Interpreting GPS info of exif data from photo in python
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}
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, 'r') 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
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}
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