我读了this post并天真地试图显示一个灰度图像,其中包含从.mat
格式的矩阵中随机抽取的100个手写数字,其中包含5000行,每行对应一个手写数字。我使用以下代码:
library(R.matlab)
data <- readMat('data.mat')
X = data$X
X = X[sample(nrow(X), size=100, replace=F),]
par(mar = rep(0, 4))
image(X, axes = FALSE, col = grey(seq(0, 1, length = 256)))
由于事实证明矩阵中的每个400元素行向量提供了像素的20 x 20像素图像的信息,因此结果并不好。因此,如果我们想要将它们并排显示,则必须先验地设置各个数字的单独显示图像的高度和宽度。这就是我的编码“技能”不足的地方。
我保存了dataset here。需要说明的是,图片位于R str(data)
列表的 X 对象中:
List of 2
$ X: num [1:5000, 1:400] 0 0 0 0 0 0 0 0 0 0 ...
$ y: num [1:5000, 1] 10 10 10 10 10 10 10 10 10 10 ...
- attr(*, "header")=List of 3
..$ description: chr "MATLAB 5.0 MAT-file, Platform: PCWIN64, Created on: Sat May 28 18:21:58 2016 "
..$ version : chr "5"
..$ endian : chr "little"
如果我在用X
矩阵构建20 x 20
矩阵后仅限于显示矩阵X = data$X
X = X[sample(nrow(X), size = 1 ,replace=F),]
X = matrix(X, nrow = 20, byrow= T)
par(mar = rep(0, 4))
image(X, axes = FALSE, col = grey(seq(0, 1, length = 256)))
的一行,我会更接近目标:
function [h, display_array] = displayData(X, example_width)
%DISPLAYDATA Display 2D data in a nice grid
% [h, display_array] = DISPLAYDATA(X, example_width) displays 2D data
% stored in X in a nice grid. It returns the figure handle h and the
% displayed array if requested.
% Set example_width automatically if not passed in
if ~exist('example_width', 'var') || isempty(example_width)
example_width = round(sqrt(size(X, 2)));
end
% Gray Image
colormap(gray);
% Compute rows, cols
[m n] = size(X);
example_height = (n / example_width);
% Compute number of items to display
display_rows = floor(sqrt(m));
display_cols = ceil(m / display_rows);
% Between images padding
pad = 1;
% Setup blank display
display_array = - ones(pad + display_rows * (example_height + pad), ...
pad + display_cols * (example_width + pad));
% Copy each example into a patch on the display array
curr_ex = 1;
for j = 1:display_rows
for i = 1:display_cols
if curr_ex > m,
break;
end
% Copy the patch
% Get the max value of the patch
max_val = max(abs(X(curr_ex, :)));
display_array(pad + (j - 1) * (example_height + pad) + (1:example_height), ...
pad + (i - 1) * (example_width + pad) + (1:example_width)) = ...
reshape(X(curr_ex, :), example_height, example_width) / max_val;
curr_ex = curr_ex + 1;
end
if curr_ex > m,
break;
end
end
% Display Image
h = imagesc(display_array, [-1 1]);
% Do not show axis
axis image off
drawnow;
end
在matlab中将矩阵数据解析为图像的代码是:
$firebaseObject
导致:
答案 0 :(得分:1)
这将矩阵行重新整形为块。不能保证非常高效,但对于像你的例子中的100x400矩阵来说要快得多。
# get 100 random rows of X
X <- data$X
X <- X[sample(nrow(X), size=100, replace=FALSE),]
# allocate empty image matrix (200 by 200 pixels)
Z <- matrix(rep(0, length(X)), nrow=200)
# fill empty image matrix
for (row in 0:9) {
rmin <- 1 + (row)*20
for (col in 0:9) {
cmin <- 1 + (col)*20
Z[rmin:(rmin+19), cmin:(cmin+19)] <- X[row * 10 + col + 1,]
}
}
# plot (after rotating matrix 90 degrees)
image(t(apply(Z, 2, rev)))