).我使用了分支定界算法来解决这个问题,使我能够获得我的“短名单”中的点,因此,以后可能可以将该部分转移到另一个平台/语言。感谢这在实际数据集上非常有效。我可以将组的数量扩展到8个,没有任何问题。我可以用Rglpk稍微提高速度,但最多只需要8秒,所以没有必
).我使用了分支定界算法来解决这个问题,使我能够获得我的“短名单”中的点,因此,以后可能可以将该部分转移到另一个平台/语言。感谢这在实际数据集上非常有效。我可以将组的数量扩展到8个,没有任何问题。我可以用Rglpk稍微提高速度,但最多只需要8秒,所以没有必,r,coordinates,grouping,distance,euclidean-distance,R,Coordinates,Grouping,Distance,Euclidean Distance,).我使用了分支定界算法来解决这个问题,使我能够获得我的“短名单”中的点,因此,以后可能可以将该部分转移到另一个平台/语言。感谢这在实际数据集上非常有效。我可以将组的数量扩展到8个,没有任何问题。我可以用Rglpk稍微提高速度,但最多只需要8秒,所以没有必要。此后,行数开始呈指数级增加,继续下去也不再实际。我还想感谢您早期的建议和指导。干杯! # A tibble: 61 x 3 indexR x y <dbl> <dbl> <dbl&
).我使用了分支定界算法来解决这个问题,使我能够获得我的“短名单”中的点,因此,以后可能可以将该部分转移到另一个平台/语言。感谢这在实际数据集上非常有效。我可以将组的数量扩展到8个,没有任何问题。我可以用Rglpk稍微提高速度,但最多只需要8秒,所以没有必要。此后,行数开始呈指数级增加,继续下去也不再实际。我还想感谢您早期的建议和指导。干杯!
# A tibble: 61 x 3
indexR x y
<dbl> <dbl> <dbl>
1 1 837 924
2 1 464 661
3 1 838 132
4 1 245 882
5 1 1161 604
6 1 1185 504
7 1 853 870
8 1 1048 859
9 1 1044 514
10 1 141 938
# ... with 51 more rows
#dput provided at bottom of this post
> df$dummy = 1
> df %>%
+ full_join(df, c("dummy" = "dummy")) %>%
+ full_join(df, c("dummy" = "dummy")) %>%
+ filter(indexR.x != indexR.y & indexR.x != indexR & indexR.y != indexR) %>%
+ mutate(dist =
+ ((.$x - .$x.x)^2 + (.$y- .$y.x)^2)^.5 +
+ ((.$x - .$x.y)^2 + (.$y- .$y.y)^2)^.5 +
+ ((.$x.x - .$x.y)^2 + (.$y.x- .$y.y)^2)^.5,
+ dist = round(dist, digits = 0)) %>%
+ arrange(dist) %>%
+ filter(dist == min(dist))
# A tibble: 6 x 11
indexR.x x.x y.x dummy indexR.y x.y y.y indexR x y dist
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 638 324 1 2 592 250 3 442 513 664
2 1 638 324 1 3 442 513 2 592 250 664
3 2 592 250 1 1 638 324 3 442 513 664
4 2 592 250 1 3 442 513 1 638 324 664
5 3 442 513 1 1 638 324 2 592 250 664
6 3 442 513 1 2 592 250 1 638 324 664
structure(list(indexR = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 3, 3), x = c(836.65, 464.43, 838.12, 244.68, 1160.86,
1184.52, 853.4, 1047.96, 1044.2, 141.06, 561.01, 1110.74, 123.4,
1087.24, 827.83, 100.86, 140.07, 306.5, 267.83, 1118.61, 155.04,
299.52, 543.5, 782.25, 737.1, 1132.14, 659.48, 871.78, 1035.33,
867.81, 192.94, 1167.8, 1099.59, 1097.3, 1089.78, 1166.59, 703.33,
671.64, 346.49, 440.89, 126.38, 638.24, 972.32, 1066.8, 775.68,
591.86, 818.75, 953.63, 1104.98, 1050.47, 722.43, 1022.17, 986.38,
1133.01, 914.27, 725.15, 1151.52, 786.08, 1024.83, 246.52, 441.53
), y = c(923.68, 660.97, 131.61, 882.23, 604.09, 504.05, 870.35,
858.51, 513.5, 937.7, 838.47, 482.69, 473.48, 171.78, 774.99,
792.46, 251.26, 757.95, 317.71, 401.93, 326.32, 725.89, 98.43,
414.01, 510.16, 973.61, 445.33, 504.54, 669.87, 598.75, 225.27,
789.45, 135.31, 935.51, 270.38, 241.19, 595.05, 401.25, 160.98,
