假设我有文件：
famcal.2012000000000000000000625502.stellar.blu
famcal.2012000000000000000000625502.stellar.grn
famcal.2012000000000000000000625502.stellar.red
famcal.2012000000000000000000625502.stellar.nir
famcal.2012000000000000000000625503.stellar.blu
famcal.2012000000000000000000625503.stellar.grn
famcal.2012000000000000000000625503.stellar.red
famcal.2012000000000000000000625503.stellar.nir
famcal.2012000000000000000000625504.stellar.blu
famcal.2012000000000000000000625504.stellar.grn
famcal.2012000000000000000000625504.stellar.red
famcal.2012000000000000000000625504.stellar.nir
famcal.2012000000000000000000625505.stellar.blu
famcal.2012000000000000000000625505.stellar.grn
famcal.2012000000000000000000625505.stellar.red
famcal.2012000000000000000000625505.stellar.nir
famcal.2012000000000000000000625506.stellar.blu
famcal.2012000000000000000000625506.stellar.grn
famcal.2012000000000000000000625506.stellar.red
famcal.2012000000000000000000625506.stellar.nir
famcal.2012000000000000000000625507.stellar.blu
famcal.2012000000000000000000625507.stellar.grn
famcal.2012000000000000000000625507.stellar.red
famcal.2012000000000000000000625507.stellar.nir
我希望能够按此顺序使用所有这些文件，以生成新的矩阵。
基本上，一段代码可以按照以下顺序执行：
625502
625503
625504
625505
625506
625507
并且对于这些数字集中的每一个都按以下顺序执行：
blu
grn
red
nir
我现在正在使用代码从每个文件中提取小矩阵，然后将它们保存到新矩阵中的新位置。例如：
a = 2000;
star_block = repmat(a,[41,78]); %populates matrix with 2000
这会创建一个填充2000的矩阵。我想用上面文件的输出替换此矩阵的某些部分。
star_block(2:10,4:12) = 625502 blu
star_block(12:20,4:12) = 625502 grn
star_block(22:30,4:12) = 625502 red
star_block(32:40,4:12) = 625502 nir
star_block(2:10,17:25) = 625503 blu
star_block(12:20,17:25) = 625503 grn
star_block(22:30,17:25) = 625503 red
star_block(32:40,17:25) = 625503 nir
star_block(2:10,30:38) = 625504 blu
star_block(12:20,30:38) = 625504 grn
star_block(22:30,30:38) = 625504 red
star_block(32:40,30:38) = 625504 nir
star_block(2:10,43:51) = 625505 blu
star_block(12:20,43:51) = 625505 grn
star_block(22:30,43:51) = 625505 red
star_block(32:40,43:51) = 625505 nir
star_block(2:10,56:64) = 625506 blu
star_block(12:20,56:64) = 625506 grn
star_block(22:30,56:64) = 625506 red
star_block(32:40,56:64) = 625506 nir
star_block(2:10,69:77) = 625507 blu
star_block(12:20,69:77) = 625507 grn
star_block(22:30,69:77) = 625507 red
star_block(32:40,69:77) = 625507 nir
修改
band_files = dir([star_path '/*.blu']);
for i=1:length(band_files)
blue01 = band_files(1).name;
green01 = strrep(blue01, 'blu', 'grn');
red01 = strrep(blue01, 'blu', 'red');
nir01 = strrep(blue01, 'blu', 'nir');
blue02 = band_files(2).name;
green02 = strrep(blue02, 'blu', 'grn');
red02 = strrep(blue02, 'blu', 'red');
nir02 = strrep(blue02, 'blu', 'nir');
blue03 = band_files(3).name;
green03 = strrep(blue03, 'blu', 'grn');
red03 = strrep(blue03, 'blu', 'red');
nir03 = strrep(blue03, 'blu', 'nir');
blue04 = band_files(4).name;
green04 = strrep(blue04, 'blu', 'grn');
red04 = strrep(blue04, 'blu', 'red');
nir04 = strrep(blue04, 'blu', 'nir');
blue05 = band_files(5).name;
green05 = strrep(blue05, 'blu', 'grn');
red05 = strrep(blue05, 'blu', 'red');
nir05 = strrep(blue05, 'blu', 'nir');
blue06 = band_files(6).name;
green06 = strrep(blue06, 'blu', 'grn');
red06 = strrep(blue06, 'blu', 'red');
nir06 = strrep(blue06, 'blu', 'nir');
end
blue1 = [star_path '\' blue01];
