Interpreting GPS info of exif data from photo in python

Question

I am writing a small program to get the GPS info of a iphone jpg photo.

The library I am using is the PIL in python. Now I am able to get the GPSInfo, which is something like:

{1: 'N', 
 2: ((1, 1), (20, 1), (5365, 100)), 
 3: 'E', 
 4: ((103, 1), (41, 1), (1052, 100)), 
 5: 0, 
 6: (43, 1), 
 7: ((15, 1), (32, 1), (7, 1)), 
 16: 'T', 
 17: (77473, 452), 
 29: '2013:10:25'}

How can I interpret this? And I notice the tag is not continuous, so is there any cheating sheet which I can refer to in order to get a better understanding of all the number tags and what they mean? Thank you!

UPDATES

Sorry, I have figured it out. In the PIL lib, there is a GPSTAGS.get() function which can help me decode the key in gps info. Thank you guys!

gpsinfo = {}
for key in exif['GPSInfo'].keys():
    decode = ExifTags.GPSTAGS.get(key,key)
    gpsinfo[decode] = exif['GPSInfo'][key]
print gpsinfo

and here is the result

{'GPSTimeStamp': ((15, 1), (32, 1), (7, 1)), 
 'GPSImgDirectionRef': 'T', 
 'GPSImgDirection': (77473, 452), 
 'GPSLongitude': ((103, 1), (41, 1), (1052, 100)), 
 'GPSLatitudeRef': 'N', 29: '2013:10:25', 
 'GPSAltitude': (43, 1), 
 'GPSLatitude': ((1, 1), (20, 1), (5365, 100)), 
 'GPSLongitudeRef': 'E', 
 'GPSAltitudeRef': 0}

Can you describe what the tuples are in the `'GPSLatitude'` and `'GPSLongitude'` values are? — mnky9800n, Jun 09 '15 at 14:11
@mnky9800n 103 degree, 41 minute, 1052 **centi**second, 103+41/60+1052/(3600*100) — fujianjin6471, Sep 29 '19 at 08:22
i.e. these are (iterator, denominator) tuples, 103/1 = 103°, 41/1 = 41', 1052/100 = 10.52'' — khaz, Apr 23 '22 at 00:29

score 18 · Answer 1 · edited Sep 24 '21 at 19:46

Use exifread module.

Here is a very helpful gist

import exifread as ef


# barrowed from 
# https://gist.github.com/snakeye/fdc372dbf11370fe29eb 
def _convert_to_degress(value):
    """
    Helper function to convert the GPS coordinates stored in the EXIF to degress in float format
    :param value:
    :type value: exifread.utils.Ratio
    :rtype: float
    """
    d = float(value.values[0].num) / float(value.values[0].den)
    m = float(value.values[1].num) / float(value.values[1].den)
    s = float(value.values[2].num) / float(value.values[2].den)

    return d + (m / 60.0) + (s / 3600.0)


def getGPS(filepath):
    '''
    returns gps data if present other wise returns empty dictionary
    '''
    with open(filepath, 'rb') as f:
        tags = ef.process_file(f)
        latitude = tags.get('GPS GPSLatitude')
        latitude_ref = tags.get('GPS GPSLatitudeRef')
        longitude = tags.get('GPS GPSLongitude')
        longitude_ref = tags.get('GPS GPSLongitudeRef')
        if latitude:
            lat_value = _convert_to_degress(latitude)
            if latitude_ref.values != 'N':
                lat_value = -lat_value
        else:
            return {}
        if longitude:
            lon_value = _convert_to_degress(longitude)
            if longitude_ref.values != 'E':
                lon_value = -lon_value
        else:
            return {}
        return {'latitude': lat_value, 'longitude': lon_value}
    return {}


file_path = 'file path of the file'    
gps = getGPS(file_path)
print gps

`float()`'s and `.0`'s no longer needed under Python3. – Gringo Suave May 15 '22 at 18:44 — Gringo Suave, May 15 '22 at 18:44

Pedro Lobito · Answer 2 · 2022-12-05T22:05:31.843

17

Late answer, but as of 2022 you can use GPSPhoto, i.e.:

from GPSPhoto import gpsphoto
# Get the data from image file and return a dictionary
data = gpsphoto.getGPSData('IMG_20181224_201933.jpg')
print(data['Latitude'], data['Longitude'])

Output:

38.71615498471598 -9.148730635643007

Installation:

pip3 install piexif
pip3 install gpsphoto

edited Dec 05 '22 at 22:05

answered Dec 25 '18 at 02:06

Pedro Lobito

94,083
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4

FYI I had to also `pip install piexif` to get this to work. Thanks! – BruceWayne May 29 '19 at 17:30
Hi Pedro, do you have any idea about how that works with a url? I got an issue when I try with the new url of the image I've saved from my models.py -> TypeError: 'NoneType' object is not subscriptable – Louis Oct 17 '20 at 11:06
1

@Louis if by _"url"_ you mean an url of a photo, you probably have to download the photo and process it locally. – Pedro Lobito Oct 24 '20 at 17:26
1

Thanks Pedo for giving me an answer. I've already found the solution – Louis Oct 24 '20 at 17:28
@PedroLobito Will the package work with satellite images? – Aizzaac Jul 29 '21 at 14:16
Absolutely weird package, without link to git and without correct dependencies – Oleg Nov 29 '21 at 22:05

score 8 · Answer 3 · edited Jan 23 '18 at 03:06

OP, has already posted a solution using PIL. If you wants to just get GPS info from Python, you can get it by using exifread

Install package using pip

$ pip install exifread

and get GPS data

In [10]: import exifread

In [11]: tags = exifread.process_file(open('./tests/demo-project/content/test.jpg', 'rb'))                                              

In [12]: geo = {i:tags[i] for i in tags.keys() if i.startswith('GPS')}

In [13]: geo
Out[13]: 
{'GPS GPSAltitude': (0x0006) Ratio=186188/239 @ 898,
 'GPS GPSAltitudeRef': (0x0005) Byte=0 @ 722,
 'GPS GPSDate': (0x001D) ASCII=2015:12:06 @ 954,
 'GPS GPSDestBearing': (0x0018) Ratio=43771/526 @ 946,
 'GPS GPSDestBearingRef': (0x0017) ASCII=T @ 806,
 'GPS GPSImgDirection': (0x0011) Ratio=43771/526 @ 938,
 'GPS GPSImgDirectionRef': (0x0010) ASCII=T @ 782,
 'GPS GPSLatitude': (0x0002) Ratio=[46, 3803/100, 0] @ 850,
 'GPS GPSLatitudeRef': (0x0001) ASCII=N @ 674,
 'GPS GPSLongitude': (0x0004) Ratio=[13, 2429/100, 0] @ 874,
 'GPS GPSLongitudeRef': (0x0003) ASCII=E @ 698,
 'GPS GPSSpeed': (0x000D) Ratio=139/50 @ 930,
 'GPS GPSSpeedRef': (0x000C) ASCII=K @ 758,
 'GPS GPSTimeStamp': (0x0007) Ratio=[10, 37, 33] @ 906,
 'GPS Tag 0x001F': (0x001F) Ratio=30 @ 966}

Interpreting GPS info of exif data from photo in python

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