Some problems when solving moore neighborhood - python-2.7

here is my code, I don't know how to solve it...
def count_neighbours(grid, row, col):
neighbor_rule = ((-1, -1), (-1, 0), (-1, 1), (0, -1),
(0, 1), (1, -1), (1, 0), (1, 1))
chip = 0
for each_loc in neighbor_rule:
n_row = row + each_loc[0]
n_col = col + each_loc[1]
if 0 <= n_row < len(gird) and 0 <= n_col < len(gird[0]):
if gird[n_row][n_col] == 1:
chip += 1
else:
continue
return chip

Related

How iterate over a 2x2 matrix in Python

Im trying to recreate Conways game of life where the average color of the surrounding cells will be the color of the new dead cell created, although I'm having issues trying to count the surrounding neighbors of a certain cell in order to determine if that cell should be dead or alive.
def count_neighbors(self, i, j):
neighbors = []
r_sum = 0
g_sum = 0
b_sum = 0
''' The range(-1, 2) in the for loop allows the loop to check the cells in the positions
relative to the current cell, which are the eight cells surrounding it.'''
for x in range(-1, 2):
for y in range(-1, 2):
if (x, y) == (0, 0):
continue
if 0 <= int(i) + x < len(self._board) and 0 <= int(j) + y < len(self._board):
if self._board[i + x][j + y] != (0, 0, 0):
neighbors.append(self._board[i + x][j + y])
r_sum += self._board[i + x][j + y][0]
g_sum += self._board[i + x][j + y][1]
b_sum += self._board[i + x][j + y][2]
num_neighbors = len(neighbors)
if num_neighbors == 0:
return 0, (0, 0, 0)
return num_neighbors, (r_sum // num_neighbors, g_sum // num_neighbors, b_sum // num_neighbors)

Evenly distribute values into array

I have a fixed size boolean array of size 8. The default value of all elements in the array is false. There will be a number of truth values to fill between 1-8.
I want to distribute the truth values as far away from one another as possible. I also wish to be able to randomize the configuration. In this scenario the array wraps around so position 7 is "next to" position 0 in the array.
here are some examples for fill values. I didn't include all possibilities, but hopefully it gets my point across.
1: [1, 0, 0, 0, 0, 0, 0, 0] or [0, 1, 0, 0, 0, 0, 0, 0]
2: [1, 0, 0, 0, 1, 0, 0, 0] or [0, 1, 0, 0, 0, 1, 0, 0]
3: [1, 0, 0, 1, 0, 0, 1, 0] or [0, 1, 0, 0, 1, 0, 0, 1]
4: [1, 0, 1, 0, 1, 0, 1, 0] or [0, 1, 0, 1, 0, 1, 0, 1]
5: [1, 1, 0, 1, 1, 0, 1, 0]
6: [1, 1, 0, 1, 1, 1, 0, 1]
7: [1, 1, 1, 1, 1, 1, 1, 0]
8: [1, 1, 1, 1, 1, 1, 1, 1]
The closest solution I have come up with so far hasn't quite produced the results I'm looking for...
I seek to write it in c++ but here is a little pseudo-code of my algorithm so far...
not quite working out how I wanted
truths = randBetween(1, 8)
values = [0,0,0,0,0,0,0,0]
startPosition = randBetween(0, 7) //starting index
distance = 4
for(i = 0; i < truths; i++) {
pos = i + startPosition + (i * distance)
values[pos % 8] = 1
}
this is an example output from my current code. those marked with a star are incorrect.
[0, 0, 0, 0, 1, 0, 0, 0]
[0, 1, 0, 0, 1, 0, 0, 0]*
[0, 1, 0, 0, 1, 0, 1, 0]
[0, 1, 0, 1, 1, 0, 1, 0]*
[1, 1, 0, 1, 1, 0, 1, 0]
[1, 1, 0, 1, 1, 1, 1, 0]*
[1, 1, 1, 1, 1, 1, 1, 0]
[1, 1, 1, 1, 1, 1, 1, 1]
I'm looking for a simple way to distribute the truth values evenly throughout the array without having to code for special cases.

