Reading a 16 bit RAW image from Grasshopper3 (Point Grey Research)

Hi,

I have been trying to read a 16 bit RAW image, captured using a Grasshopper3 Camera (Point Grey Research), in MATLAB. All images were captured using the Flycapture2 software. The image itself is 1384x1036 and has a memory stride of 2768 (since it's a 16 bit image). After a lot of reading in various parts of the web, I figured that RAW images are supposed to be considered as a binary stream of data. So with the following code i should be able to read the RAW image:

    row = 1036; col = 1384;
    fin = fopen('raw13.raw','r');
    ima = fread(fin, [col*2 row],'uint8');
    temp = zeros(col,row);
    j=1;
    for i=1:2:col*2-1
    temp(j,:) = ima(i,:) + ima(i+1,:)*2^8; %The first element is the lower 8bits and the second element is the higher 8bits
    j = j+1;
    end
    imshow(temp',[0 2^16-1])

This is not creating the image I want and there seems to be a lot of overlap and a lot of data missing in the reconstructed image. A I don't think I am doing anything wrong. Is there anyway of knowing what the exact size of the RAW file would be and does anyone else have previous experience with this camera? Would be helpful if someone could help with the file format.

Thank You.

[Guillaume]

So, from your comment, the problem is that there's a bug in the manufacturer's software and your code is fine.

I just want to point out that you could simplify it a lot by reading the image straight as uint16:

row = 1036; col = 1384;
fin = fopen('raw13.raw','r', 'l'); %'l' optional on little-endian platform (windows)
ima = fread(fin, [col row], '*uint16'); %read as uint16 and store as uint16
fclose(fin);
imshow(fin); %because ima is uint16, you don't need to specify the range

Answer 1

[Guillaume]

I've been able to read RAW images successfully using multibandread. Now, one RAW image format is different from another RAW image format, so there's no guarantee it will work for you, but give it a try.

However, are you sure that the image is truly 16 bits. A quick search on the web, would indicate your camera is either 12 or 14 bits. In which case, it's possible that reading as 16 bits is wrong. Possibly try:

img = multibandread('filename', [1036 1384 1], 'ubit12=>uint16', 'bsq', 'ieee-be')

Ultimately, you need to refer to either the software documentation or the camera documentation to know the details of their RAW format. The content of the file is supposed to be a dump of the sensor data without any processing.

[Image Analyst]

He said the manufacturer admitted that the raw files being written out in the current version are corrupt, but fortunately the pgm files are okay.

[Guillaume]

Oh! I missed that.

Answer 2

[Image Analyst]

Why not just save the image as a standard format like PNG? As far as I know all run of the mill machine vision cameras will let you save your images in a variety of standard formats. Are you absolutely sure raw is the ONLY format? I got some images from someone who used a Point Grey camera and they were in TIFF format.

Can you attach one of the raw images?

[Mathews John]

I need the RAW format for some further processing prior to this. I have already made a code in MATLAB to save images as a TIFF from the camera, but I really want the RAW format. You can download the raw file from this link :

RAW image

[Image Analyst]

Is there any header and do you know the format/layout of it? Or at least how many bytes it is?

[Image Analyst]

The link Witek gave says "The Portable Grey Map (.pgm) file format can be used to correctly save 16bpp images." so why not use that? This is one that imread() can understand:

>> imformats

EXT       ISA     INFO        READ      WRITE     ALPHA  DESCRIPTION                     
-----------------------------------------------------------------------------------------
bmp       isbmp   imbmpinfo   readbmp   writebmp  0      Windows Bitmap
cur       iscur   imcurinfo   readcur             1      Windows Cursor resources
fts fits  isfits  imfitsinfo  readfits            0      Flexible Image Transport System
gif       isgif   imgifinfo   readgif   writegif  0      Graphics Interchange Format
hdf       ishdf   imhdfinfo   readhdf   writehdf  0      Hierarchical Data Format
ico       isico   imicoinfo   readico             1      Windows Icon resources
j2c j2k   isjp2   imjp2info   readjp2   writej2c  0      JPEG 2000 (raw codestream)
jp2       isjp2   imjp2info   readjp2   writejp2  0      JPEG 2000 (Part 1)
jpf jpx   isjp2   imjp2info   readjp2             0      JPEG 2000 (Part 2)
jpg jpeg  isjpg   imjpginfo   readjpg   writejpg  0      Joint Photographic Experts Group
pbm       ispbm   impnminfo   readpnm   writepnm  0      Portable Bitmap
pcx       ispcx   impcxinfo   readpcx   writepcx  0      Windows Paintbrush
pgm       ispgm   impnminfo   readpnm   writepnm  0      Portable Graymap
png       ispng   impnginfo   readpng   writepng  1      Portable Network Graphics
pnm       ispnm   impnminfo   readpnm   writepnm  0      Portable Any Map
ppm       isppm   impnminfo   readpnm   writepnm  0      Portable Pixmap
ras       isras   imrasinfo   readras   writeras  1      Sun Raster
tif tiff  istif   imtifinfo   readtif   writetif  0      Tagged Image File Format
xwd       isxwd   imxwdinfo   readxwd   writexwd  0      X Window Dump

[Mathews John]

I will look into .pgm then. I contacted tech support and they said the .raw files doesn't have any headers but will be 14bit data (from the ADC) and a 2 bit padding. With this new information, I'll try tweaking my own code a bit and see what happens

[Image Analyst]

Maybe try

ima = fread(fin, [col row],'*uint16');

[Mathews John]

That is also not working. I tried the .pgm and that's fine. Another thing that I noticed is I am able to read much more than what should be there. Technically there should be 1036*1384*2 bytes of uint8 data. However, I am able to read much more than that. I don't understand why this is so.

[Image Analyst]

Well, what can I say? Just extract/crop out the stuff you need and throw away the stuff you don't need.

[Mathews John]

Hey! Thanks for all the help. I talked to tech support thoroughly. It seems the raw file produced from the version of the software I am using is um... corrupted I suppose. The code I am using was perfect and worked perfectly when I got the raw file from a previous version.

Thank You so much anyways :)

Answer 3

[Witek Jachimczyk]

Indeed, knowing the format precisely is the key. I found a little bit about it here:

http://www.ptgrey.com/KB/10065

but it still needs to be understood thoroughly. If it's headerless, you might be able to use vision.BinaryFileReader from the Computer Vision System toolbox:

http://www.mathworks.com/help/vision/ref/vision.binaryfilereader-class.html?refresh=true

HTH,

Witek