Does xtend support two-or more dimensional arrays? If yes: How can i create one and use it later on? I want to store Strings in these arrays and pass it to the files.
Here is my solution that creates a matrix of integers (rows x cols).
The only disadvantage is that each row is allocated separately in a loop.
#Pure
static def int[][] newIntArrayOfSize(int nrow, int ncol) {
newArrayOfSize(nrow).map[ newIntArrayOfSize(ncol) ]
}
Demo:
// allocate 3 rows by 2 columns
val m = newIntArrayOfSize(3, 2)
// notice that indexes are starting from 0
// here setting value 5 on row=2, col=1
m.get(2).set(1, 5)
for(row : 0 .. 2) {
for(col : 0..1) {
print(m.get(row).get(col))
print("\t")
}
println()
}
Easy, just import the Guava lib into your XTend script. The Guava lib has multidimenional MultiSet in it.
Related
we as 3 friends try to solve minimum spanning tree with coflicts problem using r. In solving this question, we read files in .txt format that contain for ex.
"1 2 5
2 4 6" etc. which indicates from node 1 to 2, there exists an edge with weight 5 and
"1 2 2 4" etc. which indicates there's a conflict relationship between the edges 1-2 and 2-4. To continue, we have to form an nxn conflict matrix in which we will store 0's if there exist no conflict relation between the edges or 1 if there exist a conflict relation. For this purpose, we developed a 3-for loop for(i in 1:dim(edges_read)[1]){
for(i in 1:dim(edges_read)[1]){
for(k in 1:dim(edges_read)[1]){
for(t in 1:dim(conflicts)[1]){
if(all(conflicts[t,] == c(edges_read[i,1], edges_read[i,2],
edges_read[k,1], edges_read[k,2]) )){
conflictmatrix[i,k] <- 1
}
}
}
}
However, R cannot get us a solution and this for loops take very long times. How can we solve this situation? Thanks for further assistance
As you have discovered, for() loops are not fast in R. There are faster approaches, but it's hard to provide examples without data. Please use something like dput(edges_read) and dput(conflicts) to provide a small example of the data.
As one example, you could implement the for loops in the Rcpp package for speed improvement. Based on the code in your question, you could re-implement the 3-loop code sort of like this:
Rcpp::cppFunction('NumericVector MSTC_nxn_Cpp(NumericMatrix edges_read, NumericMatrix conflicts){
int n = edges_read.nrow(); //output matrix size (adjust to what you need)
int m = conflicts.nrow(); //output matrix size (adjust to what you need)
NumericMatrix conflictmatrix( n , m ); //the output matrix
for(int i=0;i<n;i++){ //your i loop
for(int k=0;k<n;k++){ // your k loop
double te = edges_read( i, 0 ); //same as edges_read[i,1]
double tf = edges_read( i, 1 ); //same as edges_read[i,2]
double tg = edges_read( k, 0 ); //same as edges_read[k,1]
double th = edges_read( k, 1 ); //same as edges_read[k,2]
NumericVector w = NumericVector::create(te,tf,tg,th); //this could probably be more simple
for(int t=0;t<m;t++){ //your t loop
NumericVector v = conflicts( t , _ ); // same as conflicts[t,]
LogicalVector r; //vector for checking if conflicts and edges are the same
for(int p=0; p<4; p++){ //loop to check logic
r[p]=v[p]==w[p]; //True / False stored
};
int q = r.size();
for (int ii = 0; ii < q; ++ii) { //similar to all() This code could be simplified!
if (!r[ii]) {false;}
else{conflictmatrix[i,k] = 1;}}
}}}
return conflictmatrix; //your output
}')
#Then run the function
MSTC_nxn_Cpp(edges_read, conflicts )
The problem is as such:
given an array of N numbers, find two numbers in the array such that they will have a range(max - min) value of K.
for example:
input:
5 3
25 9 1 6 8
output:
9 6
So far, what i've tried is first sorting the array and then finding two complementary numbers using a nested loop. However, because this is a sort of brute force method, I don't think it is as efficient as other possible ways.