Check this out:
#include <cassert>
#include <vector>
#include <iostream>
#include <iomanip>
/**
* Generate an even spaced pattern of ones
* #param arr destination vector of ints
* #param onescnt the requested number of ones
*/
static inline
void gen(std::vector<int>& arr, size_t onescnt) {
const size_t len = arr.size();
const size_t zeroscnt = len - onescnt;
size_t ones = 1;
size_t zeros = 1;
for (size_t i = 0; i < len; ++i) {
if (ones * zeroscnt < zeros * onescnt) {
ones++;
arr[i] = 1;
} else {
zeros++;
arr[i] = 0;
}
}
}
static inline
size_t count(const std::vector<int>& arr, int el) {
size_t cnt = 0;
for (size_t i = 0; i < arr.size(); ++i) {
cnt += arr[i] == el;
}
return cnt;
}
static inline
void gen_print(size_t len, size_t onescnt) {
std::vector<int> arr(len);
gen(arr, onescnt);
std::cout << "gen_printf(" << std::setw(2) << len << ", " << std::setw(2) << onescnt << ") = {";
for (size_t i = 0; i < len; ++i) {
std::cout << arr[i] << ",";
}
std::cout << "}\n";
assert(count(arr, 1) == onescnt);
}
int main() {
for (int i = 0; i <= 8; ++i) {
gen_print(8, i);
}
for (int i = 0; i <= 30; ++i) {
gen_print(30, i);
}
return 0;
}
Generates:
gen_printf( 8, 0) = {0,0,0,0,0,0,0,0,}
gen_printf( 8, 1) = {0,0,0,0,0,0,0,1,}
gen_printf( 8, 2) = {0,0,0,1,0,0,0,1,}
gen_printf( 8, 3) = {0,1,0,0,1,0,0,1,}
gen_printf( 8, 4) = {0,1,0,1,0,1,0,1,}
gen_printf( 8, 5) = {1,0,1,1,0,1,0,1,}
gen_printf( 8, 6) = {1,1,0,1,1,1,0,1,}
gen_printf( 8, 7) = {1,1,1,1,1,1,0,1,}
gen_printf( 8, 8) = {1,1,1,1,1,1,1,1,}
gen_printf(30, 0) = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,}
gen_printf(30, 1) = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,}
gen_printf(30, 2) = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,}
gen_printf(30, 3) = {0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,}
gen_printf(30, 4) = {0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,}
gen_printf(30, 5) = {0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,}
gen_printf(30, 6) = {0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,}
gen_printf(30, 7) = {0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,1,}
gen_printf(30, 8) = {0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,}
gen_printf(30, 9) = {0,0,1,0,0,1,0,0,0,1,0,0,1,0,0,1,0,0,0,1,0,0,1,0,0,1,0,0,0,1,}
gen_printf(30, 10) = {0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,}
gen_printf(30, 11) = {0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,0,1,}
gen_printf(30, 12) = {0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,}
gen_printf(30, 13) = {0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,}
gen_printf(30, 14) = {0,1,0,1,0,1,0,1,0,1,0,1,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,0,1,}
gen_printf(30, 15) = {0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,}
gen_printf(30, 16) = {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,}
gen_printf(30, 17) = {1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,0,1,}
gen_printf(30, 18) = {1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,}
gen_printf(30, 19) = {1,0,1,1,0,1,1,0,1,0,1,1,0,1,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,}
gen_printf(30, 20) = {1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,}
gen_printf(30, 21) = {1,1,0,1,1,0,1,1,0,1,1,1,0,1,1,0,1,1,0,1,1,1,0,1,1,0,1,1,0,1,}
gen_printf(30, 22) = {1,1,0,1,1,1,0,1,1,1,0,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,0,1,}
gen_printf(30, 23) = {1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,}
gen_printf(30, 24) = {1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,}
gen_printf(30, 25) = {1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,}
gen_printf(30, 26) = {1,1,1,1,1,1,0,1,1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,1,1,1,0,1,}
gen_printf(30, 27) = {1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,}
gen_printf(30, 28) = {1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,}
gen_printf(30, 29) = {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,}
gen_printf(30, 30) = {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,}
#edit - better evenly spaced pattern.
Explanation:
So let's take an array of 8 ints and we want to have 5 ones. The ideal ratio of (ones / zeros) in a sequence with 8 elements and 5 ones, well would be (5 / 3). We will never approach such ratio, but we can try.
The idea is to loop through the array and remember the number of ones and zeros we have written in the array. If the ratio of (written ones / written zeros) is lower then the destination ratio (ones / zeros) we want to achieve, we need to put a one to the sequence. Otherwise we put zero in the sequence. The ratio changes and we make the decision next time. The idea is to pursue the ideal ratio of ones per zeros in each slice of the array.

