How to extract jpg EXIF metadata from a folder in chronological order

Question

I am currently writing a script to extract EXIF GPS data from a folder of jpg images. I am using os.scandir to extract the entries from the folder but from my understanding os.scandir opens the files in an arbitrary way. I need the images to be opened in chronological order by filename. Below is my current code which works as intended however it does not open the images in the correct order. The files within my image folder are named chronologically like so: "IMG_0097, IMG_0098" etc.

#!/usr/bin/python

import os, exif, folium

def convert_lat(coordinates, ref):
    latCoords = coordinates[0] + coordinates[1] / 60 + coordinates[2] / 3600
    if ref == 'W' or ref == 'S':
        latCoords = -latCoords
    return latCoords

coordList=[]
map = folium.Map(location=[51.50197125069916, -0.14000860301423912], zoom_start = 16)

from exif import Image
with os.scandir('gps/') as entries:
    try:
        for entry in entries:   
            img_path = 'gps/'+entry.name
            with open (img_path, 'rb') as src:
                img = Image(src)            
                if img.has_exif:
                    latCoords = (convert_lat(img.gps_latitude, img.gps_latitude_ref))
                    longCoords = (convert_lat(img.gps_longitude, img.gps_longitude_ref))
                    coord = [latCoords, longCoords]
                    coordList.append(coord)
                    folium.Marker(coord, popup=str(coord)).add_to(map)
                    folium.PolyLine(coordList, color =" red", weight=2.5, opacity=1).add_to(map)
                    print(img_path)
                    print(coord)
                else:
                    print (src.name,'has no EXIF information')
    except:
        print(img_path)
        print("error occured")

map.save(outfile='/home/jamesdean/Desktop/Python scripts/map.html')
print ("Map generated successfully")

Answer 1

You can sort a list using the built-in sorted function, Paul made an interesting point and simply sorting without any arguments will work just as fine:

a = ["IMG_0097.jpg", "IMG_0085.jpg", "IMG_0043.jpg", "IMG_0098.jpg", "IMG_0099.jpg", "IMG_0100.jpg"] 
sorted_list = sorted(a)
print(sorted_list)

Output:

['IMG_0043.jpg', 'IMG_0085.jpg', 'IMG_0097.jpg', 'IMG_0098.jpg', 'IMG_0099.jpg', 'IMG_0100.jpg']

In your case you can do:

for entry in sorted(entries):