Determine if photos have been rotated to portrait orientation instead of normal landscape orientation

for i in *; do identify $i | awk '{split($3,a,"x"); if (a[2]>a[1]) print $1;}'; done
Most people take photos in landscape orientation (wider than it is tall). Sometimes though you turn the camera sideways to capture a narrow/tall subject. Assuming you then manually rotate those picture files 90 degrees for proper viewing on screen or photo frame, you now have a mix of orientations in your photos directory. This command will print out the names of all the photos in the current directory whose vertical resolution is larger than its horizontal resolution (i.e. portrait orientation). You can then take that list of files and deal with them however you need to, like re-rotating back to landscape for consistent printing with all the others. This command requires the "identify" command from the ImageMagick command-line image manipulation suite. Sample output from identify: identify PICT2821.JPG PICT2821.JPG JPEG 1536x2048 1536x2048+0+0 8-bit DirectClass 688KB 0.016u 0:00.006
Sample Output
PICT0010.JPG
PICT0021.JPG
PICT0022.JPG
PICT0045.JPG
...

What Others Think

Why the superfluous cut? :-) for i in *.jpg; do identify $i | awk '{split($3,a,"x"); if (a[2]>a[1]) print $1;}'; done ... works exactly the same way.
wejn · 225 weeks and 5 days ago
Good point, I've removed the cut. It was there because my initial iterations didn't use awk; once I added in the awk it didn't occur to me that I could remove the cut. Thanks.
dmmst19 · 225 weeks and 5 days ago

What do you think?

Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?

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