A simple way to do this would be to round the ideal fractional positions.
truths = randBetween(1, 8)
values = [0,0,0,0,0,0,0,0]
offset = randBetween(0, 8 * truths - 1)
for(i = 0; i < truths; i++) {
pos = (offset + (i * 8)) / truths
values[pos % 8] = 1
}

This is an application of Bresenham's line-drawing algorithm. I use it not because it's fast on old hardware, but it places true values exactly.
#include <iostream>
#include <stdexcept>
#include <string>
#include <random>
int main(int argc, char **argv) {
try {
// Read the argument.
if(argc != 2) throw std::invalid_argument("one argument");
int dy = std::stoi(argv[1]);
if(dy < 0 || dy > 8) throw std::out_of_range("[0..8]");
int values[8] = {0};
// https://en.wikipedia.org/wiki/Bresenham%27s_line_algorithm
int dx = 8;
int delta = 2 * dy - dx; // Balance the line. Permute it up later.
for(int x = 0; x < dx; x++) {
if(delta > 0) {
values[x] = 1;
delta -= 2 * dx;
}
delta += 2 * dy;
}
for(int x = 0; x < dx; x++)
std::cout << (x ? ", " : "") << values[x];
std::cout << std::endl;
// Rotate the number by a random amount.
// I'm sure there is an easier way to do this.
// https://stackoverflow.com/questions/7560114/random-number-c-in-some-range
std::random_device rd; // obtain a random number from hardware
std::mt19937 eng(rd()); // seed the generator
std::uniform_int_distribution<> distr(0, dx - 1);
int rotate = distr(eng);
bool first = true;
int x = rotate;
do {
std::cout << (first ? "" : ", ") << values[x];
first = false;
x = (x + 1) % dx;
} while(x != rotate);
std::cout << std::endl;
} catch(const std::exception &e) {
std::cerr << "Something went wrong: " << e.what() << std::endl;
return 1;
}
return 0;
}
Once you have an exact solution, rotate it by a random amount.
0, 1, 0, 0, 1, 0, 1, 0
1, 0, 0, 1, 0, 0, 1, 0

You need to calculate distance dynamically. One element is clear, that can reside at arbitrary location
2 elements is clear, too, distance needs to be 4.
4 elements need a distance of 2
8 elements a distance of 1
More difficult are numbers that don't divide the array:
3 requires a distance of 2.66.
5 requires a distance of 1.6
7 requires a distance of 0.875
Errm... In general, if you have a distance of X.Y, you will have to place some of the elements at distances of X and some at distances of X + 1. X is simple, it will be the result of an integer division: 8 / numberOfElements. The remainder will determine how often you will have to switch to X + 1: 8 % numberOfElements. For 3, this will result in 2, too, so you will have 1x distance of 2 and 2x distance of 3:
[ 1 0 1 0 0 1 0 0 ]
2 3 3 (distance to very first 1)
For 5, you'll get: 8/5 = 1, 8%5 = 3, so: 2x distance of 1, 3x distance of 2
[ 1 1 1 0 1 0 1 0 ]
1 1 2 2 2
For 7 you'll get: 8/7 = 1, 8%7 = 1, so: 7x distance of 1, 1x distance of 2
[ 1 1 1 1 1 1 1 0 ]
1 1 1 1 1 1 2
That will work for arbitrary array length L:
L/n = minimum distance
L%n = number of times to apply minimum distance
L-L%n = number of times to apply minimum distance + 1
Mathematical metrics won't reveal any difference between first applying all smaller distances then all larger ones, human sense for aesthetics, though, might prefer if you alternate between larger and smaller as often as possible – or you apply the algorithm recursively (for larger array length), to get something like 2x2, 3x3, 2x2, 3x3 instead of 4x2 and 6x3.

