I am working with the R programming language.
In the "datasets" library in R, there is a data set called "eurodist" that contains the distance between each combination of cities :
library(datasets)
This data set can be then converted into a "matrix":
eurodist = as.matrix(eurodist)
Athens Barcelona Brussels Calais Cherbourg Cologne Copenhagen Geneva Gibraltar Hamburg Hook of Holland Lisbon Lyons Madrid Marseilles Milan Munich Paris Rome Stockholm Vienna
Athens 0 3313 2963 3175 3339 2762 3276 2610 4485 2977 3030 4532 2753 3949 2865 2282 2179 3000 817 3927 1991
Barcelona 3313 0 1318 1326 1294 1498 2218 803 1172 2018 1490 1305 645 636 521 1014 1365 1033 1460 2868 1802
Brussels 2963 1318 0 204 583 206 966 677 2256 597 172 2084 690 1558 1011 925 747 285 1511 1616 1175
Calais 3175 1326 204 0 460 409 1136 747 2224 714 330 2052 739 1550 1059 1077 977 280 1662 1786 1381
Cherbourg 3339 1294 583 460 0 785 1545 853 2047 1115 731 1827 789 1347 1101 1209 1160 340 1794 2196 1588
Cologne 2762 1498 206 409 785 0 760 1662 2436 460 269 2290 714 1764 1035 911 583 465 1497 1403 937
Copenhagen 3276 2218 966 1136 1545 760 0 1418 3196 460 269 2971 1458 2498 1778 1537 1104 1176 2050 650 1455
Geneva 2610 803 677 747 853 1662 1418 0 1975 1118 895 1936 158 1439 425 328 591 513 995 2068 1019
Gibraltar 4485 1172 2256 2224 2047 2436 3196 1975 0 2897 2428 676 1817 698 1693 2185 2565 1971 2631 3886 2974
Hamburg 2977 2018 597 714 1115 460 460 1118 2897 0 550 2671 1159 2198 1479 1238 805 877 1751 949 1155
Hook of Holland 3030 1490 172 330 731 269 269 895 2428 550 0 2280 863 1730 1183 1098 851 457 1683 1500 1205
Lisbon 4532 1305 2084 2052 1827 2290 2971 1936 676 2671 2280 0 1178 668 1762 2250 2507 1799 2700 3231 2937
Lyons 2753 645 690 739 789 714 1458 158 1817 1159 863 1178 0 1281 320 328 724 471 1048 2108 1157
Madrid 3949 636 1558 1550 1347 1764 2498 1439 698 2198 1730 668 1281 0 1157 1724 2010 1273 2097 3188 2409
Marseilles 2865 521 1011 1059 1101 1035 1778 425 1693 1479 1183 1762 320 1157 0 618 1109 792 1011 2428 1363
Milan 2282 1014 925 1077 1209 911 1537 328 2185 1238 1098 2250 328 1724 618 0 331 856 586 2187 898
Munich 2179 1365 747 977 1160 583 1104 591 2565 805 851 2507 724 2010 1109 331 0 821 946 1754 428
Paris 3000 1033 285 280 340 465 1176 513 1971 877 457 1799 471 1273 792 856 821 0 1476 1827 1249
Rome 817 1460 1511 1662 1794 1497 2050 995 2631 1751 1683 2700 1048 2097 1011 586 946 1476 0 2707 1209
Stockholm 3927 2868 1616 1786 2196 1403 650 2068 3886 949 1500 3231 2108 3188 2428 2187 1754 1827 2707 0 2105
Vienna 1991 1802 1175 1381 1588 937 1455 1019 2974 1155 1205 2937 1157 2409 1363 898 428 1249 1209 2105 0
My Question: Suppose I have 6 cities and the Longitude/Latitude for each of these cities :
data_1 = data.frame(id = c(1,2,3), long = rnorm(3, -74, 1 ), lat = rnorm(3, 40, 1 ))
data_2 = data.frame(id = c(4,5,6), long = rnorm(3, -78, 1 ), lat = rnorm(3, 42, 1 ))
final_data = rbind(data_1, data_2)
final_data$names <- c("city_1", "city_2", "city_3", "city_4", "city_5", "city_6")
id long lat names
1 1 -75.28447 40.21079 city_1
2 2 -73.29385 40.09104 city_2
3 3 -75.12737 38.88355 city_3
4 4 -79.42325 42.61917 city_4
5 5 -77.82508 41.11707 city_5
6 6 -77.62831 39.94935 city_6
I can also make a similar matrix for these cities that contains the distance between each pair of cities:
library(geosphere)
N <- nrow(final_data)
dists <- outer(seq_len(N), seq_len(N), function(a,b) {
geosphere::distHaversine(final_data[a,2:3], final_data[b,2:3]) # Notes 1, 2
})
D <- as.matrix(dists)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 0.0 169895.7 148361.1 437239.3 237056.7 201742.0
[2,] 169895.7 0.0 207068.8 584183.9 399577.9 369814.4
[3,] 148361.1 207068.8 0.0 551356.0 338698.3 245620.3
[4,] 437239.3 584183.9 551356.0 0.0 213326.6 332955.7
[5,] 237056.7 399577.9 338698.3 213326.6 0.0 131051.7
[6,] 201742.0 369814.4 245620.3 332955.7 131051.7 0.0
How can I make my matrix look the same way as the "eurodist" matrix?
