GPS coordinate from jpeg file metadata

I'm working on a geo location progect and I need to plot some gps coordinate obtained from Jpeg pictures taken with a smartphone. Using info = imfinfo(filename) I get the information from the jpeg file but they are only ''visual'' information, so i would like to know if there's function to utilize the info i obtain from imfinfo to write a gps variable. To be clear:

imglatitude=0
info=imfinfo('file.jpg')
info =
scalar structure containing the fields:
  Filename = C:\Users\...\file.jpg
  FileModDate = 28-Apr-2015 17:20:16
  FileSize =  3129532
  Format = JPEG
  GPSInfo =
    scalar structure containing the fields:
     GPSLatitudeRef = N
     GPSLongitudeRef = E
     GPSAltitude =  44
     GPSLatitude =
         xx.xxx
         xx.xxx
         xx.xxx
     GPSLongitude =
         xx.xxx
         xx.xxx
         xx.xxx
imglatitude=*GPSLatitude obtanined from imfinfo*

I've got no clue about how to get this gps info out from the imfinfo function! Thanks in advance for your help

[Souka kiw]

can you give me picture to test it ?

Answer 1

[Amy Haskins]

The data returned from imfinfo is a structure and the fields can be accessed individually. If you have access to the Mapping Toolbox then with a little bit of effort you can use the wmmarker function display your image on a map.

Example: Use a sample image from the Image Processing Toolbox which contains geographic data.

handsInfo = imfinfo('hands1.jpg');

Display the GPS information

handsInfo.GPSInfo
ans = 
      GPSLatitudeRef: 'N'
         GPSLatitude: [42 17 57.0600]
     GPSLongitudeRef: 'W'
        GPSLongitude: [71 21 5.8200]
      GPSAltitudeRef: 0
         GPSAltitude: 69.7069
        GPSTimeStamp: [16 48 35]
  GPSImgDirectionRef: 'T'
     GPSImgDirection: 291.9459
        GPSDateStamp: '2014:03:11'

Note that the latitude and longitude are given in Degrees, Minutes, and Seconds (DMS). We will need to convert them to decimal degrees.

lat = dms2degrees(handsInfo.GPSInfo.GPSLatitude);
lon = dms2degrees(handsInfo.GPSInfo.GPSLongitude);

The GPSLatitudeRef and GPSLongitudeRef fields determine if the lat and lon should be positive or negative.

if strcmp(handsInfo.GPSInfo.GPSLatitudeRef,'S')
    lat = -1 * lat;
end
if strcmp(handsInfo.GPSInfo.GPSLongitudeRef,'W')
    lon = -1 * lon;
end

Display the photo on a webmap to put it in context. This particualar image is small, but for a larger image you might want to use imresize to create a smaller thumbnail.

wmmarker(lat, lon, 'OverlayName', 'Hands1', 'Description', ...
    'Photo taken at The Mathworks, Inc', 'Icon', 'hands1.jpg')

[Matteo Breda]

One more thing, I get

 lat = dms2degrees(Info.GPSInfo.GPSLatitude);
 Error using dms2degrees>validateInput (line 70)
 DMS input array must be N-by-3.
 Error in dms2degrees (line 53)
 validateInput(dms)

but dms is allready N-by-3 (degrees, minutes and seconds) so i don't get why matlab is showing this error

[Amy Haskins]

It's possible judging by the format of your output above that your image has the data stored in a column vector (3x1) instead of a row vector (1x3) like I used in my example. If that is the case, then try using the transpose.

 lat = dms2degrees(Info.GPSInfo.GPSLatitude');

[Jesus Garcia]

Just put [] to group the 3 vectors and it will work:

lat = dms2degrees([handsInfo.GPSInfo.GPSLatitude]);

lon = dms2degrees([handsInfo.GPSInfo.GPSLongitude])

Adrian

[Adam Danz]

Thanks for this great demo @Amy Haskins!

I found that some of my image files contained GPS data that had a value of 60 in seconds or minutes which causes an error in dms2degrees where minutes and seconds must be <60.

To get around this, I wrapped dms2degrees in a helper function that corrected the coordinates containing 60. I'll share it here in case someone else (or my future self) needs it.

lat = dms2degreesCorrection(handsInfo.GPSInfo.GPSLatitude);
lon = dms2degreesCorrection(handsInfo.GPSInfo.GPSLongitude);


function dg = dms2degreesCorrection(dms)
% Some image files can have values of 60 for seconds and minutes but this
% errors in MATLAB's dms2degrees.  It's corrected here. dms is a 1x3
% vector. dg is a scalar.
if dms(3) >= 60
    dms(2) = dms(2) + floor(dms(3)/60);
    dms(3) = mod(dms(3),60);
end
if dms(2) >= 60
    dms(1) = dms(1) + floor(dms(2)/60);
    dms(2) = mod(dms(2),60);
end
dg = dms2degrees(dms);
end