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I cannot come up with idiomatic Mathematica code to solve this simple problem:

  1. take: image and desired aspect ratio
    1. get image dimensions
    2. upsize the dimensions to desired aspect ratio
  2. return{{left, right}, {bottom, up}} for ImagePad

https://www.instagram.com/p/CMxh05OH4Ss/ (⬅️ needed the answer for these)

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    $\begingroup$ for (1) you can use ImageAspectRatio $\endgroup$
    – kglr
    Mar 23 at 22:48
  • $\begingroup$ @kglr I know but I am struggling to get the second argument to ImagePad calculated o__O $\endgroup$
    – Cetin Sert
    Mar 23 at 23:02
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I have come up with this:

pad // ClearAll
pad[i_Image, r_] := pad[ImageDimensions@i, r]
pad[d_, r_] := Module[{n, w, h},
  n = Max[d]/Max[r]*Min[r];
  {w, h} = Clip[(n - d)/2, {0, \[Infinity]}];
  {w, h} {{1, 1}, {1, 1}} // Round
  ]

To answer the request for an example photo:

i = "https://i.stack.imgur.com/3GmI2.png" // Import
r = {4, 5} (* {w, h} = instagram portrait aspect ratio *)
ImagePad[i, i~pad~r, "Fixed"]

Before padding enter image description here

After padding enter image description here

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    $\begingroup$ do you mind adding an example input(s) showing the connection to image aspect ratio? $\endgroup$
    – kglr
    Mar 24 at 0:07
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$\begingroup$ 
ClearAll[pad2]
pad2[i_Image, ar_] := Module[{d = ImageDimensions @ i},
  Round[{#, #}/2] & /@ Ramp[Max[d] Min[ar, 1/ar] - d]]

ImagePad[i, pad2[i, 4/5], "Fixed"]

enter image description here

ImagePad[#, pad2[#, 4/5], "Fixed"] & @ ImageRotate[i, Pi/2]

enter image description here

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    $\begingroup$ This does feel more idiomatic! Sweet use of Ramp and 4/5. Obviously, I was not posting the above receipt to Instagram but (helping post) a watercolor painting by my wife 💖: instagram.com/p/CMxh05OH4Ss $\endgroup$
    – Cetin Sert
    Mar 24 at 11:42
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    $\begingroup$ @CetinSert, wonderful pictures! $\endgroup$
    – kglr
    Mar 24 at 11:57
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Would the following do?

pad[im_Image, r : {w_, h_}] := 
 Module[{idim = ImageDimensions[im], 
   ratim = Divide @@ ImageDimensions[im], rat = Divide @@ r, pad},
  If[ratim < rat
   , pad = Round[ (idim[[2]] rat - idim[[1]])/2]; 
   ImagePad[im, {{pad, pad}, {0, 0}}, "Fixed"]
   , pad = Round[ (idim[[1]] /rat - idim[[2]])/2]; 
   ImagePad[im, {{0, 0}, {pad, pad}}, "Fixed"]
   ]
  ]

t = pad[i, {4, 5}]
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  • $\begingroup$ I guess so (not a fan of the explicit If though, feels too C-ish) but I also believe something a lot shorter than both of our answers must be possible with a language like Mathematica. $\endgroup$
    – Cetin Sert
    Mar 24 at 11:13
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    $\begingroup$ It is sort of a game to produce the shortest code. But it is often very cryptical and the ideas behind are well hidden. $\endgroup$
    – Daniel Huber
    Mar 24 at 16:39

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