11

So i have an image lets say small.png and a bigger image big.png .The small image occurs 2 times in the bigger image and i wish to find its location.

I tried using numpy and Image but got error

'JpegImageFile' object has no attribute '__getitem__'.I had png format earlier and it gave the same error.

Is there any other module or way to get this done.I am open to any .

The code as of now which throws error is

import numpy
from PIL import Image
here = Image.open(r"/Users/vks/Desktop/heren.jpg")
big = Image.open(r"/Users/vks/Desktop/Settingsn.jpg")
print here,big
herear = numpy.asarray(here)
bigar = numpy.asarray(big)
herearx = herear.shape[0]
hereary = herear.shape[1]

bigarx = bigar.shape[0]
bigary = bigar.shape[1]

print herearx , hereary , bigarx, bigary

stopx = bigarx - herearx + 1
stopy = bigary - hereary + 1

for x in range(0, stopx):
    for y in range(0, stopy):
        x2 = x+ herearx
        y2 = y + hereary
        pic = big[y:y2, x:x2] ===> error thrown here
        test = pic = here
        if test.all():
            print x,y
else:
    print "None"

There was another cv2 module but it just doesnt get installed on my mac. pip install cv2 fails saying on package found.

enter image description here

enter image description here

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    OpenCV indeed has a set of functions for this: Template Matching (docs)
    – jedwards
    Commented Jul 20, 2016 at 6:45
  • It will give coordinates. I don't own a mac unfortunately. On windows, I downloaded a precompiled binary and on linux I built it and the bindings (much more annoying). This site may help? In any case, on the line you indicate throwing an error, should you be using bigar instead of big?
    – jedwards
    Commented Jul 20, 2016 at 6:49
  • I see what you're trying to do, but I think the lines should read something like: pic = bigar[y:y2, x:x2]; test = (pic == herear) (note big and here were both changed to their numpy array versions (so you can use your indexing and equivalence testing) and the second equal sign on the next line was replaced with a double equal sign for equivalence testing, not assignment)
    – jedwards
    Commented Jul 20, 2016 at 6:55
  • @Divakar, while the question essentially boils down to finding a matching subarray, it's hardly seems like a duplicate, especially because the issue in the poster's code wasn't actually restricted to just this.
    – jedwards
    Commented Jul 20, 2016 at 7:51
  • 1
    @Divakar No, I agree it will. My point was that there were issues in his code that exceeded just what was outlined in the dupe target. Without my answer, just referencing the dupe target wouldn't have solved his problems. I agree referencing it would be useful, but as far as it being a duplicate, I disagree. It'd be like saying it's a duplicate of this. It's part of the solution, yes, but not a duplicate. The point is somewhat moot, however, as I think it was satisfactorily (IMHO) answered before being closed.
    – jedwards
    Commented Jul 20, 2016 at 8:00

2 Answers 2

11

The following code works for me:

import numpy
from PIL import Image

here = Image.open(r"eye.png")
big  = Image.open(r"lenna.png")

herear = numpy.asarray(here)
bigar  = numpy.asarray(big)

hereary, herearx = herear.shape[:2]
bigary,  bigarx  = bigar.shape[:2]

stopx = bigarx - herearx + 1
stopy = bigary - hereary + 1

for x in range(0, stopx):
    for y in range(0, stopy):
        x2 = x + herearx
        y2 = y + hereary
        pic = bigar[y:y2, x:x2]
        test = (pic == herear)
        if test.all():
            print(x,y)

Output: 312 237

Graphically:

enter image description here

Test Images Used

lenna.png

Lenna

eye.png

Eye

Note: It's important that you use a lossless image format when you create the smaller, cropped version (PNG is lossless, JPEG is most usually lossy). If you use a lossy format, you risk the pixel values being close, but not identical. The above code based off yours will only find exact pixel-by-pixel matches. The OpenCV template matching functions are a bit more flexible in this regard. That is not to say you couldn't modify your code to work just as well, you could. But as it stands, the code in this answer has that limitation.