C++ Points of Vertices in Cuboid (Bitwise AND)

I'm trying to calculate the points in a cuboid given its centre (which is a Vector3) and the lengths of the sides along the x, y and z axis. I found the following on math.stackexchange.com: https://math.stackexchange.com/questions/107778/simplest-equation-for-drawing-a-cube-based-on-its-center-and-or-other-vertices which says I can use the following formulae:
The constructor for the World class is:
World::World(Vector3 o, float d1, float d2, float d3) : origin(o)
{
// If we consider an edge length to be d, we need to find r such that
// 2r = d in order to calculate the positions of each vertex in the world.
float r1 = d1 / 2,
r2 = d2 / 2,
r3 = d3 / 2;
for (int i = 0; i < 8; i++)
{
/* Sets up the vertices of the cube.
*
* #see http://bit.ly/1cc2RPG
*/
float x = o.getX() + (std::pow(-1, i&1) * r1),
y = o.getY() + (std::pow(-1, i&2) * r2),
z = o.getZ() + (std::pow(-1, i&4) * r3);
points[i] = Vector3(x, y, z);
std::cout << points[i] << "\n";
}
}
And I passing the following parameters to the constructor:
Vector3 o(0, 0, 0);
World w(o, 100.f, 100.f, 100.f);
The coordinates being output for all 8 vertices are:
(50, 50, 50)
(-50, 50, 50)
(50, 50, 50)
(-50, 50, 50)
(50, 50, 50)
(-50, 50, 50)
(50, 50, 50)
(-50, 50, 50)
Which cannot be correct. Any guidance would be very much appreciated!

The problem lies in the bitwise & inside your pow calls:
In the y and z components, they always return 0 and 2 or 4, respectively. -1^2 = -1^4 = 1, which is why the sign of these components is always positive. You could try (i&2)!=0 or (i&2) >> 1 for the y component instead. The same goes for the z component.

Change this:
float x = o.getX() + (std::pow(-1, i&1) * r1),
y = o.getY() + (std::pow(-1, i&2) * r2),
z = o.getZ() + (std::pow(-1, i&4) * r3);
To this:
float x = o.getX() + (std::pow(-1, (i ) & 1) * r1), // pow(-1, 0) == 1, pow(-1, 1) == -1
y = o.getY() + (std::pow(-1, (i >> 1) & 1) * r2), // pow(-1, 0) == 1, pow(-1, 1) == -1
z = o.getZ() + (std::pow(-1, (i >> 2) & 1) * r3); // pow(-1, 0) == 1, pow(-1, 1) == -1
Or even to this:
float x = o.getX() + (std::pow(-1, (i )) * r1), // pow(-1, {0, 2, 4, 6}) == 1, pow(-1, {1, 3, 5, 7}) == -1
y = o.getY() + (std::pow(-1, (i >> 1)) * r2), // pow(-1, {0, 2}) == 1, pow(-1, {1, 3}) == -1
z = o.getZ() + (std::pow(-1, (i >> 2)) * r3); // pow(-1, 0) == 1, pow(-1, 1) == -1
The problem is that as written even though the values you mask out identify weather or not the lengths need to be negated. They are not in the correct place value to get the desired properties from the exponentiation of -1.
Rewriting the code as I have above will solve this issue, however it would be more readable and in general more permanent just to unroll the loop and manually write if each one is an addition or subtraction without using the pow function.

Finding Neighbourhood in Matrices [duplicate]

This question already has an answer here:
Conway's Game of Life, counting neighbors [closed]
(1 answer)
Closed 9 years ago.
I am working on project containing cellular automat methods. What I am trying to figure is how to write function helping to find all the neighbours in a 2d array.
for example i ve got size x size 2d array [size = 4 here]
[x][n][ ][n]
[n][n][ ][n]
[ ][ ][ ][ ]
[n][n][ ][n]
Field marked as x [0,0 index] has neighbours marked as [n] -> 8 neighbours. What Im trying to do is to write a function which can find neighbours wo writting tousands of if statements
Does anybody have an idea how to do it ?
thanks

For the neighbours of element (i,j) in NxM matrix:
int above = (i-1) % N;
int below = (i+1) % N;
int left = (j-1) % M;
int right = (j+1) % M;
decltype(matrix[0][0]) *indices[8];
indices[0] = & matrix[above][left];
indices[1] = & matrix[above][j];
indices[2] = & matrix[above][right];
indices[3] = & matrix[i][left];
// Skip matrix[i][j]
indices[4] = & matrix[i][right];
indices[5] = & matrix[below][left];
indices[6] = & matrix[below][j];
indices[7] = & matrix[below][right];

