GPS coordinate from jpeg file metadata

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Matteo Breda
Matteo Breda 2015-5-18
评论： Adam Danz 2024-3-23
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

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采纳的回答

Amy Haskins
Amy Haskins 2015-5-18
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')
  4 个评论
显示 2更早的评论
Jesus Garcia
Jesus Garcia 2017-10-30
Just put [] to group the 3 vectors and it will work:
lat = dms2degrees([handsInfo.GPSInfo.GPSLatitude]);
lon = dms2degrees([handsInfo.GPSInfo.GPSLongitude])
Adrian
Adam Danz
Adam Danz 2024-3-23
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

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