计算R中每组的凸包

Question

I have a following data set:

structure(list(time = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 
3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L), 
x = c(40.8914337158203, 20.0796813964844, 13.9093618392944, 
17.1513957977295, 18.5109558105469, 40.7868537902832, 19.9750995635986, 
13.804780960083, 16.8376483917236, 18.4063758850098, 40.6822700500488, 
19.7659358978271, 13.7001991271973, 16.6284866333008, 18.3017921447754, 
40.5776901245117, 19.66135597229, 13.5956182479858, 16.3147411346436, 
18.1972122192383, 40.5776901245117, 19.5567722320557, 13.4910354614258, 
16.1055774688721, 17.9880485534668), y = c(0.603550314903259, 
-8.24852085113525, 9.65680503845215, -19.0118350982666, 6.43787002563477, 
0.704141974449158, -8.34911251068115, 9.75739574432373, -19.2130165100098, 
6.43787002563477, 0.704141974449158, -8.44970417022705, 9.75739574432373, 
-19.5147914886475, 6.43787002563477, 0.704141974449158, -8.65088748931885, 
9.85798835754395, -19.8165683746338, 6.33727836608887, 0.704141974449158, 
-8.85207080841064, 9.85798835754395, -20.1183433532715, 6.33727836608887
), object = c(1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 
2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L)), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -25L), .Names = c("time", 
"x", "y", "object"))

Now, I would like to calculate a convex hull (using chull function) for each value of time and store it within the same dataset (as I would like to make a plot with ggplot2 then). I can use chull for each time value using with

chull(filter(data_sample, time == 1)$x, filter(data_sample, time == 1)$y)

which returns a vector of 4 3 1 . So I thought that I can group by time firstly and calculate convex hull points within groups with something like

data_sample %>% group_by(time) %>% summarise(pts = chull(data_sample$x, data_sample$y))

The problem is that I cannot store a vector in a row. Storing each of vertices in separate column would be an option, but the following

data_sample %>% group_by(time) %>% summarise(pt1 = chull(data_sample$x, data_sample$y)[1])

doesn't give reasonable results. So my questions are: 1. How can I store a vector for each row within one column? I have read that tibbles can actually have a list column, but how can I create that in my case? 2. What's wrong with my attempt to calculate chull within each group?

• (extra question, if I may) Why actually data_sample %>% filter(time == 1) %>% chull(.$x, .$y) doesn't work? Is this because chull is not design to work with pipes and dplyr ?

Answer 1

Since chull is giving you indices on the original data, you probably want to preserve the coordinates as you go, which means you probably should not be using summarize . I suggest you go with the "nested" concept as done with tidyr . The first step is nesting your data:

library(tidyr)
data_sample %>%
  group_by(time) %>%
  nest()
# # A tibble: 5 × 2
#    time             data
#   <int>           <list>
# 1     1 <tibble [5 × 3]>
# 2     2 <tibble [5 × 3]>
# 3     3 <tibble [5 × 3]>
# 4     4 <tibble [5 × 3]>
# 5     5 <tibble [5 × 3]>

From here, it's just a matter of calculating the hull (which will return a vector of indices) and then output the relevant rows, in the order provided. This will benefit from the map functions provided by purrr :

library(purrr)
data_sample %>%    data_sample %>%
  group_by(time) %>%
  nest() %>%
  mutate(
    hull = map(data, ~ with(.x, chull(x, y))),
    out = map2(data, hull, ~ .x[.y,,drop=FALSE])
  )
# # A tibble: 5 × 4
#    time             data      hull              out
#   <int>           <list>    <list>           <list>
# 1     1 <tibble [5 × 3]> <int [3]> <tibble [3 × 3]>
# 2     2 <tibble [5 × 3]> <int [3]> <tibble [3 × 3]>
# 3     3 <tibble [5 × 3]> <int [3]> <tibble [3 × 3]>
# 4     4 <tibble [5 × 3]> <int [3]> <tibble [3 × 3]>
# 5     5 <tibble [5 × 3]> <int [3]> <tibble [3 × 3]>

(You should be able to get away with putting both assignments into a single mutate . I

From here, you can turn it into the coordinates you need by removing now-unnecessary columns and unnesting:

data_sample %>%
  group_by(time) %>%
  nest() %>%
  mutate(
    hull = map(data, ~ with(.x, chull(x, y))),
    out = map2(data, hull, ~ .x[.y,,drop=FALSE])
  ) %>%
  select(-data) %>%
  unnest()
# # A tibble: 15 × 5
#     time  hull        x           y object
#    <int> <int>    <dbl>       <dbl>  <int>
# 1      1     4 17.15140 -19.0118351      4
# 2      1     3 13.90936   9.6568050      3
# 3      1     1 40.89143   0.6035503      1
# 4      2     4 16.83765 -19.2130165      4
# 5      2     3 13.80478   9.7573957      3
# 6      2     1 40.78685   0.7041420      1
# 7      3     4 16.62849 -19.5147915      4
# 8      3     3 13.70020   9.7573957      3
# 9      3     1 40.68227   0.7041420      1
# 10     4     4 16.31474 -19.8165684      4
# 11     4     3 13.59562   9.8579884      3
# 12     4     1 40.57769   0.7041420      1
# 13     5     4 16.10558 -20.1183434      4
# 14     5     3 13.49104   9.8579884      3
# 15     5     1 40.57769   0.7041420      1

(I kept hull here for demonstration purposes; you probably can select(-data, -hull) above since you'll have what you need, especially if redundant with object .)

For your last question, you could have done either one of these:

filter(data_sample, time == 1) %>%
  with(., chull(x, y))
with(filter(data_sample, time == 1), chull(x, y))

Answer 2

You can simply pass chull function inside a list:

df <- df %>% 
  group_by(time) %>% 
  mutate(chull_val = list(chull(x,y)))

Answer 3

If you don't want to work with list columns*, you may consider using (the more flexible) data.table .

library(data.table)
setDT(d)
d[d[ , .I[chull(x, y)], by = time]$V1]

Explanation: convert your data to a data.table ( setDT(d) ). For each time ( by = time ), calculate the chull indices and select the corresponding rows ( .I ) (see here ).

---

If you want to plot the chull polygons, you need to add the first index to close the polygon.

d2 <- d[ , {

  # for each time (by = time):
  # compute the indices lying on the convex hull  
  ix <- chull(x, y)

  # use indices to select data of each subset (.SD)
  # possibly also add the first coordinate to close the polygon for plotting   
  .SD[c(ix, ix[1])]}, by = time]


# plot chull and original polygons
library(ggplot2) 
ggplot(d2, aes(x, y, fill = factor(time))) +
  geom_polygon(alpha = 0.2) +
  geom_polygon(data = d, alpha = 0.2)

---

*Related dplyr issues: Summarising verbs with variable-length outputs , Optional parameter to control length of summarise .