778.86, 192.17, 323.76, 361.08, 444.92, 354, 249.57, 301.64,
375.75, 440.03, 428.79, 276.5, 408.84, 381.14, 459.14, 370.26,
304.05, 439.14, 339.91, 435.85, 759.42, 513.37)), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -61L), .Names = c("indexR",
"x", "y"))
df$id <- row.names(df) # to create ID's for the points
df2 <- merge(df, df, by = NULL ) # the first cross join
df3 <- merge(df2, df, by = NULL) # the second cross join
# eliminating rows where the points are of the same indexR
df3 <- df3[df3$indexR.x != df3$indexR.y & df3$indexR.x != df3$indexR
& df3$indexR.y != df3$indexR,]
## calculating the total distance
df3$total_distance <- ((df3$x - df3$x.x)^2 + (df3$y- df3$y.x)^2)^.5 +
((df3$x - df3$x.y)^2 + (df3$y- df3$y.y)^2)^.5 +
((df3$x.x - df3$x.y)^2 + (df3$y.x- df3$y.y)^2)^.5
## minimum distance
df3[which.min(df3$total_distance),]
indexR.x x.x y.x id.x indexR.y x.y y.y id.y indexR x y id total_distance
155367 3 441.53 513.37 61 2 591.86 249.57 46 1 638.24 323.76 42 664.3373
opt.closest <- function(df) {
# Compute every pair of indices
library(dplyr)
pairs <- as.data.frame(t(combn(nrow(df), 2))) %>%
mutate(G1=df$indexR[V1], G2=df$indexR[V2]) %>%
filter(G1 != G2) %>%
mutate(dist = sqrt((df$x[V1]-df$x[V2])^2+(df$y[V1]-df$y[V2])^2))
# Compute a few convenience values
n <- nrow(df)
nP <- nrow(pairs)
groups <- sort(unique(df$indexR))
nG <- length(groups)
gpairs <- combn(groups, 2)
nGP <- ncol(gpairs)
# Solve the optimization problem
obj <- c(pairs$dist, rep(0, n))
constr <- rbind(cbind(diag(nP), -outer(pairs$V1, seq_len(n), "==")),
cbind(diag(nP), -outer(pairs$V2, seq_len(n), "==")),
cbind(diag(nP), -outer(pairs$V1, seq_len(n), "==") - outer(pairs$V2, seq_len(n), "==")),
cbind(matrix(0, nG, nP), outer(groups, df$indexR, "==")),
cbind((outer(gpairs[1,], pairs$G1, "==") &
outer(gpairs[2,], pairs$G2, "==")) |
(outer(gpairs[2,], pairs$G1, "==") &
outer(gpairs[1,], pairs$G2, "==")), matrix(0, nGP, n)))
dir <- rep(c("<=", ">=", "="), c(2*nP, nP, nG+nGP))
rhs <- rep(c(0, -1, 1), c(2*nP, nP, nG+nGP))
library(lpSolve)
mod <- lp("min", obj, constr, dir, rhs, all.bin=TRUE)
which(tail(mod$solution, n) == 1)
}
df[opt.closest(df),]
# A tibble: 3 x 3
# indexR x y
# <dbl> <dbl> <dbl>
# 1 1 638.24 323.76
# 2 2 591.86 249.57
# 3 3 441.53 513.37
make.dataset <- function(n, nG) {
set.seed(144)
data.frame(indexR = sample(seq_len(nG), n, replace=T), x = rnorm(n), y=rnorm(n))
}
df100 <- make.dataset(100, 7)
system.time(opt.closest(df100))
# user system elapsed
# 11.536 2.656 15.407
df200 <- make.dataset(200, 7)
system.time(opt.closest(df200))
# user system elapsed
# 187.363 86.454 323.167