incorrect input size of image
16 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
new_user
am 21 Dez. 2021
Kommentiert: Image Analyst
am 21 Dez. 2021
images = imageDatastore('call', 'IncludeSubfolders',true, 'LabelSource','foldernames');
[Training_image, Validation_image] = splitEachLabel(images, 0.6); %split images in 75, 25% use 'randomized' also
Input_Layer_Size = net.Layers(1).InputSize(1:2); % (1:2 = 1st 2 elemnts of input size), input layer size stored in this variable (Input_layer_size)
Resized_Training_image = augmentedImageDatastore(Input_Layer_Size, Training_image, 'ColorPreprocessing','gray2rgb');
net = trainNetwork(Resized_Training_image, New_Network, Training_Options) % training the network
trainall=[]
for i=1:2200
i
a=readimage(Resized_Training_image,i);
[m n c]=size(a);m
Error using DAGNetwork/activations (line 262)
The spatial dimension sizes [169 300 3] of the input images to layer 'input_1' must be greater than or equal to the corresponding
minimum input dimension sizes of the layer [224 224 3].
%% what should i do to insert the resized taining image in the for loop operation.
0 Kommentare
Akzeptierte Antwort
Image Analyst
am 21 Dez. 2021
Replace
a=readimage(Resized_Training_image,i);
[m n c]=size(a);m
with
rgbImage = readimage(Resized_Training_image, i);
[rows, columns, numberOfColorChannels] = size(rgbImage);
fprintf('Original image #%d has %d rows, %d columns, and %d color channels.\n', ...
i, rows, columns, numberOfColorChannels)
if rows ~= 224 || columns ~= 224
% Need to resize laterally to 224 by 224.
rgbImage = imresize(rgbImage, [224, 224]);
end
if numberOfColorChannels == 1
% It's gray scale. Need to convert to color.
rgbImage = ind2rgb(rgbImage, gray(256));
end
3 Kommentare
Image Analyst
am 21 Dez. 2021
I don't know. It's difficult for me to reproduce without your images and your folder structure. But the error is now in readimage() which comes before any of my resizing code, so the error lies in your code. Maybe try reading the image this way instead
fullFileName = images.Files{i};
a = imread(fullFileName);
and again I ask you to use descriptive filenames. No one like looking at code that is an alphabet soup of single letter variables. It's maybe not so bad with a very short program but once you get dozens of lines, it can get confusing knowing what each letter actually represents without searching back over the code.
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Image Data Workflows finden Sie in Help Center und File Exchange
Produkte
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!