Suppose you are in cell (i, j). Then, on an infinite grid, your neighbors should be [(i-1, j-1), (i-1,j), (i-1, j+1), (i, j-1), (i, j+1), (i+1, j-1), (i+1, j), (i+1, j+1)].
However, since the grid is finite some of the above values will get outside the bounds. But we know modular arithmetic: 4 % 3 = 1 and -1 % 3 = 2. So, if the grid is of size n, m you only need to apply %n, %m on the above list to get the proper list of neighbors: [((i-1) % n, (j-1) % m), ((i-1) % n,j), ((i-1) % n, (j+1) % m), (i, (j-1) % m), (i, (j+1) % m), ((i+1) % n, (j-1) % m), ((i+1) % n, j), ((i+1) % n, (j+1) % m)]
That works if your coordinates are between 0 and n and between 0 and m. If you start with 1 then you need to tweak the above by doing a -1 and a +1 somewhere.
For your case n=m=4 and (i, j) = (0, 0). The first list is [(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0), (1, 1)]. Applying the modulus operations you get to [(3, 3), (3, 0), (3, 1), (0, 3), (0, 1), (1, 3), (1, 0), (1, 1)] which are exactly the squares marked [n] in your picture.

Add and subtract one from the coordinates, in all possible permutations. Results outside the boundaries wrap around (e.g. -1 becomes 3 and 4 becomes 0). Just a couple of simple loops needed basically.
Something like
// Find the closest neighbours (one step) from the coordinates [x,y]
// The max coordinates is max_x,max_y
// Note: Does not contain any error checking (for valid coordinates)
std::vector<std::pair<int, int>> getNeighbours(int x, int y, int max_x, int max_y)
{
std::vector<std::pair<int, int>> neighbours;
for (int dx = -1; dx <= 1; ++dx)
{
for (int dy = -1; dy <= 1; ++dy)
{
// Skip the coordinates [x,y]
if (dx == 0 && dy == 0)
continue;
int nx = x + dx;
int ny = y + dy;
// If the new coordinates goes out of bounds, wrap them around
if (nx < 0)
nx = max_x;
else if (nx > max_x)
nx = 0;
if (ny < 0)
ny = max_y;
else if (ny > max_y)
ny = 0;
// Add neighbouring coordinates to result
neighbours.push_back(std::make_pair(nx, ny));
}
}
return neighbours;
}
Example use for you:
auto n = getNeighbours(0, 0, 3, 3);
for (const auto& p : n)
std::cout << '[' << p.first << ',' << p.second << "]\n";
Prints out
[3,3]
[3,0]
[3,1]
[0,3]
[0,1]
[1,3]
[1,0]
[1,1]
which is the correct answer.