green1 = [star_path '\' green01];
nir1 = [star_path '\' nir01];
red1 = [star_path '\' red01];
%blue
fid=fopen(blue1,'rb'); % opens the file for reading
fseek (fid, 800, -1);% Skip past header, which is 800 bytes long
blue = fread(fid, [3552,6536], '*uint16', 'ieee-be');%reads in and converts .RAS file
blue_chip = extract_region_fwd(blue);
star_block(2:10,4:12) = blue_chip;%inserts star chip into main matrix
%green
fid=fopen(green1,'rb'); % opens the file for reading
fseek (fid, 800, -1);% Skip past header, which is 800 bytes long
green = fread(fid, [3552,6536], '*uint16', 'ieee-be');%reads in and converts .RAS file
green_chip = extract_region_fwd(green);
star_block(12:20,4:12) = green_chip;%inserts star chip into main matrix
%red
fid=fopen(red1,'rb'); % opens the file for reading
fseek (fid, 800, -1);% Skip past header, which is 800 bytes long
red = fread(fid, [3552,6536], '*uint16', 'ieee-be');%reads in and converts .RAS file
red_chip = extract_region_fwd(red);
star_block(22:30,4:12) = red_chip;%inserts star chip into main matrix
%nir
fid=fopen(nir1,'rb'); % opens the file for reading
fseek (fid, 800, -1);% Skip past header, which is 800 bytes long
nir = fread(fid, [3552,6536], '*uint16', 'ieee-be');%reads in and converts .RAS file
nir_chip = extract_region_fwd(nir);
star_block(32:40,4:12) = nir_chip;%inserts star chip into main matrix
答案 0 :(得分：1)
您可以生成这些数字，但更可靠的方法是使用dir然后通过返回的结构访问文件：
blu_files = dir('*.blu');
for i=1:length(blu_files)
blu_file = blu_files(i).name;
grn_file = strrep(blu_file, 'blu', 'grn');
red_file = strrep(blu_file, 'blu', 'red');
nir_file = strrep(blu_file, 'blu', 'nir');
% load files ...
end
将相同的程序放入函数中。
根据您在问题编辑中发布的代码示例，以下功能是有意义的：
<强> apply_chips.m
function [] = apply_chips(blue, bx, by, green, gx, gy, red, rx, ry, nir, nx, ny)
%insert star chips into main matrix
star_block(bx, by) = read_block(blue );
star_block(gx, gy) = read_block(green);
star_block(rx, ry) = read_block(red );
star_block(nx, ny) = read_block(nir );
<强> read_block.m
function [block] = read_block(filename)
fid = fopen(filename, 'rb'); % opens the file for reading
fseek(fid, 800, -1); % Skip past header, which is 800 bytes long
data = fread(fid, [3552,6536], '*uint16', 'ieee-be'); % reads in and converts .RAS file
block = extract_region_fwd(data);
您现在应该能够：
ranges.blue.x = { 2:10, 2:10, 2:10, 2:10, 2:10, 2:10};
ranges.blue.y = { 4:12, 17:25, 30:38, 43:51, 56:64, 69:77};
ranges.green.x = {12:20, 12:20, 12:20, 12:20, 12:20, 12:20};
ranges.green.y = { 4:12, 17:25, 30:38, 43:51, 56:64, 69:77};
ranges.red.x = {22:30, 22:30, 22:30, 22:30, 22:30, 22:30};
ranges.red.y = { 4:12, 17:25, 30:38, 43:51, 56:64, 69:77};
ranges.nir.x = {32:40, 32:40, 32:40, 32:40, 32:40, 32:40};
ranges.nir.y = { 4:12, 17:25, 30:38, 43:51, 56:64, 69:77};
blu_files = dir('*.blu');
for i=1:length(blu_files)
blu_file = blu_files(i).name;
grn_file = strrep(blu_file, 'blu', 'grn');
red_file = strrep(blu_file, 'blu', 'red');
nir_file = strrep(blu_file, 'blu', 'nir');
apply_chips([star_path '\' blu_file], ranges.blue.x{i} , ranges.blue.y{i} , ...
[star_path '\' grn_file], ranges.green.x{i}, ranges.green.y{i}, ...
[star_path '\' red_file], ranges.red.x{i} , ranges.red.y{i} , ...
[star_path '\' nir_file], ranges.nir.x{i} , ranges.nir.y{i} , ...);
end
答案 1 :(得分：0)
这显示了你可以用数组做什么，但是不应该太难以把它变成矩阵