I had thought of the following way to do this:
colnames(dists) <- c("city_1", "city_2", "city_3", "city_4", "city_5", "city_6")
rownames(dists) <- c("city_1", "city_2", "city_3", "city_4", "city_5", "city_6")
city_1 city_2 city_3 city_4 city_5 city_6
city_1 0.0 169895.7 148361.1 437239.3 237056.7 201742.0
city_2 169895.7 0.0 207068.8 584183.9 399577.9 369814.4
city_3 148361.1 207068.8 0.0 551356.0 338698.3 245620.3
city_4 437239.3 584183.9 551356.0 0.0 213326.6 332955.7
city_5 237056.7 399577.9 338698.3 213326.6 0.0 131051.7
city_6 201742.0 369814.4 245620.3 332955.7 131051.7 0.0
In the end, I would like to use the above matrix as input for a customized Travelling Salesman Problem (R: Customizing the Travelling Salesman Problem) - e.g. Try to find the optimal path when you are forced to start at "city 4" and the third city should be "city 5":
D <- dists
transformMatrix <- function(fixed_points, D){
if(length(fixed_points) == 0) return(D)
p <- integer(nrow(D))
pos <- match(names(fixed_points), colnames(D))
p[fixed_points] <- pos
p[-fixed_points] <- sample(setdiff(seq_len(nrow(D)), pos))
D[p, p]
}
fixed_points <- c(
"city_4" = 1, "city_5" = 3
)
D_perm <- transformMatrix(fixed_points, D)
feasiblePopulation <- function(n, size, fixed_points){
positions <- setdiff(seq_len(n), fixed_points)
m <- matrix(0, size, n)
if(length(fixed_points) > 0){
m[, fixed_points] <- rep(fixed_points, each = size)
for(i in seq_len(size))
m[i, -fixed_points] <- sample(positions)
} else {
for(i in seq_len(size))
m[i,] <- sample(positions)
}
m
}
mutation <- function(n, fixed_points){
positions <- setdiff(seq_len(n), fixed_points)
function(obj, parent){
vec <- obj#population[parent,]
if(length(positions) < 2) return(vec)
indices <- sample(positions, 2)
replace(vec, indices, vec[rev(indices)])
}
}
fitness <- function(tour, distMatrix) {
tour <- c(tour, tour[1])
route <- embed(tour, 2)[,2:1]
1/sum(distMatrix[route])
}
popSize = 500
res <- ga(
type = "permutation",
fitness = fitness,
distMatrix = D_perm,
lower = 1,
upper = nrow(D_perm),
mutation = mutation(nrow(D_perm), fixed_points),
crossover = gaperm_pmxCrossover,
suggestions = feasiblePopulation(nrow(D_perm), popSize, fixed_points),
popSize = popSize,
maxiter = 5000,
run = 500,
pmutation = 0.2
)
colnames(D_perm)[res#solution[1,]]
This results in the following error:
Error in if (object#run >= run) break :
missing value where TRUE/FALSE needed
In addition: Warning messages:
1: In max(fitness) : no non-missing arguments to max; returning -Inf
2: In max(Fitness, na.rm = TRUE) :
no non-missing arguments to max; returning -Inf
3: In max(fitness) : no non-missing arguments to max; returning -Inf
4: In max(x, na.rm = TRUE) :
no non-missing arguments to max; returning -Inf
Is the above error because I have not made "distance matrix" (i.e. "D") properly? Is there a different way to name the columns and rows of a matrix in R?
