Is it possible to snip a certain number of pixels from a specified side of an image, without specifying the image's final width and height?

Question

It seems that ImageCrop allows one to specify the final size of the image. For example, ImageCrop[image, size, spec] crops image image to size size, and one can specify which side of the image the pixels should be taken from, using spec.

But, what if I don't know the final size size of the image? What if I just want to snip 10 pixels from the right side of the image? I might want to do this for a series of images, whose sizes are not known (at least immediately).

Dr. belisarius · Answer 1 · 2012-07-27T18:21:36.190

12

Your code would be

ImageTake[i, All, {1, -11}]

Examples:

img = Import["ExampleData/rose.gif"];
ImageTake[img, {25, -25}, {10, -10}]

Mathematica graphics

ImageTake[ExampleData[{"TestImage", "Lena"}], {180, -180}, {180, -180}]

Mathematica graphics

edited Jul 27 '12 at 18:21

answered Jul 24 '12 at 19:30

Dr. belisarius

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Thanks! OK, I need to specify size as All. Cool! – Andrew Jul 24 '12 at 19:37
@MatthiasOdisio OBOE :) – Dr. belisarius Jul 27 '12 at 18:21

image_doctor · Accepted Answer · 2012-07-24T22:16:44.723

4

Same basic solution using image take but wrapped up a little:

img = Import["ExampleData/rose.gif"];

Crop[img_, {{dxl_, dxr_}, {dyl_, dyu_}}] := ImageTake[img, {dyl, -dyu}, {dxl, -dxr}]

Where dxl, dxr, dyl, dyu represent the x lext, x right, y lower and y upper sizes of crop.

Crop[img,{{10, 10}, {25, 25}}]

Removes 10 pixels from the left and right edges and 25 from the bottom and top.

You can then apply the same trim a group of images held in a list, by the following

images={im1,im2,im3,im4,...,imn};

Crop[#,{{0,10},{5,0}}]&/@images

edited Jul 24 '12 at 22:16

answered Jul 24 '12 at 19:50

image_doctor

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When ImageTake is provided with negative arguments, it counts down from ImageDimensions, like Part – Dr. belisarius Jul 24 '12 at 20:11
@belisarius Thanks for that, I didn't really pick that up from the docs. Which reduces this pretty much to the same as your solution :) – image_doctor Jul 24 '12 at 22:15
From the help -n ... last n elements :) – Dr. belisarius Jul 24 '12 at 23:16
@Belisarius It must be the way I read them :) I took, ImageTake[image,-n] gives an image consisting of the last n rows of image to mean you got rows N-n+1 to N, not, 0 to N-n. As in -3 gives you the last 3 rows, not everything but the last 3 rows. – image_doctor Jul 24 '12 at 23:22
Well, you know: Mma docs have been written by 10^7 monkeys – Dr. belisarius Jul 24 '12 at 23:28
@belisarius It's the technique I apply to write most of my code ;) – image_doctor Jul 24 '12 at 23:38

score 1 · Answer 3 · answered Jul 27 '12 at 18:13

As an alternative to ImageTake, ImagePad will do the job using the syntax ImagePad[img, {{left, right}, {bottom, top}}]:

img = Import["ExampleData/rose.gif"];
In[10]:= ImagePad[img, {{0, -10}, {0, 0}}] == ImageTake[img, All, {1, -11}]
Out[10]= True

Note the difference between taking up to the last minus 10th column and removing 10 columns.

Is it possible to snip a certain number of pixels from a specified side of an image, without specifying the image's final width and height?

3 Answers3

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