3

我在一个正方形中有很多点。我想将正方形划分为许多小矩形并检查每个矩形中有多少点，即我想计算点的联合概率分布。我正在报告几种常识方法，使用循环并且效率不高：

% Data
N = 1e5;    % number of points
xy = rand(N, 2);    % coordinates of points
xy(randi(2*N, 100, 1)) = 0;    % add some points on one side
xy(randi(2*N, 100, 1)) = 1;    % add some points on the other side
xy(randi(N, 100, 1), :) = 0;    % add some points on one corner
xy(randi(N, 100, 1), :) = 1;    % add some points on one corner
inds= unique(randi(N, 100, 1)); xy(inds, :) = repmat([0 1], numel(inds), 1);    % add some points on one corner
inds= unique(randi(N, 100, 1)); xy(inds, :) = repmat([1 0], numel(inds), 1);    % add some points on one corner

% Intervals for rectangles
K1 = ceil(sqrt(N/5));    % number of intervals along x
K2 = K1;    % number of intervals along y
int_x = [0:(1 / K1):1, 1+eps];    % intervals along x
int_y = [0:(1 / K2):1, 1+eps];    % intervals along y

% First approach
tic
count_cells = zeros(K1 + 1, K2 + 1);
for k1 = 1:K1+1
  inds1 = (xy(:, 1) >= int_x(k1)) & (xy(:, 1) < int_x(k1 + 1));
  for k2 = 1:K2+1
    inds2 = (xy(:, 2) >= int_y(k2)) & (xy(:, 2) < int_y(k2 + 1));
    count_cells(k1, k2) = sum(inds1 .* inds2);
  end
end
toc
% Elapsed time is 46.090677 seconds.

% Second approach
tic
count_again = zeros(K1 + 2, K2 + 2);
for k1 = 1:K1+1
  inds1 = (xy(:, 1) >= int_x(k1));
  for k2 = 1:K2+1
    inds2 = (xy(:, 2) >= int_y(k2));
    count_again(k1, k2) = sum(inds1 .* inds2);
  end
end
count_again_fix = diff(diff(count_again')');
toc
% Elapsed time is 22.903767 seconds.

% Check: the two solutions are equivalent
all(count_cells(:) == count_again_fix(:))

如何在时间、内存和可能避免循环方面更有效地做到这一点？

编辑 -->我也刚刚找到了这个，这是迄今为止找到的最好的解决方案：

tic
count_cells_hist = hist3(xy, 'Edges', {int_x int_y});
count_cells_hist(end, :) = []; count_cells_hist(:, end) = [];
toc
all(count_cells(:) == count_cells_hist(:))
% Elapsed time is 0.245298 seconds.

但它需要统计工具箱。

编辑 --> chappjc 建议的测试解决方案

tic
xcomps = single(bsxfun(@ge,xy(:,1),int_x));
ycomps = single(bsxfun(@ge,xy(:,2),int_y));
count_again = xcomps.' * ycomps; %' 143x143 = 143x1e5 * 1e5x143
count_again_fix = diff(diff(count_again')');
toc
% Elapsed time is 0.737546 seconds.
all(count_cells(:) == count_again_fix(:))
4

3 回答 3

2

我写了一个简单的 mex 函数，当 N 很大时它工作得很好。这当然是作弊，但仍然...

功能是

#include "mex.h"

void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[])
{
    unsigned long int hh, ctrl;       /*  counters                       */
    unsigned long int N, m, n;        /*  size of matrices               */
    unsigned long int *xy;            /*  data                           */
    unsigned long int *count_cells;   /*  joint frequencies              */
    /*  matrices needed */
    mxArray *count_cellsArray;

/*  Now we need to get the data */
    if (nrhs == 3) {
        xy = (unsigned long int*) mxGetData(prhs[0]);
        N = (unsigned long int) mxGetM(prhs[0]);
        m = (unsigned long int) mxGetScalar(prhs[1]);
        n = (unsigned long int) mxGetScalar(prhs[2]);
    }

/*  Then build the matrices for the output */
    count_cellsArray = mxCreateNumericMatrix(m + 1, n + 1, mxUINT32_CLASS, mxREAL);
    count_cells = mxGetData(count_cellsArray);
    plhs[0] = count_cellsArray;

    hh = 0; /* counter for elements of xy */
    /* for all points from 1 to N */
    for(hh=0; hh<N; hh++) {
        ctrl = (m + 1) * xy[N + hh] + xy[hh];
        count_cells[ctrl] = count_cells[ctrl] + 1;
    }
}

它可以保存在文件“joint_dist_points_2D.c”中，然后编译：

mex joint_dist_points_2D.c

并检查一下：

% Data
N = 1e7;    % number of points
xy = rand(N, 2);    % coordinates of points
xy(randi(2*N, 1000, 1)) = 0;    % add some points on one side
xy(randi(2*N, 1000, 1)) = 1;    % add some points on the other side
xy(randi(N, 1000, 1), :) = 0;    % add some points on one corner
xy(randi(N, 1000, 1), :) = 1;    % add some points on one corner
inds= unique(randi(N, 1000, 1)); xy(inds, :) = repmat([0 1], numel(inds), 1);    % add some points on one corner
inds= unique(randi(N, 1000, 1)); xy(inds, :) = repmat([1 0], numel(inds), 1);    % add some points on one corner

% Intervals for rectangles
K1 = ceil(sqrt(N/5));    % number of intervals along x
K2 = ceil(sqrt(N/7));    % number of intervals along y
int_x = [0:(1 / K1):1, 1+eps];    % intervals along x
int_y = [0:(1 / K2):1, 1+eps];    % intervals along y

% Use Statistics Toolbox: hist3
tic
count_cells_hist = hist3(xy, 'Edges', {int_x int_y});
count_cells_hist(end, :) = []; count_cells_hist(:, end) = [];
toc
% Elapsed time is 4.414768 seconds.