libsvm (C++) Always Outputting Same Prediction

I have implemented an OpenCV/C++ wrapper for libsvm. When doing a grid-search for SVM parameters (RBF kernel), the prediction always returns the same label. I have created artificial data sets which have very easily separable data (and tried predicting data I just trained on) but still, it returns the same label.
I have used the MATLAB implementation of libsvm and achieved high accuracy on the same data set. I must be doing something wrong with setting up the problem but I've gone through the README many times and I can't quite find the issue.
Here is how I set up the libsvm problem, where data is an OpenCV Mat:
const int rowSize = data.rows;
const int colSize = data.cols;
this->_svmProblem = new svm_problem;
std::memset(this->_svmProblem,0,sizeof(svm_problem));
//dynamically allocate the X matrix...
this->_svmProblem->x = new svm_node*[rowSize];
for(int row = 0; row < rowSize; ++row)
this->_svmProblem->x[row] = new svm_node[colSize + 1];
//...and the y vector
this->_svmProblem->y = new double[rowSize];
this->_svmProblem->l = rowSize;
for(int row = 0; row < rowSize; ++row)
{
for(int col = 0; col < colSize; ++col)
{
//set the index and the value. indexing starts at 1.
this->_svmProblem->x[row][col].index = col + 1;
double tempVal = (double)data.at<float>(row,col);
this->_svmProblem->x[row][col].value = tempVal;
}
this->_svmProblem->x[row][colSize].index = -1;
this->_svmProblem->x[row][colSize].value = 0;
//add the label to the y array, and feature vector to X matrix
double tempVal = (double)labels.at<float>(row);
this->_svmProblem->y[row] = tempVal;
}
}/*createProblem()*/
Here is how I set up the parameters, where svmParams is my own struct for C/Gamma and such:
this->_svmParameter = new svm_parameter;
std::memset(this->_svmParameter,0,sizeof(svm_parameter));
this->_svmParameter->svm_type = svmParams.svmType;
this->_svmParameter->kernel_type = svmParams.kernalType;
this->_svmParameter->C = svmParams.C;
this->_svmParameter->gamma = svmParams.gamma;
this->_svmParameter->nr_weight = 0;
this->_svmParameter->eps = 0.001;
this->_svmParameter->degree = 1;
this->_svmParameter->shrinking = 0;
this->_svmParameter->probability = 1;
this->_svmParameter->cache_size = 100;
I use the provided param/problem checking function and no errors are returned.
I then train as such:
this->_svmModel = svm_train(this->_svmProblem, this->_svmParameter);
And then predict like so:
float pred = (float)svm_predict(this->_svmModel, x[i]);
If anyone could point out where I'm going wrong here I would greatly appreciate it. Thank you!
EDIT:
Using this code I printed the contents of the problem
for(int i = 0; i < rowSize; ++i)
{
std::cout << "[";
for(int j = 0; j < colSize + 1; ++j)
{
std::cout << " (" << this->_svmProblem->x[i][j].index << ", " << this->_svmProblem->x[i][j].value << ")";
}
std::cout << "]" << " <" << this->_svmProblem->y[i] << ">" << std::endl;
}
Here is the output:
[ (1, -1) (2, 0) (-1, 0)] <1>
[ (1, -0.92394) (2, 0) (-1, 0)] <1>
[ (1, -0.7532) (2, 0) (-1, 0)] <1>
[ (1, -0.75977) (2, 0) (-1, 0)] <1>
[ (1, -0.75337) (2, 0) (-1, 0)] <1>
[ (1, -0.76299) (2, 0) (-1, 0)] <1>
[ (1, -0.76527) (2, 0) (-1, 0)] <1>
[ (1, -0.74631) (2, 0) (-1, 0)] <1>
[ (1, -0.85153) (2, 0) (-1, 0)] <1>
[ (1, -0.72436) (2, 0) (-1, 0)] <1>
[ (1, -0.76485) (2, 0) (-1, 0)] <1>
[ (1, -0.72936) (2, 0) (-1, 0)] <1>
[ (1, -0.94004) (2, 0) (-1, 0)] <1>
[ (1, -0.92756) (2, 0) (-1, 0)] <1>
[ (1, -0.9688) (2, 0) (-1, 0)] <1>
[ (1, 0.05193) (2, 0) (-1, 0)] <3>
[ (1, -0.048488) (2, 0) (-1, 0)] <3>
[ (1, 0.070436) (2, 0) (-1, 0)] <3>
[ (1, 0.15191) (2, 0) (-1, 0)] <3>
[ (1, -0.07331) (2, 0) (-1, 0)] <3>
[ (1, 0.019786) (2, 0) (-1, 0)] <3>
