. You can use full_join to generate all of the combinations, then calculate the distances. Takes less than a second on my machine with your sample data.
library(dplyr)
points %>%
full_join(points, by = c("combination" = "combination")) %>%
mutate(distance = (longitude.x - longitude.y)^2 + (latitude.x - latitude.y)^2) %>%
group_by(latitude.x, longitude.x, combination) %>%
summarise(total = sum(distance)) %>%
select(Distance = total, X = latitude.x, Y = longitude.x, combination) %>%
arrange(combination)
`summarise()` regrouping output by 'latitude.x', 'longitude.x' (override with `.groups` argument)
# A tibble: 15 x 4
# Groups: X, Y [4]
Distance X Y combination
<dbl> <dbl> <dbl> <chr>
1 0.00000025 0.199 -0.0015 Combination1
2 0.00000025 0.199 -0.001 Combination1
3 0 0.201 -0.0015 Combination2
4 0 0.201 -0.001 Combination3
5 0.00000466 0.199 -0.0015 Combination4
6 0.00000491 0.199 -0.001 Combination4
7 0.00000907 0.201 -0.0015 Combination4
8 0.00000491 0.199 -0.0015 Combination5
9 0.00000466 0.199 -0.001 Combination5
10 0.00000907 0.201 -0.001 Combination5
11 0.00000025 0.201 -0.0015 Combination6
12 0.00000025 0.201 -0.001 Combination6
13 0.00000907 0.199 -0.0015 Combination7
14 0.00000466 0.201 -0.0015 Combination7
15 0.00000491 0.201 -0.001 Combination7
In this sample set, Combinations 2 and 3 have a total distance of 0 because there is only one point in them.