I have lat/long data of two animals tracked in Western Australia and I'd like to find their home ranges using adehabitatHR.
library(sp)
library(rgdal)
library(raster)
library(adehabitatHR)
library(sf)
quolls<-read.csv("quolls.csv")
head(quolls)
Latitude Longitude animal_ID
1 -22.62271 117.1247 1
2 -22.62286 117.1246 1
3 -22.62192 117.1223 1
4 -22.62021 117.1224 1
5 -22.61989 117.1244 1
6 -22.62022 117.1260 1
But the home range estimates of each animal are obviously too small.
I think the EPSG must be wrong but after a very long time looking I still can't find the right one.
Can anyone point me in the right direction please?
# make a SpatialPoints dataframe without a CRS
quolls2 <- quolls
quoll.latlong<-data.frame(x=quolls2$Longitude,y=quolls2$Latitude)
coordinates(quolls2) <- quoll.latlong
# add crs
proj4string(quolls2) <- CRS(SRS_string = "EPSG:4283")
mcp<-mcp(quolls2[,7],percent=95,unout = c("ha"))
mcp
Home range for animal 1 is 1.217428e-08 and animal 2 is 6.253689e-08.
And likewise with kernel density estimation;
quoll_ud <- adehabitatHR::kernelUD(quolls2[7],grid = 450)
quoll_hr <- adehabitatHR::getverticeshr(quoll_ud, 99)
print(quoll_hr)
which estimates animal 1 at 2.36917592701502e-08 and animal 2 at 1.16018636413173e-07.
Just stumbled across the answer.. it's EPSG 28350.
I got it to work in the end by abandoning the raw lats and longs and instead importing a shapefile I had of the animal data with st_read.
Then st_transform to 28350.
Then as mcp accepts only SpatialPoints, I converted the object with
as(obj, "Spatial").
Related
I have a dataframe with 3 columns. Each of the columns contains some "labels". I want to study a correspondence between the labels of the three columns. As such, I created 3 heatmaps, for each pair of columns, that shows the number of times a pair of labels has appeared.
For example:
colA | colB | colC
dog car USA
cat plane Germany
fish truck Spain
eagle bike France
dog car USA
eagle train UK
A heat map of the first two columns above is:
dog 2 0 0 0 0
cat 0 1 0 0 0
fish 0 0 1 0 0
eagle 0 0 0 1 1
car plane truck bike train
Now, in the same manner, I can create the other two heatmaps. My question is, can I combine two of them (for example keeping the horizontal axis the same and adding two vertical axes, for the other two columns) create a heatmap that includes the entire triplet correspondence?
Sorry if my question seems a bit vague, but I am trying to see if there are ways of visualizing a three-way correspondence in the style of a heatmap.
this link may help you:
Combine multiple heatmaps in matplotlib
one of the answers from above link:
There are a few options to present 2 datasets together:
Options 1 - draw a heatmap of the difference of 2 datasets (or ratio, whatever is more appropriate in your case)
pcolor(D2-D1)
and then present several of these comparison figures.
Option 2 - present 1 dataset as pcolor, and another as countour:
pcolor(D1)
contour(D2)
If you really need to show N>2 datasets together, I would go with contour or contourf:
contourf(D1,cmap='Blues')
contourf(D2,cmap='Reds', alpha=0.66)
contourf(D2,cmap='Reds', alpha=0.33)
example output of 3 contourf commands
or
contour(D1,cmap='Blues')
contour(D2,cmap='Reds')
contour(D2,cmap='Reds')
example output of 3 contour commands
unfortunately, simiar alpha tricks do not work well with pcolor.
You’re given a chess board with dimension n x n. There’s a king at the bottom right square of the board marked with s. The king needs to reach the top left square marked with e. The rest of the squares are labeled either with an integer p (marking a point) or with x marking an obstacle. Note that the king can move up, left and up-left (diagonal) only. Find the maximum points the king can collect and the number of such paths the king can take in order to do so.
Input Format
The first line of input consists of an integer t. This is the number of test cases. Each test case contains a number n which denotes the size of board. This is followed by n lines each containing n space separated tokens.
Output Format
For each case, print in a separate line the maximum points that can be collected and the number of paths available in order to ensure maximum, both values separated by a space. If e is unreachable from s, print 0 0.