[ (1, -0.072793) (2, 0) (-1, 0)] <3>
[ (1, 0.16157) (2, 0) (-1, 0)] <3>
[ (1, -0.057188) (2, 0) (-1, 0)] <3>
[ (1, -0.11187) (2, 0) (-1, 0)] <3>
[ (1, 0.15886) (2, 0) (-1, 0)] <3>
[ (1, -0.0701) (2, 0) (-1, 0)] <3>
[ (1, -0.17816) (2, 0) (-1, 0)] <3>
[ (1, 0.12305) (2, 0) (-1, 0)] <3>
[ (1, 0.058615) (2, 0) (-1, 0)] <3>
[ (1, 0.80203) (2, 0) (-1, 0)] <1>
[ (1, 0.734) (2, 0) (-1, 0)] <1>
[ (1, 0.9072) (2, 0) (-1, 0)] <1>
[ (1, 0.88061) (2, 0) (-1, 0)] <1>
[ (1, 0.83903) (2, 0) (-1, 0)] <1>
[ (1, 0.86604) (2, 0) (-1, 0)] <1>
[ (1, 1) (2, 0) (-1, 0)] <1>
[ (1, 0.77988) (2, 0) (-1, 0)] <1>
[ (1, 0.8578) (2, 0) (-1, 0)] <1>
[ (1, 0.79559) (2, 0) (-1, 0)] <1>
[ (1, 0.99545) (2, 0) (-1, 0)] <1>
[ (1, 0.78376) (2, 0) (-1, 0)] <1>
[ (1, 0.72177) (2, 0) (-1, 0)] <1>
[ (1, 0.72619) (2, 0) (-1, 0)] <1>
[ (1, 0.80149) (2, 0) (-1, 0)] <1>
[ (1, 0.092327) (2, -1) (-1, 0)] <2>
[ (1, 0.019054) (2, -1) (-1, 0)] <2>
[ (1, 0.15287) (2, -1) (-1, 0)] <2>
[ (1, -0.1471) (2, -1) (-1, 0)] <2>
[ (1, -0.068182) (2, -1) (-1, 0)] <2>
[ (1, -0.094567) (2, -1) (-1, 0)] <2>
[ (1, -0.17071) (2, -1) (-1, 0)] <2>
[ (1, -0.16646) (2, -1) (-1, 0)] <2>
[ (1, -0.030421) (2, -1) (-1, 0)] <2>
[ (1, 0.094346) (2, -1) (-1, 0)] <2>
[ (1, -0.14408) (2, -1) (-1, 0)] <2>
[ (1, 0.090025) (2, -1) (-1, 0)] <2>
[ (1, 0.043706) (2, -1) (-1, 0)] <2>
[ (1, 0.15065) (2, -1) (-1, 0)] <2>
[ (1, -0.11751) (2, -1) (-1, 0)] <2>
[ (1, -0.02324) (2, 1) (-1, 0)] <2>
[ (1, 0.0080356) (2, 1) (-1, 0)] <2>
[ (1, -0.17752) (2, 1) (-1, 0)] <2>
[ (1, 0.011135) (2, 1) (-1, 0)] <2>
[ (1, -0.029063) (2, 1) (-1, 0)] <2>
[ (1, 0.15398) (2, 1) (-1, 0)] <2>
[ (1, 0.097746) (2, 1) (-1, 0)] <2>
[ (1, 0.01018) (2, 1) (-1, 0)] <2>
[ (1, 0.015592) (2, 1) (-1, 0)] <2>
[ (1, -0.062793) (2, 1) (-1, 0)] <2>
[ (1, 0.014444) (2, 1) (-1, 0)] <2>
[ (1, -0.1205) (2, 1) (-1, 0)] <2>
[ (1, -0.18011) (2, 1) (-1, 0)] <2>
[ (1, 0.010521) (2, 1) (-1, 0)] <2>
[ (1, 0.036914) (2, 1) (-1, 0)] <2>
Here, the data is printed in the format [ (index, value)...] label.
The artificial dataset I created just has 3 classes, all which are easily separable with a non-linear decision boundary. Each row is a feature vector (observation), with 2 features (x coord, y coord). Libsvm asks to terminate each vector with a -1 label, so I do.
EDIT2:
This edit pertains to my C and Gamma values used for training, as well as data scaling. I normally data between 0 and 1 (as suggested here: http://www.csie.ntu.edu.tw/~cjlin/papers/guide/guide.pdf). I will scale this fake dataset as well and try again, although I used this same exact dataset with the MATLAB implementation of libsvm and it could separate this unscaled data with 100% accuracy.
For C and Gamma, I also use the values recommended in the guide. I create two vectors and use a double nested loop to try all combinations:
std::vector<double> CList, GList;
double baseNum = 2.0;
for(double j = -5; j <= 15; j += 2) //-5 and 15
CList.push_back(pow(baseNum,j));
for(double j = -15; j <= 3; j += 2) //-15 and 3
GList.push_back(pow(baseNum,j));
And the loop looks like:
for(auto CIt = CList.begin(); CIt != CList.end(); ++CIt) //for all C's
{
double C = *CIt;
for(auto GIt = GList.begin(); GIt != GList.end(); ++GIt) //for all gamma's
{
double gamma = *GIt;
svmParams.svmType = C_SVC;
svmParams.kernalType = RBF;
svmParams.C = C;
svmParams.gamma = gamma;
......training code etc..........
EDIT3:
Since I keep referencing MATLAB, I will show the accuracy differences. Here is a heat map of the accuracy libsvm yields:
And here is the accuracy map MATLAB yields using the same parameters and same C/Gamma grid:
Here is the code used to generate the C/Gamma lists, and how I train:
CList = 2.^(-15:2:15);%(-5:2:15);
