Today we are going to kick-off a three part series on **calculating the size of objects in images** along with **measuring the distances between them**.

These tutorials have been some of the most *heavily requested *lessons on the PyImageSearch blog. I’m super excited to get them underway — *and I’m sure you are too.*

However, before we start learning how to measure the size (and not to mention, the distance between) objects in images, we first need to talk about something…

A little over a year ago, I wrote one my favorite tutorials on the PyImageSearch blog: **How to build a kick-ass mobile document scanner in just 5 minutes**. Even though this tutorial is over a year old, its *still* one of the most popular blog posts on PyImageSearch.

Building our mobile document scanner was predicated on our ability to **apply a 4 point cv2.getPerspectiveTransform with OpenCV**, enabling us to obtain a top-down, birds-eye-view of our document.

However. Our perspective transform has a * deadly flaw* that makes it unsuitable for use in production environments.

You see, there are cases where the pre-processing step of arranging our four points in top-left, top-right, bottom-right, and bottom-left order can return* incorrect results!*

**To learn more about this bug, and how to squash it, keep reading.**

Looking for the source code to this post?

Jump right to the downloads section.

## Ordering coordinates clockwise with Python and OpenCV

The goal of this blog post is two-fold:

- The
**primary purpose**is to learn how to arrange the*(x, y)*-coordinates associated with a rotated bounding box in top-left, top-right, bottom-right, and bottom-left order. Organizing bounding box coordinates in such an order is a prerequisite to performing operations such as perspective transforms or matching corners of objects (such as when we compute the distance between objects). - The
**secondary purpose**is to address a subtle, hard-to-find bug in the order_points method of the imutils package. By resolving this bug, our order_points function will no longer be susceptible to a debilitating bug.

All that said, let’s get this blog post started by reviewing the original, flawed method at ordering our bounding box coordinates in clockwise order.

### The original (flawed) method

Before we can learn how to arrange a set of bounding box coordinates in (1) clockwise order and more specifically, (2) a top-left, top-right, bottom-right, and bottom-left order, we should first review the order_points method detailed in the original 4 point getPerspectiveTransform blog post.

I have renamed the (flawed) order_points method to order_points_old so we can compare our original and updated methods. To get started, open up a new file and name it order_coordinates.py :

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 |
# import the necessary packages from __future__ import print_function from imutils import perspective from imutils import contours import numpy as np import argparse import imutils import cv2 def order_points_old(pts): # initialize a list of coordinates that will be ordered # such that the first entry in the list is the top-left, # the second entry is the top-right, the third is the # bottom-right, and the fourth is the bottom-left rect = np.zeros((4, 2), dtype="float32") # the top-left point will have the smallest sum, whereas # the bottom-right point will have the largest sum s = pts.sum(axis=1) rect[0] = pts[np.argmin(s)] rect[2] = pts[np.argmax(s)] # now, compute the difference between the points, the # top-right point will have the smallest difference, # whereas the bottom-left will have the largest difference diff = np.diff(pts, axis=1) rect[1] = pts[np.argmin(diff)] rect[3] = pts[np.argmax(diff)] # return the ordered coordinates return rect |

**Lines 2-8** handle importing our required Python packages for this example. We’ll be using the
imutils package later in this blog post, so if you don’t already have it installed, be sure to install it via
pip :

1 |
$ pip install imutils |

Otherwise, if you *do* have
imutils installed, you should upgrade to the latest version (which has the updated
order_points implementation):

1 |
$ pip install --upgrade imutils |

**Line 10** defines our
order_points_old function. This method requires only a single argument, the set of points that we are going to arrange in top-left, top-right, bottom-right, and bottom-left order; although, as we’ll see, this method has some flaws.

We start on **Line 15** by defining a NumPy array with shape
(4, 2) which will be used to store our set of four *(x, y)*-coordinates.

Given these
pts , we add the *x* and *y* values together, followed by finding the smallest and largest sums (**Lines 19-21**). These values give us our top-left and bottom-right coordinates, respectively.