Sample Input
3
3
e 2 3
2 x 2
1 2 s
3
e 1 2
1 x 1
2 1 s
3
e 1 1
x x x
1 1 s
Sample Output
7 1
4 2
0 0
Constraints
1 <= t <= 100
2 <= n <= 200
1 <= p <= 9
I think this problem could be solved using dynamic-programing. We could use dp[i,j] to calculate the best number of points you can obtain by going from the right bottom corner to the i,j position. We can calculate dp[i,j], for a valid i,j, based on dp[i+1,j], dp[i,j+1] and dp[i+1,j+1] if this are valid positions(not out of the matrix or marked as x) and adding them the points obtained in the i,j cell. You should start computing from the bottom right corner to the left top, row by row and beginning from the last column.
For the number of ways you can add a new matrix ways and use it to store the number of ways.
This is an example code to show the idea:
dp[i,j] = dp[i+1,j+1] + board[i,j]
ways[i,j] = ways[i+1,j+1]
if dp[i,j] < dp[i+1,j] + board[i,j]:
dp[i,j] = dp[i+1,j] + board[i,j]
ways[i,j] = ways[i+1,j]
elif dp[i,j] == dp[i+1,j] + board[i,j]:
ways[i,j] += ways[i+1,j]
# check for i,j+1
This assuming all positions are valid.
The final result is stored in dp[0,0] and ways[0,0].
Brief Overview:
This problem can be solved through recursive method call, starting from nn till it reaches 00 which is the king's destination.
For the detailed explanation and the solution for this problem,check it out here -> https://www.callstacker.com/detail/algorithm-1
I'm trying to create 20 unique cards with numbers, but I struggle a bit.. So basically I need to create 20 unique matrices 3x3 having numbers 1-10 in first column, numbers 11-20 in the second column and 21-30 in the third column.. Any ideas? I'd prefer to have it done in r, especially as I don't know Visual Basic. In excel I know how to generate the cards, but not sure how to ensure they are unique..
It seems to be quite precise and straightforward to me. Anyway, i needed to create 20 matrices that would look like :
[,1] [,2] [,3]
[1,] 5 17 23
[2,] 8 18 22
[3,] 3 16 24
Each of the matrices should be unique and each of the columns should consist of three unique numbers ( the 1st column - numbers 1-10, the 2nd column 11-20, the 3rd column - 21-30).
Generating random numbers is easy, though how to make sure that generated cards are unique?Please have a look at the post that i voted for as an answer - as it gives you thorough explanation how to achieve it.
(N.B. : I misread "rows" instead of "columns", so the following code and explanation will deal with matrices with random numbers 1-10 on 1st row, 11-20 on 2nd row etc., instead of columns, but it's exactly the same just transposed)
This code should guarantee uniqueness and good randomness :
library(gtools)
# helper function
getKthPermWithRep <- function(k,n,r){
k <- k - 1
if(n^r< k){
stop('k is greater than possibile permutations')
}
v <- rep.int(0,r)
index <- length(v)
while ( k != 0 )
{
remainder<- k %% n
k <- k %/% n
v[index] <- remainder
index <- index - 1
}
return(v+1)
}
# get all possible permutations of 10 elements taken 3 at a time
# (singlerowperms = 720)
allperms <- permutations(10,3)
singlerowperms <- nrow(allperms)
# get 20 random and unique bingo cards
cards <- lapply(sample.int(singlerowperms^3,20),FUN=function(k){
perm2use <- getKthPermWithRep(k,singlerowperms,3)
m <- allperms[perm2use,]
m[2,] <- m[2,] + 10
m[3,] <- m[3,] + 20
return(m)
# if you want transpose the result just do:
# return(t(m))
})
Explanation
(disclaimer tl;dr)
To guarantee both randomness and uniqueness, one safe approach is generating all the possibile bingo cards and then choose randomly among them without replacements.
To generate all the possible cards, we should :
generate all the possibilities for each row of 3 elements
get the cartesian product of them
Step (1) can be easily obtained using function permutations of package gtools (see the object allPerms in the code). Note that we just need the permutations for the first row (i.e. 3 elements taken from 1-10) since the permutations of the other rows can be easily obtained from the first by adding 10 and 20 respectively.
Step (2) is also easy to get in R, but let's first consider how many possibilities will be generated. Step (1) returned 720 cases for each row, so, in the end we will have 720*720*720 = 720^3 = 373248000 possible bingo cards!
