imread, reshape and zeros function syntax problems

Question

Manu Chaudhary 2022-1-16

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https://ww2.mathworks.cn/matlabcentral/answers/1629730-imread-reshape-and-zeros-function-syntax-problems

评论： Manu Chaudhary 2022-1-16

I am completely new to matlab. Studing some sample codes to learn matlab. I have some basic questions in the below code:

input_image_filename= './Images/Image_64x64.jpg';   % It is a 64x64  colored image
input_im_3D= imread(input_image_filename);    

What is the return type of input_im_3D ? What I observe that it is storing some huge data in the variable input_im_3D?

n_pix= 12288;
x = double(reshape(input_im_3D, [n_pix 1]));

What I observe is that reshape function is putting data in the single column of a excel sheet till 12288.

I am really not able to understand the working of reshape function? Return type of reshape function ? What changes double put in the results? What does the syntax of [n_data 1] means ?

n_data= 16384 ;
x = [x; zeros(n_data - n_pix , 1)];

Please also explain me the syntax of zeros?

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Answer 1

Walter Roberson 2022-1-16

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https://ww2.mathworks.cn/matlabcentral/answers/1629730-imread-reshape-and-zeros-function-syntax-problems#answer_875290

在 MATLAB Online 中打开

imread() returns the data type stored in the file. The most common data type stored in image files is uint8(), but there are other possibilities. For example,

basename = tempname();
fn1 = basename + "logical.png";
fn3 = basename + "uint16.png";
img1 = rand([48 64]) < 0.5;
img3 = randi([0 65535], [48 64 3], 'uint16');
imwrite(img1, fn1);
imwrite(img3, fn3);
back1 = imread(fn1);
back3 = imread(fn3);
whos back1 back3
  Name        Size              Bytes  Class      Attributes

  back1      48x64               3072  logical              
  back3      48x64x3            18432  uint16               
isequal(img1, back1)
ans = logical
   1
isequal(img3, back3)
ans = logical
   1

True floating point images are possible, but writing one out takes more work for demonstration takes more work than I care to bother with at the moment.

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Walter Roberson 2022-1-16

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The return type of reshape() is exactly the same as the original data object. When reshape() is applied to any of the primitive data types (such as numeric or cell), then reshape() creates a new "view" on the data in which the data has exactly the same type as before, but is organized in the specified number of rows and columns without changing the memory organization

Suppose that you had the array

A = [1 2; 3 4; 5 6]
A = 3×2
     1     2
     3     4
     5     6

then unlike in C or C++, the internal implementation does not work by storing each row or column separately. Instead, the data is stored as a continuous stream in memory. The organization is to go "down" each column in turn and then move on to the next column. So the actual in-memory order of the data for A would be

as consecutive memory locations.

MATLAB knows what size of array this stream of memory represents because it has a separate header that gives the dimensions... [3 2] (three rows, two columns).

Now suppose you take the same internal data pointer, but give it a new header that says that the array size is [1 6] -- one row, six columns. The data does not need to move around in memory, just the header changes.

reshape(A, [1, 6])

same data type, same internal data pointer, but the "interpretation" of the data is now as

1 3 5 2 4 6

Walter Roberson 2022-1-16

The data returned by imread() was very likely uint8 . uint8 is a series of 8-bit unsigned integers, possible values 0, 1, 2, 3, 4, ... up to 255, all as 8-bit patterns such as 01011010 internally (that would correspond to integer decimal 90).

When you call double() then a representation conversion is done that attempts to preserve value as best possible, but using a diffferent data type -- in particular, as IEEE 754 Double Precision, which is a 64 bit floating point representation that is notably more complicated, but can also represent a much fine range of values. double() does not just use the same bits and re-interpret them: it finds the nearest double-precision value that represents the uint8 value. In this example it would create the double precision value 90.0 (which happens to have bit pattern 0000000000000000000000000000000000000000100000000101011001000000 internally)

But... you have to be careful. Functions such as image() and imshow() and imwrite() expect that if an array is double precision, that the image information is represented in the range 0.0 to 1.0 and that any value below 0.0 is to be treated as 0.0 and any value above 1.0 is to be treated as 1.0 . There sometimes are good reasons to convert integer image data to the nearest floating point equivalent... but if you do so you have to be careful to later convert back to integer data type for image processing purposes.

Walter Roberson 2022-1-16

在 MATLAB Online 中打开

Remember that you started with a 64 x 64 x 3 matrix (RGB image), and you shaped that as a column of data, and then you added 4096 zeros to the end of the column. Now you are reshaping that result to 64 x 64 x 4.

But 4096 = 64 x 64, so what you have effectively done is taken the original 64 x 64 x 3 array, and added another layer of zeros on in the third dimension, giving you a 64 x 64 x 4 result. The same result could have been achieved by

y_data = input_im_3D;
y_data(:,:,4) = 0;   %add a 4th layer that is all 0

After that, selecting y_data(1:64, 1:64, 1:3) would be the same as y_data(:,:,1:3)... which would be the data with that layer of 0 removed, giving you back the original data.

It is not clear why any of this is being done.

Manu Chaudhary 2022-1-16

Trying to test quantum image processing application. Image converted to Textfile data of 16384 length-> Normalized -> Given Input to quantum circuit simulator -> Result taken out -> Image regenerated using matlab. Testing the basic input output image functionality of matlab.

Thank you Walter for this great help.

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