We then take the difference between the *x* and *y* values, where the top-right point will have the smallest difference and the bottom-left will have the largest distance (**Lines 26-28**).

Finally, **Line 31** returns our ordered *(x, y)*-coordinates to our calling function.

So all that said, can you spot the flaw in our logic?

I’ll give you a hint:

**What happens when the sum or difference of the two points is the ****same ?**

In short, tragedy.

If either the sum array s or the difference array diff have the same values, we are at risk of choosing the incorrect index, which causes a cascade affect on our ordering.

Selecting the wrong index implies that we chose the incorrect point from our pts list. And if we take the incorrect point from pts , then our clockwise top-left, top-right, bottom-right, bottom-left ordering will be be destroyed.

So how can we address this problem and ensure that it doesn’t happen?

To handle this problem, we need to devise a better order_points function using more sound mathematic principles. And that’s exactly what we’ll cover in the next section.

### A better method to order coordinates clockwise with OpenCV and Python

Now that we have looked at a *flawed* version of our
order_points function, let’s review an *updated, correct* implementation.

The implementation of the order_points function we are about to review can be found in the imutils package; specifically, in the perspective.py file. I’ve included the exact implementation in this blog post as a matter of completeness:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 |
# import the necessary packages from scipy.spatial import distance as dist import numpy as np import cv2 def order_points(pts): # sort the points based on their x-coordinates xSorted = pts[np.argsort(pts[:, 0]), :] # grab the left-most and right-most points from the sorted # x-roodinate points leftMost = xSorted[:2, :] rightMost = xSorted[2:, :] # now, sort the left-most coordinates according to their # y-coordinates so we can grab the top-left and bottom-left # points, respectively leftMost = leftMost[np.argsort(leftMost[:, 1]), :] (tl, bl) = leftMost # now that we have the top-left coordinate, use it as an # anchor to calculate the Euclidean distance between the # top-left and right-most points; by the Pythagorean # theorem, the point with the largest distance will be # our bottom-right point D = dist.cdist(tl[np.newaxis], rightMost, "euclidean")[0] (br, tr) = rightMost[np.argsort(D)[::-1], :] # return the coordinates in top-left, top-right, # bottom-right, and bottom-left order return np.array([tl, tr, br, bl], dtype="float32") |

Again, we start off on **Lines 2-4** by importing our required Python packages. We then define our
order_points function on **Line 6** which requires only a single parameter — the list of
pts that we want to order.

**Line 8** then sorts these
pts based on their *x-*values. Given the sorted
xSorted list, we apply array slicing to grab the two left-most points along with the two right-most points (**Lines 12 and 13**).

The
leftMost points will thus correspond to the *top-left* and *bottom-left* points while
rightMost will be our *top-right* and *bottom-right* points — **the trick is to figure out which is which.**

Luckily, this isn’t too challenging.

If we sort our
leftMost points according to their *y-*value, we can derive the top-left and bottom-left points, respectively (**Lines 18 and 19**).

Then, to determine the bottom-right and bottom-left points, we can apply a bit of geometry.

Using the top-left point as an anchor, we can apply the Pythagorean theorem and compute the Euclidean distance between the top-left and rightMost points. By the definition of a triangle, the hypotenuse will be the largest side of a right-angled triangle.

Thus, by taking the top-left point as our anchor, the bottom-right point will have the largest Euclidean distance, allowing us to extract the bottom-right and top-right points (**Lines 26 and 27**).

Finally, **Line 31** returns a NumPy array representing our ordered bounding box coordinates in top-left, top-right, bottom-right, and bottom-left order.