Generate all of them is not practical since the occupied memory would be huge, thus we need to find a way to get 20 random elements in this big range of possibilities without actually keeping them in memory.
The solution comes from the function getKthPermWithRep, which, given an index k, it returns the k-th permutation with repetition of r elements taken from 1:n (note that in this case permutation with repetition corresponds to the cartesian product).
e.g.
# all permutations with repetition of 2 elements in 1:3 are
permutations(n = 3, r = 2,repeats.allowed = TRUE)
# [,1] [,2]
# [1,] 1 1
# [2,] 1 2
# [3,] 1 3
# [4,] 2 1
# [5,] 2 2
# [6,] 2 3
# [7,] 3 1
# [8,] 3 2
# [9,] 3 3
# using the getKthPermWithRep you can get directly the k-th permutation you want :
getKthPermWithRep(k=4,n=3,r=2)
# [1] 2 1
getKthPermWithRep(k=8,n=3,r=2)
# [1] 3 2
Hence now we just choose 20 random indexes in the range 1:720^3 (using sample.int function), then for each of them we get the corresponding permutation of 3 numbers taken from 1:720 using function getKthPermWithRep.
Finally these triplets of numbers, can be converted to actual card rows by using them as indexes to subset allPerms and get our final matrix (after, of course, adding +10 and +20 to the 2nd and 3rd row).
Bonus
Explanation of getKthPermWithRep
If you look at the example above (permutations with repetition of 2 elements in 1:3), and subtract 1 to all number of the results you get this :
> permutations(n = 3, r = 2,repeats.allowed = T) - 1
[,1] [,2]
[1,] 0 0
[2,] 0 1
[3,] 0 2
[4,] 1 0
[5,] 1 1
[6,] 1 2
[7,] 2 0
[8,] 2 1
[9,] 2 2
If you consider each number of each row as a number digit, you can notice that those rows (00, 01, 02...) are all the numbers from 0 to 8, represented in base 3 (yes, 3 as n). So, when you ask the k-th permutation with repetition of r elements in 1:n, you are also asking to translate k-1 into base n and return the digits increased by 1.
Therefore, given the algorithm to change any number from base 10 to base n :
changeBase <- function(num,base){
v <- NULL
while ( num != 0 )
{
remainder = num %% base # assume K > 1
num = num %/% base # integer division
v <- c(remainder,v)
}
if(is.null(v)){
return(0)
}
return(v)
}
you can easily obtain getKthPermWithRep function.
One 3x3 matrix with the desired value range can be generated with the following code:
mat <- matrix(c(sample(1:10,3), sample(11:20,3), sample(21:30, 3)), nrow=3)
Furthermore, you can use a for loop to generate a list of 20 unique matrices as follows:
for (i in 1:20) {
mat[[i]] <- list(matrix(c(sample(1:10,3), sample(11:20,3), sample(21:30,3)), nrow=3))
print(mat[[i]])
}
Well OK I may fall on my face here but I propose a checksum (using Excel).
This is a unique signature for each bingo card which will remain invariate if the order of numbers within any column is changed without changing the actual numbers. The formula is
=SUM(10^MOD(A2:A4,10)+2*10^MOD(B2:B4,10)+4*10^MOD(C2:C4,10))
where the bingo numbers for the first card are in A2:C4.
The idea is to generate a 10-digit number for each column, then multiply each by a constant and add them to get the signature.
So here I have generated two random bingo cards using a standard formula from here plus two which are deliberately made to be just permutations of each other.
Then I check if any of the signatures are duplicates using the formula
=MAX(COUNTIF(D5:D20,D5:D20))
which shouldn't given an answer more than 1.
In the unlikely event that there were duplicates, then you would just press F9 and generate some new cards.
All formulae are array formulae and must be entered with CtrlShiftEnter
Here is an inelegant way to do this. Generate all possible combinations and then sample without replacement. These are permutations, combinations: order does matter in bingo
library(dplyr)
library(tidyr)
library(magrittr)
generate_samples = function(n) {
first = data_frame(first = (n-9):n)
first %>%
merge(first %>% rename(second = first)) %>%
merge(first %>% rename(third = first)) %>%
sample_n(20)
}
suffix = function(df, suffix)
df %>%
setNames(names(.) %>%
paste0(suffix))
generate_samples(10) %>% suffix(10) %>%
bind_cols(generate_samples(20) %>% suffix(20)) %>%
bind_cols(generate_samples(30) %>% suffix(30)) %>%
rowwise %>%
do(matrix = t(.) %>% matrix(3)) %>%
use_series(matrix)
We are investigating migrating a prototype into SQL Server (azure).