% Use mex function
tic
xy2 = uint32(floor(xy ./ repmat([1 / K1, 1 / K2], N, 1)));
count_cells = joint_dist_points_2D(xy2, uint32(K1), uint32(K2));
toc
% Elapsed time is 0.586855 seconds.

% Check: the two solutions are equivalent
all(count_cells_hist(:) == count_cells(:))
于 2013-11-03T12:19:08.373 回答
0

改进有问题的代码

您的循环（和嵌套的点积）可以通过bsxfun矩阵乘法消除，如下所示：

xcomps = bsxfun(@ge,xy(:,1),int_x);
ycomps = bsxfun(@ge,xy(:,2),int_y);
count_again = double(xcomps).'*double(ycomps); %' 143x143 = 143x1e5 * 1e5x143
count_again_fix = diff(diff(count_again')');

乘法步骤完成了 中的 AND 和求和sum(inds1 .* inds2)，但没有遍历密度矩阵。 编辑：如果您使用single而不是double，执行时间几乎减半，但请务必将您的答案转换double为其余代码所需的内容。在我的电脑上，这大约需要0.5 秒。

注意：rot90(count_again/size(xy,1),2)你有一个 CDF，rot90(count_again_fix/size(xy,1),2)你有一个 PDF。

使用累加数组

另一种方法是accumarray在我们对数据进行分箱后制作联合直方图。

以int_x, int_y, K1,xy等开头：

% take (0,1) data onto [1 K1], following A.Dondas approach for easy comparison
ii = floor(xy(:,1)*(K1-eps))+1; ii(ii<1) = 1; ii(ii>K1) = K1;
jj = floor(xy(:,2)*(K1-eps))+1; jj(jj<1) = 1; jj(jj>K1) = K1;

% create the histogram and normalize
H = accumarray([ii jj],ones(1,size(ii,1)));
PDF = H / size(xy,1); % for probabilities summing to 1

在我的电脑上，这大约需要0.01 秒。

输出与 A.Donda 的从稀疏转换为完整 ( full(H)) 相同。虽然，正如他 A.Donda 指出的那样，将尺寸设置为K1x是正确的K1，而不是count_again_fixOPs 代码中的大小为K1+1x K1+1。

要获得 CDF，我相信您可以简单地应用于cumsumPDF 的每个轴。

于 2013-11-02T19:49:51.290 回答
0

chappjc 的回答和使用hist3都很好，但是由于我前段时间碰巧想要这样的东西，但由于某种原因没有找到hist3我自己写的，我想我会把它贴在这里作为奖励。它用于sparse进行实际计数并将结果作为稀疏矩阵返回，因此它对于处理不同模式相距很远的多峰分布可能很有用 - 或者对于没有统计工具箱的人来说可能很有用。

对francesco数据的应用：

K1 = ceil(sqrt(N/5));
[H, xs, ys] = hist2d(xy(:, 1), xy(:, 2), [K1 K1], [0, 1 + eps, 0, 1 + eps]);

使用输出参数调用该函数只返回结果，而不制作彩色图。

这是功能：

函数 [H, xs, ys] = hist2d(x, y, n, ax)

% plot 2d-histogram as an image
%
% hist2d(x, y, n, ax)
% [H, xs, ys] = hist2d(x, y, n, ax)
%
% x:    data for horizontal axis
% y:    data for vertical axis
% n:    how many bins to use for each axis, default is [100 100]
% ax:   axis limits for the plot, default is [min(x), max(x), min(y), max(y)]
% H:    2d-histogram as a sparse matrix, indices 1 & 2 correspond to x & y
% xs:   corresponding vector of x-values
% ys:   corresponding vector of y-values
%
% x and y have to be column vectors of the same size. Data points
% outside of the axis limits are allocated to the first or last bin,
% respectively. If output arguments are given, no plot is generated;
% it can be reproduced by "imagesc(ys, xs, H'); axis xy".


% defaults
if nargin < 3
    n = [100 100];
end
if nargin < 4
    ax = [min(x), max(x), min(y), max(y)];
end

% parameters
nx = n(1);
ny = n(2);
xl = ax(1 : 2);
yl = ax(3 : 4);

% generate histogram
i = floor((x - xl(1)) / diff(xl) * nx) + 1;
i(i < 1) = 1;
i(i > nx) = nx;
j = floor((y - yl(1)) / diff(yl) * ny) + 1;
j(j < 1) = 1;
j(j > ny) = ny;
H = sparse(i, j, ones(size(i)), nx, ny);

% generate axes
xs = (0.5 : nx) / nx * diff(xl) + xl(1);
ys = (0.5 : ny) / ny * diff(yl) + yl(1);

% possibly plot
if nargout == 0
    imagesc(ys, xs, H')
    axis xy
    clear H xs ys
end
于 2013-11-03T12:01:18.600 回答