### Testing our coordinate ordering implementations

Now that we have both the *original* and *updated* versions of
order_points , let’s continue the implementation of our
order_coordinates.py script and give them both a try:

33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 |
# construct the argument parse and parse the arguments ap = argparse.ArgumentParser() ap.add_argument("-n", "--new", type=int, default=-1, help="whether or not the new order points should should be used") args = vars(ap.parse_args()) # load our input image, convert it to grayscale, and blur it slightly image = cv2.imread("example.png") gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY) gray = cv2.GaussianBlur(gray, (7, 7), 0) # perform edge detection, then perform a dilation + erosion to # close gaps in between object edges edged = cv2.Canny(gray, 50, 100) edged = cv2.dilate(edged, None, iterations=1) edged = cv2.erode(edged, None, iterations=1) |

**Lines 33-37** handle parsing our command line arguments. We only need a single argument,
--new , which is used to indicate whether or not the *new* or the *original*
order_points function should be used. We’ll default to using the *original* implementation.

From there, we load example.png from disk and perform a bit of pre-processing by converting the image to grayscale and smoothing it with a Gaussian filter.

We continue to process our image by applying the Canny edge detector, followed by a dilation + erosion to close any gaps between outlines in the edge map.

After performing the edge detection process, our image should look like this:

As you can see, we have been able to determine the outlines/contours of the objects in the image.

Now that we have the outlines of the edge map, we can apply the
cv2.findContours function to actually *extract* the outlines of the objects:

50 51 52 53 54 55 56 57 58 |
# find contours in the edge map cnts = cv2.findContours(edged.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE) cnts = imutils.grab_contours(cnts) # sort the contours from left-to-right and initialize the bounding box # point colors (cnts, _) = contours.sort_contours(cnts) colors = ((0, 0, 255), (240, 0, 159), (255, 0, 0), (255, 255, 0)) |

We then sort the object contours from left-to-right, which isn’t a requirement, but makes it easier to view the output of our script.

The next step is to loop over each of the contours individually:

60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 |
# loop over the contours individually for (i, c) in enumerate(cnts): # if the contour is not sufficiently large, ignore it if cv2.contourArea(c) < 100: continue # compute the rotated bounding box of the contour, then # draw the contours box = cv2.minAreaRect(c) box = cv2.cv.BoxPoints(box) if imutils.is_cv2() else cv2.boxPoints(box) box = np.array(box, dtype="int") cv2.drawContours(image, [box], -1, (0, 255, 0), 2) # show the original coordinates print("Object #{}:".format(i + 1)) print(box) |

**Line 61** starts looping over our contours. If a contour is not sufficiently large (due to “noise” in the edge detection process), we discard the contour region (**Lines 63 and 64**).

Otherwise, **Lines 68-71** handle computing the rotated bounding box of the contour (taking care to use
cv2.cv.BoxPoints [if we are using OpenCV 2.4] or
cv2.boxPoints [if we are using OpenCV 3]) and drawing the contour on the
image .

We’ll also print the original rotated bounding box so we can compare the results after we order the coordinates.

We are now ready to order our bounding box coordinates in a clockwise arrangement:

77 78 79 80 81 82 83 84 85 86 87 88 89 90 |
# order the points in the contour such that they appear # in top-left, top-right, bottom-right, and bottom-left # order, then draw the outline of the rotated bounding # box rect = order_points_old(box) # check to see if the new method should be used for # ordering the coordinates if args["new"] > 0: rect = perspective.order_points(box) # show the re-ordered coordinates print(rect.astype("int")) print("") |

**Line 81** applies the *original* (i.e., flawed)
order_points_old function to arrange our bounding box coordinates in top-left, top-right, bottom-right, and bottom-left order.

If the
--new 1 flag has been passed to our script, then we’ll apply our *updated*
order_points function (**Lines 85 and 86**).

Just like we printed the *original bounding box* to our console, we’ll also print the *ordered points* so we can ensure our function is working properly.

Finally, we can visualize our results:

92 93 94 95 96 97 98 99 100 101 102 103 |
# loop over the original points and draw them for ((x, y), color) in zip(rect, colors): cv2.circle(image, (int(x), int(y)), 5, color, -1) # draw the object num at the top-left corner cv2.putText(image, "Object #{}".format(i + 1), (int(rect[0][0] - 15), int(rect[0][1] - 15)), cv2.FONT_HERSHEY_SIMPLEX, 0.55, (255, 255, 255), 2) # show the image cv2.imshow("Image", image) cv2.waitKey(0) |

We start looping over our (hopefully) ordered coordinates on **Line 93** and draw them on our
image .