We have LineStrings that also have M values. What we would like to do is given another M value find out what its geographical location is.
To aid your visualisation, here is a real-world example:
I have a linestring that represents a flight path. Because the flight goes up and down the distance the plane has actually moved is not the same as the total length of the linestring. We have calibrated M values as a part of the linestring but need to be able to plot on it where a given event occurred. All we know about this event is its M value.
SET #g = geometry::STGeomFromText('LINESTRING(1 0 NULL 0, 2 2 NULL 5, 1 4 NULL 9, 3 6 NULL 15)', 0);
Given something like the above, what is the lat and long of a point with an M value of 8?
This should be an equivalent postgis's ST_LocateAlong
The M value is not a time, but a distance. It should be understood that this distance is arbitrary and does not directly relate to the length of the line and is calibrated against known points. This is due to the set being based on historic data that is in no way accurate by today's standards.
*Note I am not sure if I have Nulled the Z or M value. The extra parameter we are considering here is the M only.
Can someone please explain the Differential Evolution method? The Wikipedia definition is extremely technical.
A dumbed-down explanation followed by a simple example would be appreciated :)
Here's a simplified description. DE is an optimisation technique which iteratively modifies a population of candidate solutions to make it converge to an optimum of your function.
You first initialise your candidate solutions randomly. Then at each iteration and for each candidate solution x you do the following:
you produce a trial vector: v = a + ( b - c ) / 2, where a, b, c are three distinct candidate solutions picked randomly among your population.
you randomly swap vector components between x and v to produce v'. At least one component from v must be swapped.
you replace x in your population with v' only if it is a better candidate (i.e. it better optimise your function).
(Note that the above algorithm is very simplified; don't code from it, find proper spec. elsewhere instead)
Unfortunately the Wikipedia article lacks illustrations. It is easier to understand with a graphical representation, you'll find some in these slides: http://www-personal.une.edu.au/~jvanderw/DE_1.pdf .
It is similar to genetic algorithm (GA) except that the candidate solutions are not considered as binary strings (chromosome) but (usually) as real vectors. One key aspect of DE is that the mutation step size (see step 1 for the mutation) is dynamic, that is, it adapts to the configuration of your population and will tend to zero when it converges. This makes DE less vulnerable to genetic drift than GA.
Answering my own question...
Overview
The principal difference between Genetic Algorithms and Differential Evolution (DE) is that Genetic Algorithms rely on crossover while evolutionary strategies use mutation as the primary search mechanism.
DE generates new candidates by adding a weighted difference between two population members to a third member (more on this below).
If the resulting candidate is superior to the candidate with which it was compared, it replaces it; otherwise, the original candidate remains unchanged.
Definitions
The population is made up of NP candidates.
Xi = A parent candidate at index i (indexes range from 0 to NP-1) from the current generation. Also known as the target vector.
Each candidate contains D parameters.
Xi(j) = The jth parameter in candidate Xi.
Xa, Xb, Xc = three random parent candidates.
Difference vector = (Xb - Xa)
F = A weight that determines the rate of the population's evolution.
Ideal values: [0.5, 1.0]
CR = The probability of crossover taking place.
Range: [0, 1]
Xc` = A mutant vector obtained through the differential mutation operation. Also known as the donor vector.
Xt = The child of Xi and Xc`. Also known as the trial vector.
Algorithm
For each candidate in the population
for (int i = 0; i<NP; ++i)
Choose three distinct parents at random (they must differ from each other and i)
do
{
a = random.nextInt(NP);
} while (a == i)
do
{
b = random.nextInt(NP);
} while (b == i || b == a);
do
{
c = random.nextInt(NP);
} while (c == i || c == b || c == a);
(Mutation step) Add a weighted difference vector between two population members to a third member
Xc` = Xc + F * (Xb - Xa)
(Crossover step) For every variable in Xi, apply uniform crossover with probability CR to inherit from Xc`; otherwise, inherit from Xi. At least one variable must be inherited from Xc`
int R = random.nextInt(D);
for (int j=0; j < D; ++j)
{
double probability = random.nextDouble();
if (probability < CR || j == R)
Xt[j] = Xc`[j]
else
Xt[j] = Xi[j]
}
(Selection step) If Xt is superior to Xi then Xt replaces Xi in the next generation. Otherwise, Xi is kept unmodified.