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt(), k = sc.nextInt();
int[] arr = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
Arrays.sort(arr);
int count = 0;
int a, b;
for(int i = 0; i < n; i++) {
for(int j = i; j < n; j++) {
if(Math.max(arr[i], arr[j]) - Math.min(arr[i], arr[j]) == k) {
a = arr[i];
b = arr[j];
}
}
}
System.out.println(a + " " + b);
}
}
Much appreciated if the solution was in code (any language).
Here is code in Python 3 that solves your problem. This should be easy to understand, even if you do not know Python.
This routine uses your idea of sorting the array, but I use two variables left and right (which define two places in the array) where each makes just one pass through the array. So other than the sort, the time efficiency of my code is O(N). The sort makes the entire routine O(N log N). This is better than your code, which is O(N^2).
I never use the inputted value of N, since Python can easily handle the actual size of the array. I add a sentinel value to the end of the array to make the inner short loops simpler and quicker. This involves another pass through the array to calculate the sentinel value, but this adds little to the running time. It is possible to reduce the number of array accesses, at the cost of a few more lines of code--I'll leave that to you. I added input prompts to aid my testing--you can remove those to make my results closer to what you seem to want. My code prints the larger of the two numbers first, then the smaller, which matches your sample output. But you may have wanted the order of the two numbers to match the order in the original, un-sorted array--if that is the case, I'll let you handle that as well (I see multiple ways to do that).
# Get input
N, K = [int(s) for s in input('Input N and K: ').split()]
arr = [int(s) for s in input('Input the array: ').split()]
arr.sort()
sentinel = max(arr) + K + 2
arr.append(sentinel)
left = right = 0
while arr[right] < sentinel:
# Move the right index until the difference is too large
while arr[right] - arr[left] < K:
right += 1
# Move the left index until the difference is too small
while arr[right] - arr[left] > K:
left += 1
# Check if we are done
if arr[right] - arr[left] == K:
print(arr[right], arr[left])
break
I have a matrix of the form
vector<vector<int>> K
which has size NxN. How can i swap two elements of this vector say K[i]k[j] with K[n-j][i]?
In general, how can i swap two elements of an 2D vector?
Because vector's [] operator returns a reference, a std::swap() will work. For example:
swap(K[i][j], K[n-i][i]);
The general swapping could look something like this
void swap(int& a, int& b)
{
int temp = a;
a = b;
b = tmp;
}
Then you call it with
swap(K[i][j], K[n - j][i]);
Or you can just call std::swap as #Jeffrey suggests
I have a groovy collections which is an array, containing value starting from 0 through 'n'. I need to find a particular array index when a series of conditions occured. And,I do not need to scan through every value of the array but can jump across pre-defined intervals. For example, look for the condition for every 10 values in the array. Can someone tell me a way to do this?
For example, I want to do somehting like this below
def alltimes = [0 . . . . . 10000]
def end_time = 10000
def time = 0
while(time <= end_time)
{
// check the condition for alltimes[time]
if(condition_satisfied){
println "condition satisfied at time ${time}"
break
}
time = time + 50
}
When i explored all available methods of array, i did not find any one which can allow to jump variables instead of just one as in methods each, eachwithindex.
Seems like I need to use metaclass and create a new method?
You can use find for this:
def allTimes = 0..10000
Closure<Boolean> checkCondition = { all, single ->
single > 300
}
(0..10000).step( 50 ).find { time -> checkCondition( allTimes, time ) }
Which is ripe for currying:
def allTimes = 0..10000
Closure<Boolean> checkCondition = { all, single ->
single > 300
}
(0..10000).step( 50 ).find checkCondition.curry( allTimes )
I found this project on GitHub; it was the only search term returned for "nimrod matrix". I took the bare bones of it and changed it a little bit so that it compiled without errors, and then I added the last two lines to build a simple matrix, and then output a value, but the "getter" function isn't working for some reason. I adapted the instructions for adding properties found here, but something isn't right.