GList = 2.^(-15:2:15);%(-15:2:3);
cmd = ['-q -s 0 -t 2 -c ', num2str(C), ' -g ', num2str(gamma)];
model = ovrtrain(yTrain,xTrain,cmd);
EDIT4
As a sanity check, I reformatted my fake scaled dataset to conform to the dataset used by libsvm's Unix/Linux terminal API. I trained and predicted using a C/Gamma found in in the MATLAB accuracy map. The prediction accuracy was 100%. Thus I am absolutely doing something wrong in the C++ implementation.
EDIT5
I loaded the model trained from the Linux terminal into my C++ wrapper class. I then tried predicting the same exact dataset used for training. The accuracy in C++ was still awful! However, I'm very close to narrowing the source of the problem. If MATLAB/Linux both agree in terms of 100% accuracy, and the model it produces has already been proven to yield 100% accuracy on the same dataset that was trained on, and now my C++ wrapper class shows poor performance with the verified model... there are three possible situations:
The method I use to transform cv::Mats into the svm_node* it requires for prediction has a problem in it.
The method I use to predict labels has a problem in it.
BOTH 2 and 3!
The code to really inspect now is how I create the svm_node. Here it is again:
svm_node** LibSVM::createNode(INPUT const cv::Mat& data)
{
const int rowSize = data.rows;
const int colSize = data.cols;
//dynamically allocate the X matrix...
svm_node** x = new svm_node*[rowSize];
if(x == NULL)
throw MLInterfaceException("Could not allocate SVM Node Array.");
for(int row = 0; row < rowSize; ++row)
{
x[row] = new svm_node[colSize + 1]; //+1 here for the index-terminating -1
if(x[row] == NULL)
throw MLInterfaceException("Could not allocate SVM Node.");
}
for(int row = 0; row < rowSize; ++row)
{
for(int col = 0; col < colSize; ++col)
{
double tempVal = data.at<double>(row,col);
x[row][col].value = tempVal;
}
x[row][colSize].index = -1;
x[row][colSize].value = 0;
}
return x;
} /*createNode()*/
And prediction:
cv::Mat LibSVM::predict(INPUT const cv::Mat& data)
{
if(this->_svmModel == NULL)
throw MLInterfaceException("Cannot predict; no model has been trained or loaded.");
cv::Mat predMat;
//create the libsvm representation of data
svm_node** x = this->createNode(data);
//perform prediction for each feature vector
for(int i = 0; i < data.rows; ++i)
{
double pred = svm_predict(this->_svmModel, x[i]);
predMat.push_back<double>(pred);
}
//delete all rows and columns of x
for(int i = 0; i < data.rows; ++i)
delete[] x[i];
delete[] x;
return predMat;
}
EDIT6:
For those of you tuning in at home, I trained a model (using optimal C/Gamma found in MATLAB) in C++, saved it to file, and then tried predicting on the training data via Linux terminal. It scored 100%. Something is wrong with my prediction. o_0
EDIT7:
I found the issue finally. I had tremendous bug-tracking help in finding it. I printed the contents of the svm_node** 2D array used for prediction. It was a subset of the createProblem() method. There was a piece of it that I failed to copy + paste over to the new function. It was the index of a given feature; it was never written. There should have been 1 more line:
x[row][col].index = col + 1; //indexing starts at 1
And the prediction works fine now.

It would be useful to see your gamma value, since your data is not normalized that would make a huge difference.
The gamma in libsvm is inversely to the hypersphere radius, so if those spheres are too small with respect to the input range, everything will be activated always and then the model would output always the same value.
So, the two recommendations would be 1) Scale your input values to the range [-1,1]. 2) Play with the gamma values.