Thanks!
Note : If anyone knows another way to solve this constraint Travelling Salesman Problem with custom cities using the Genetic Algorithm in R (e.g. different objective function, different way to specify constraints, etc.), please let me know. I am open to different ways to solving this problem!
That’s not the problem. The error says the it encountered code:
if (object#run >= run) break
… and either object#run or run had length 0 which the if function cannot handle gracefully. It may be an error in the ga function itself or in the arguments to it.
To address the direct question about how to make the distance matrix look like the example in eurodist: There is a dimnames attribute for matrices. You need to assign a list with a rownames and a colnames value in it and assign that list to the dimnames attribute.
dimnames(D) <- list(rownames=final_data$names,
colnames=final_data$names)
Then when you run your code you get an error from the ga(...) call:
Error in gaperm_pmxCrossover_Rcpp(object, parents) : index error
Looking at the problem setup, your population size appears much larger than needed. If you drop it down a bit to say 100 or 200, then the results begin to be computed.
popSize=200;
# now calculate a res
colnames(D_perm)[res#solution[1,]]
#[1] "city_4" "city_6" "city_5" "city_1" "city_3" "city_2"
popSize=100
colnames(D_perm)[res#solution[1,]]
#[1] "city_4" "city_6" "city_5" "city_1" "city_3" "city_2"
popSiz=20
colnames(D_perm)[res#solution[1,]]
#[1] "city_4" "city_6" "city_5" "city_1" "city_3" "city_2"
It doesn't seem "proper" that a population size larger than needed should cause an obscure error, so you might contact the package maintainer with your example (now that it has been "dressed up" properly.)
Using ts() in R with a single column csv data set, the first row/value in the column is ignored. If I use that row as a label/header, the data is read as multivariate, which throws an error.
Code looks like this:
entry_ts<-ts(data, frequency = 12, start = c(2016,7), end=c(2022,1))
plot(entry_ts)
entry_ts %>%
stl(s.window = "periodic")%>%
autoplot()
Data looks like this, in a csv file that I am viewing in Excel- no header row, just values.
950
1083
1028
967
891
918
980
875
961
986
1019
771
666
908
(with additional values...)
I get this, where the July value is showing as the second value in the vector, and because the code is looping back to select that same "first" value again to fill in the last (because it ignored the first value so it doesn't have the right number of values for the start and end dates I gave).
entry_ts
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
2016 1083 1028 967 891 918 980
2017 875 961 986 1019 771 666 908 818 815 828 744 871
2018 773 823 827 827 769 757 871 765 866 717 687 776
2019 738 876 866 990 769 844 1067 756 838 774 755 849
2020 768 831 728 737 780 773 969 961 873 761 715 599
2021 670 747 685 701 730 707 710 770 565 623 592 568
2022 1083
I tried to enter a label in the first row, but then get an error that the data is not univariate.
I'd like to define a series of variables in a for loop. (create a array as dictionary. Convert tops to d1 as shown below)
Firstly, I assign values to them (d1~d11);
then I try to define the names of these variables.
How should I call specific variables in the names() function to make it work like "names(d1)<-..."
