Histc under octave

[lucas21]

Hi everybody,

I recently start (since this morning) to work with Octave. I use to operate
with Matlab.
Anyway, I try to use my matlab code under Octave, there was a few
modifications to make, which I did, but I cant figure out the last one.
Here is the message that I have when I try to execute my code:


parse error near line 49 of file
C:\Octave\OCTAVE~1.1\share\octave\4.2.1\m\statistics\base\histc.m

  syntax error

>>>   if(nargin < 2 || nargin > 3)
       ^

error: called from
    mc3dBASF_GW2 at line 553 column 19
    MC_fluo_Main3d_BASF_GW_zemax_base_de_travail at line 301 column 194
    mc3dBASF_GW2 at line 553 column 19
    MC_fluo_Main3d_BASF_GW_zemax_base_de_travail at line 301 column 194
    mc3dBASF_GW2 at line 9 column 7
stopped in C:\Users\LZ419B9N\Desktop\Code_Lucas\mc3dBASF_GW2.m at line 553
553: MPV(lambdaIdx0un) = histc(lambdaIdx0,lambdaIdx0un);


I tried to look to make some modification to this file "histc" but nothing
worked...

Here is the histc file:

function N = histc (X, EDGES)
N = histc (X, EDGES, DIM)
[N, IDX] = histc (...)

  if(nargin < 2 || nargin > 3)
print_usage ();
  endif

  if (! isreal (x))
    error ("histc: X argument must be real-valued, not complex");
  endif

  num_edges = numel (edges);
  if (num_edges == 0)
    error ("histc: EDGES must not be empty");
  endif

  if (! isreal (edges))
    error ("histc: EDGES must be real-valued, not complex");
  else
    ## Make sure 'edges' is sorted
    edges = edges(:);
    if (! issorted (edges) || edges(1) > edges(end))
      warning ("histc: edge values not sorted on input");
      edges = sort (edges);
    endif
  endif

  nd = ndims (x);
  sz = size (x);
  if (nargin < 3)
    ## Find the first non-singleton dimension.
    (dim = find (sz > 1, 1)) || (dim = 1);
  else
    if (!(isscalar (dim) && dim == fix (dim))
        || !(1 <= dim && dim <= nd))
      error ("histc: DIM must be an integer and a valid dimension");
    endif
  endif

  nsz = sz;
  nsz(dim) = num_edges;

  ## the splitting point is 3 bins

  if (num_edges <= 3)

    ## This is the O(M*N) algorithm.

    ## Allocate the histogram
    n = zeros (nsz);

    ## Allocate 'idx'
    if (nargout > 1)
      idx = zeros (sz);
    endif

    ## Prepare indices
    idx1 = cell (1, dim-1);
    for k = 1:length (idx1)
      idx1{k} = 1:sz(k);
    endfor
    idx2 = cell (length (sz) - dim);
    for k = 1:length (idx2)
      idx2{k} = 1:sz(k+dim);
    endfor

    ## Compute the histograms
    for k = 1:num_edges-1
      b = (edges(k) <= x & x < edges(k+1));
      n(idx1{:}, k, idx2{:}) = sum (b, dim);
      if (nargout > 1)
        idx(b) = k;
      endif
    endfor
    b = (x == edges(end));
    n(idx1{:}, num_edges, idx2{:}) = sum (b, dim);
    if (nargout > 1)
      idx(b) = num_edges;
    endif

  else

    ## This is the O(M*log(N) + N) algorithm.

    ## Look-up indices.
    idx = lookup (edges, x);
    ## Zero invalid ones (including NaNs).  x < edges(1) are already zero.
    idx(! (x <= edges(end))) = 0;

    iidx = idx;

    ## In case of matrix input, we adjust the indices.
    if (! isvector (x))
      nl = prod (sz(1:dim-1));
      nn = sz(dim);
      nu = prod (sz(dim+1:end));
      if (nl != 1)
        iidx = (iidx-1) * nl;
        iidx += reshape (kron (ones (1, nn*nu), 1:nl), sz);
      endif
      if (nu != 1)
        ne =length (edges);
        iidx += reshape (kron (nl*ne*(0:nu-1), ones (1, nl*nn)), sz);
      endif
    endif

    ## Select valid elements.
    iidx = iidx(idx != 0);

    ## Call accumarray to sum the indexed elements.
    n = accumarray (iidx(:), 1, nsz);

  endif

endfunction


%!test
%! x = linspace (0, 10, 1001);
%! n = histc (x, 0:10);
%! assert (n, [repmat(100, 1, 10), 1]);

%!test
%! x = repmat (linspace (0, 10, 1001), [2, 1, 3]);
%! n = histc (x, 0:10, 2);
%! assert (n, repmat ([repmat(100, 1, 10), 1], [2, 1, 3]));

%!error histc ()
%!error histc (1)
%!error histc (1, 2, 3, 4)
%!error histc ([1:10 1+i], 2)
%!error histc (1:10, [])
%!error histc (1, 1, 3)

Thanks in advance,
Lucas



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