Here is my code so far. I'd like to use the GNU Scientific Library from within Nimrod, and I figured that this was the first logical step.
type
TMatrix*[T] = object
transposed: bool
dataRows: int
dataCols: int
data: seq[T]
proc index[T](x: TMatrix[T], r,c: int): int {.inline.} =
if r<0 or r>(x.rows()-1):
raise newException(EInvalidIndex, "matrix index out of range")
if c<0 or c>(x.cols()-1):
raise newException(EInvalidIndex, "matrix index out of range")
result = if x.transposed: c*x.dataCols+r else: r*x.dataCols+c
proc rows*[T](x: TMatrix[T]): int {.inline.} =
## Returns the number of rows in the matrix `x`.
result = if x.transposed: x.dataCols else: x.dataRows
proc cols*[T](x: TMatrix[T]): int {.inline.} =
## Returns the number of columns in the matrix `x`.
result = if x.transposed: x.dataRows else: x.dataCols
proc matrix*[T](rows, cols: int, d: openarray[T]): TMatrix[T] =
## Constructor. Initializes the matrix by allocating memory
## for the data and setting the number of rows and columns
## and sets the data to the values specified in `d`.
result.dataRows = rows
result.dataCols = cols
newSeq(result.data, rows*cols)
if len(d)>0:
if len(d)<(rows*cols):
raise newException(EInvalidIndex, "insufficient data supplied in matrix constructor")
for i in countup(0,rows*cols-1):
result.data[i] = d[i]
proc `[][]`*[T](x: TMatrix[T], r,c: int): T =
## Element access. Returns the element at row `r` column `c`.
result = x.data[x.index(r,c)]
proc `[][]=`*[T](x: var TMatrix[T], r,c: int, a: T) =
## Sets the value of the element at row `r` column `c` to
## the value supplied in `a`.
x.data[x.index(r,c)] = a
var m = matrix( 2, 2, [1,2,3,4] )
echo( $m[0][0] )
This is the error I get:
c:\program files (x86)\nimrod\config\nimrod.cfg(36, 11) Hint: added path: 'C:\Users\H127\.babel\libs\' [Path]
Hint: used config file 'C:\Program Files (x86)\Nimrod\config\nimrod.cfg' [Conf]
Hint: system [Processing]
Hint: mat [Processing]
mat.nim(48, 9) Error: type mismatch: got (TMatrix[int], int literal(0))
but expected one of:
system.[](a: array[Idx, T], x: TSlice[Idx]): seq[T]
system.[](a: array[Idx, T], x: TSlice[int]): seq[T]
system.[](s: string, x: TSlice[int]): string
system.[](s: seq[T], x: TSlice[int]): seq[T]
Thanks you guys!
I'd like to first point out that the matrix library you refer to is three years old. For a programming language in development that's a lot of time due to changes, and it doesn't compile any more with the current Nimrod git version:
$ nimrod c matrix
...
private/tmp/n/matrix/matrix.nim(97, 8) Error: ']' expected
It fails on the double array accessor, which seems to have changed syntax. I guess your attempt to create a double [][] accessor is problematic, it could be ambiguous: are you accessing the double array accessor of the object or are you accessing the nested array returned by the first brackets? I had to change the proc to the following:
proc `[]`*[T](x: TMatrix[T], r,c: int): T =
After that change you also need to change the way to access the matrix. Here's what I got:
for x in 0 .. <2:
for y in 0 .. <2:
echo "x: ", x, " y: ", y, " = ", m[x,y]
Basically, instead of specifying two bracket accesses you pass all the parameters inside a single bracket. That code generates:
x: 0 y: 0 = 1
x: 0 y: 1 = 2
x: 1 y: 0 = 3
x: 1 y: 1 = 4
With regards to finding software for Nimrod, I would like to recommend you using Nimble, Nimrod's package manager. Once you have it installed you can search available and maintained packages. The command nimble search math shows two potential packages: linagl and extmath. Not sure if they are what you are looking for, but at least they seem more fresh.