Resources
See this for an overview of the terminology
See Optimization Using Differential Evolution by Vasan Arunachalam for an explanation of the Differential Evolution algorithm
See Evolution: A Survey of the State-of-the-Art by Swagatam Das and Ponnuthurai Nagaratnam Suganthan for different variants of the Differential Evolution algorithm
See Differential Evolution Optimization from Scratch with Python for a detailed description of an implementation of a DE algorithm in python.
The working of DE algorithm is very simple.
Consider you need to optimize(minimize,for eg) ∑Xi^2 (sphere model) within a given range, say [-100,100]. We know that the minimum value is 0. Let's see how DE works.
DE is a population-based algorithm. And for each individual in the population, a fixed number of chromosomes will be there (imagine it as a set of human beings and chromosomes or genes in each of them).
Let me explain DE w.r.t above function
We need to fix the population size and the number of chromosomes or genes(named as parameters). For instance, let's consider a population of size 4 and each of the individual has 3 chromosomes(or genes or parameters). Let's call the individuals R1,R2,R3,R4.
Step 1 : Initialize the population
We need to randomly initialise the population within the range [-100,100]
G1 G2 G3 objective fn value
R1 -> |-90 | 2 | 1 | =>8105
R2 -> | 7 | 9 | -50 | =>2630
R3 -> | 4 | 2 | -9.2| =>104.64
R4 -> | 8.5 | 7 | 9 | =>202.25
objective function value is calculated using the given objective function.In this case, it's ∑Xi^2. So for R1, obj fn value will be -90^2+2^2+2^2 = 8105. Similarly it is found for all.
Step 2 : Mutation
Fix a target vector,say for eg R1 and then randomly select three other vectors(individuals)say for eg.R2,R3,R4 and performs mutation. Mutation is done as follows,
MutantVector = R2 + F(R3-R4)
(vectors can be chosen randomly, need not be in any order).F (scaling factor/mutation constant) within range [0,1] is one among the few control parameters DE is having.In simple words , it describes how different the mutated vector becomes. Let's keep F =0.5.
| 7 | 9 | -50 |
+
0.5 *
| 4 | 2 | -9.2|
+
| 8.5 | 7 | 9 |
Now performing Mutation will give the following Mutant Vector
MV = | 13.25 | 13.5 | -50.1 | =>2867.82
Step 3 : Crossover
Now that we have a target vector(R1) and a mutant vector MV formed from R2,R3 & R4 ,we need to do a crossover. Consider R1 and MV as two parents and we need a child from these two parents. Crossover is done to determine how much information is to be taken from both the parents. It is controlled by Crossover rate(CR). Every gene/chromosome of the child is determined as follows,
a random number between 0 & 1 is generated, if it is greater than CR , then inherit a gene from target(R1) else from mutant(MV).
Let's set CR = 0.9. Since we have 3 chromosomes for individuals, we need to generate 3 random numbers between 0 and 1. Say for eg, those numbers are 0.21,0.97,0.8 respectively. First and last are lesser than CR value, so those positions in the child's vector will be filled by values from MV and second position will be filled by gene taken from target(R1).
Target-> |-90 | 2 | 1 | Mutant-> | 13.25 | 13.5 | -50.1 |
random num - 0.21, => `Child -> |13.25| -- | -- |`
random num - 0.97, => `Child -> |13.25| 2 | -- |`
random num - 0.80, => `Child -> |13.25| 2 | -50.1 |`
Trial vector/child vector -> | 13.25 | 2 | -50.1 | =>2689.57
Step 4 : Selection
Now we have child and target. Compare the obj fn of both, see which is smaller(minimization problem). Select that individual out of the two for next generation
R1 -> |-90 | 2 | 1 | =>8105
Trial vector/child vector -> | 13.25 | 2 | -50.1 | =>2689.57
Clearly, the child is better so replace target(R1) with the child. So the new population will become
G1 G2 G3 objective fn value
R1 -> | 13.25 | 2 | -50.1 | =>2689.57
R2 -> | 7 | 9 | -50 | =>2500
R3 -> | 4 | 2 | -9.2 | =>104.64
R4 -> | -8.5 | 7 | 9 | =>202.25
This procedure will be continued either till the number of generations desired has reached or till we get our desired value. Hope this will give you some help.