More general version

Here, as a function, this gathers all matching coordinates and returns them as a list of (x,y) tuples:

import numpy as np
from PIL import Image

im_haystack = Image.open(r"lenna.png")
im_needle   = Image.open(r"eye.png")

def find_matches(haystack, needle):
    arr_h = np.asarray(haystack)
    arr_n = np.asarray(needle)

    y_h, x_h = arr_h.shape[:2]
    y_n, x_n = arr_n.shape[:2]

    xstop = x_h - x_n + 1
    ystop = y_h - y_n + 1

    matches = []
    for xmin in range(0, xstop):
        for ymin in range(0, ystop):
            xmax = xmin + x_n
            ymax = ymin + y_n

            arr_s = arr_h[ymin:ymax, xmin:xmax]     # Extract subimage
            arr_t = (arr_s == arr_n)                # Create test matrix
            if arr_t.all():                         # Only consider exact matches
                matches.append((xmin,ymin))

    return matches

print(find_matches(im_haystack, im_needle))

Update:

Given the images you provided, you'll notice that the way the matching is set up, it will only match one of the two here's. The top-left one is one the one you cropped for the needle image, so it matches exactly. The bottom-right image would need to match exactly pixel-for-pixel. With this implementation, even a single bit difference in one of the color channels would cause it to be ignored.

As it turns out, the two here's vary quite a bit more: enter image description here

Tested Versions:

  • Python 2.7.10, Numpy 1.11.1, PIL (Pillow) 1.1.7
  • Python 3.5.0, Numpy 1.10.4, PIL (Pillow) 1.1.7
6
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    Nice Image!!!! but i get AttributeError: 'bool' object has no attribute 'all'.... and then on first comparison i get false as value of test.then this error comes...i am using png
    – vks
    Commented Jul 20, 2016 at 7:30
  • Are you using the exact same code as I listed? What version of python / PIL / numpy are you using? I'll try to match it.
    – jedwards
    Commented Jul 20, 2016 at 7:36
  • @vks, I corrected the issue and the example code now works for the two images you provided. Do note, however, that it will only detect one of the "here"s. The second here is not an exact pixel-for-pixel match.
    – jedwards
    Commented Jul 20, 2016 at 9:22
  • I corrected my code and it works fine but why is the second not same?it looks exactly same doesnt it?
    – vks
    Commented Jul 20, 2016 at 9:23
  • As a rule of why it wouldn't match, it's because at least one of the pixels differs. In fact, it's many more.
    – jedwards
    Commented Jul 20, 2016 at 9:36
4

Here is the code using OpenCV that I research and found online.

import datetime

import numpy as np
from PIL import Image
import cv2

# sourceImagePath = ".\\COCImages\\ErrorPrompts\\AnyoneThere.png"
# needleImagePath = ".\\COCImages\\ErrorPrompts\\AnyoneThere_Search.png"
sourceImagePath = ".\\COCImages\\ErrorPrompts\\main.png"
needleImagePath = ".\\COCImages\\ErrorPrompts\\eye.png"


def start_findNeedle(sourceImagePath, needleImagePath):
    startTime = datetime.datetime.now()
    image = cv2.imread(sourceImagePath)
    # cv2.imshow('Rainforest', image)
    # cv2.waitKey(0)
    gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)

    template = cv2.imread(needleImagePath, 0)

    result = cv2.matchTemplate(gray, template, cv2.TM_SQDIFF)
    min_val, max_val, min_loc, max_loc = cv2.minMaxLoc(result)
    print("min_val: ", min_val)
    print("min_loc: ", min_loc)

    height, width = template.shape[:2]

    top_left = max_loc
    bottom_right = (top_left[0] + width, top_left[1] + height)
    cv2.rectangle(image, top_left, bottom_right, (0, 0, 255), 5)

    # cv2.imshow('Rainforest', image)
    # cv2.waitKey(0)
    # cv2.destroyAllWindows()

    endTime = datetime.datetime.now()
    totalTimeNeeded = endTime - startTime
    print("totalTimeNeeded: ", totalTimeNeeded)

    return min_loc

start_findNeedle(sourceImagePath, needleImagePath)

It runs faster than the previous code provide by @jedwards: Main Image (click here), Eye Image (click here)

Old version consumes 0:00:01.499806

My version consumes 0:00:00.009005

Note: there will a little warning shown which is libpng warning: iCCP: known incorrect sRGB profile because the image is not comes with sRGB format.

Hope this can solve and improve the time efficiency.

ADD ON : You may test with these two images too to check out the time efficiency

Main Image Main Image

Eye Image Eye Image

Update:

  1. changed from cv2.TM_CCOEFF to cv2.TM_SQDIFF, which is more accurate.

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