for (i = 1:11)
{
assign(paste("d",i,sep=""),tops[,2*i])
names(eval(parse(text=paste("d",i,sep=""))))<-tops[,2*i-1]
}
> tops[,c(1,2)]
V1 V2 V3 V4 V5 V6
1 shift 2136 shift 2211 shift 2324
2 bed 1463 k 1551 plant 1664
3 run 1338 bed 1527 run 1466
4 plant 1309 run 1504 k 1456
5 k 1294 hr 1484 bed 1390
6 hr 1285 clean 1464 hr 1366
7 check 1255 plant 1386 clean 1359
8 clean 1203 check 1261 s 1254
9 s 1052 s 1205 check 1048
10 unload 1024 start 1115 end 1028
11 chang 1023 fine 1113 fine 1020
12 fine 960 chang 1104 start 1006
13 end 924 end 1050 chang 977
14 start 905 stop 974 stop 950
15 pellet 878 pellet 915 pellet 897
16 work 866 work 907 remov 874
17 due 856 screen 900 sinter 862
18 stop 853 bwr 888 side 841
19 complet 772 side 888 due 809
20 remov 750 due 861 conveyor 792
21 requir 726 complet 841 work 777
22 sinter 711 sinter 834 north 771
23 south 710 conveyor 775 south 760
24 side 688 north 768 west 738
25 issu 682 remov 764 belt 737
26 t 675 ok 759 carri 735
27 belt 672 t 753 screen 727
28 carri 668 requir 750 stock 725
29 strand 649 unload 749 unload 719
30 conveyor 646 chute 747 chute 688
> d1
shift bed run plant k hr check clean s
2136 1463 1338 1309 1294 1285 1255 1203 1052
unload chang fine end start pellet work due stop
1024 1023 960 924 905 878 866 856 853
complet remov requir sinter south side issu t belt
772 750 726 711 710 688 682 675 672
carri strand conveyor
668 649 646
> length(d1)
[1] 30
I hope I make it clear. if not, please free to ask me
As David mentioned, don't assign 11 different variables; create a list with 11 elements. This will simplify your code considerably.
d <- lapply(1:11, function(i) tops[, 2 * i = 1])
The data in the table is given below:
Year NSW Vic. Qld SA WA Tas. NT ACT Aust.
1 1917 1904 1409 683 440 306 193 5 3 4941
2 1927 2402 1727 873 565 392 211 4 8 6182
3 1937 2693 1853 993 589 457 233 6 11 6836
4 1947 2985 2055 1106 646 502 257 11 17 7579
5 1957 3625 2656 1413 873 688 326 21 38 9640
6 1967 4295 3274 1700 1110 879 375 62 103 11799
7 1977 5002 3837 2130 1286 1204 415 104 214 14192
8 1987 5617 4210 2675 1393 1496 449 158 265 16264
9 1997 6274 4605 3401 1480 1798 474 187 310 18532
I want to plot a graph with (Year) on my x-axis and (total value) on my Y-axis. The barplot should depicting the ACT and NT value for the respective (Years).
I tried the following command:
barplot(as.matrix(r_data$ACT, r_data$NT), main="r_data", ylab="Total", beside=TRUE)
The above command showed the barplot of ACT column per year but didn't show the Bar plot of NT column.
You have to create the matrix in a different way:
barplot(as.matrix(r_data[c("ACT", "NT")]),
main="r_data", ylab="Total", beside=TRUE)
You can also use cbind instead of as.matrix and keep the rest of your original approach:
barplot(cbind(r_data$ACT, r_data$NT),
main="r_data", ylab="Total", beside=TRUE)
Here is my input, question is below:
Targets <- read.csv("miR155Aicda.csv", row.names=1, sep="", header=T)
head(Targets)
T0h T0.25h T0.5h T1h T2h T3h T6h T12h T24h T48h C0h C0.25h C0.5h C1h C2h
Aicda 785 1150 707 513 1265 3268 8294 8625 7387 4397 677 911 673 737 1782
mmu-miR-155-3p 622 548 558 1213 1195 1172 1115 1883 3257 1900 499 562 584 543 580
Targets.m <- melt(Targets)
> head(Targets.m)
variable value
1 T0h 9.616549
2 T0h 9.280771
3 T0.25h 10.167418
4 T0.25h 9.098032
5 T0.5h 9.465566
6 T0.5h 9.124121
Question: How do I add Aicda and mmu-mIR-155-2p as a variable?
I want this:
ID variable value
1 Aicda T0h 9.616549
2 mmu-miR-155-3p T0h 9.280771
3 Aicda T0.25h 10.167418
4 mmu-miR-155-3p T0.25h 9.098032
5 Aicda T0.5h 9.465566
6 mmu-miR-155-3p T0.5h 9.124121
You need to put the rownames in a column and then specify an id variable for melt:
DF <- read.table(text=" T0h T0.25h T0.5h T1h T2h T3h T6h T12h T24h T48h C0h C0.25h C0.5h C1h C2h
Aicda 785 1150 707 513 1265 3268 8294 8625 7387 4397 677 911 673 737 1782
mmu-miR-155-3p 622 548 558 1213 1195 1172 1115 1883 3257 1900 499 562 584 543 580",header=TRUE)
DF$ID <- rownames(DF)
library(reshape2)
melt(DF,id="ID")