According to the
colors list, the top-left point should be *red*, the top-right point *purple*, the bottom-right point *blue*, and finally, the bottom-left point *teal*.

Lastly, **Lines 97-103** draw the object number on our
image and display the output result.

To execute our script using the *original, flawed* implementation, just issue the following command:

1 |
$ python order_coordinates.py |

As we can see, our output is anticipated with the points ordered clockwise in a top-left, top-right, bottom-right, and bottom-left arrangement — **except for Object #6!**

**Note:*** Take a look at the output circles — notice how there isn’t a blue one?*

Looking at our terminal output for Object #6, we can see why:

Taking the sum of these coordinates we end up with:

- 520 + 255 = 775
- 491 + 226 = 717
- 520 + 197 = 717
- 549 + 226 = 775

While the difference gives us:

- 520 – 255 = 265
- 491 – 226 = 265
- 520 – 197 = 323
- 549 – 226 = 323

As you can see, **we end up with duplicate values!**

And since there are duplicate values, the argmin() and argmax() functions don’t work as we expect them to, giving us an incorrect set of “ordered” coordinates.

To resolve this issue, we can use our updated order_points function in the imutils package. We can verify that our updated function is working properly by issuing the following command:

1 |
$ python order_coordinates.py --new 1 |

This time, all of our points are ordered correctly, including Object #6:

When utilizing perspective transforms (or any other project that requires ordered coordinates), *make sure you use our updated implementation!*

## Summary

In this blog post, we started a three part series on *calculating the size of objects in images* and *measuring the distance between objects*. To accomplish these goals, we’ll need to order the 4 points associated with the rotated bounding box of each object.

We’ve already implemented such a function in a previous blog post; however, as we discovered, this implementation has a fatal flaw — it can return the *wrong coordinates* under *very specific* situations.

To resolve this problem, we defined a new, updated order_points function and placed it in the imutils package. This implementation ensures that our points are always ordered correctly.

Now that we can order our *(x, y)*-coordinates in a reliable manner, we can move on to *measuring the size of objects in an image*, which is exactly what I’ll be discussing in our next blog post.

**Be sure to signup for the PyImageSearch Newsletter by entering your email address in the form below — you won’t want to miss this series of posts!**

This is a great solution to the problem. I have run into this problem before and never was able to devise a reliable solution.

I’m glad the blog post helped, David! 🙂

Oh many thanks Adiran !! I knew you would help me out

I will try this and see if my results are better this time

Many Thanks again Adrian !!

Mahed

No problem Mahed!

Thanks for this post Adrian, but I’m a little confused about one aspect of the new algorithm.

In the new order_points function, lines 18 & 19 sort the left most coordinates in order of the y-value, which gives us the top-left & bottom-left points. That makes perfect sense to me.

My question is, why was it not just as simple for the right side points?

Rather than the Euclidean distance calculation forming the basis of the sort in lines 26 & 27, why could the right most points not have also just been sorted by their y-value to determine the top-right & bottom-right points?

I expect there is a scenario where that wont work (which is the reason for the more complicated solution), but I just cant think of what that scenario would be.

Yes, you are correct. However, after the previous bug from what looked like an innocent heuristic approach, I decided to go with the Euclidean distance, that way I always knew the bottom-right corner of the rectangle would be based on the properties of triangles (and therefore the correct corner chosen), rather than running into another scenario where the

`order_points`

function broke (and being left again to figure which heuristic broke it). Consider this more “preemptive strike” against any bugs that could arise.I think the y-coordinates order method is more stable. The Euclidean distance method will get wrong order when the object is a trapezoid. For examle, they points are (10,10),(30,10),(20,20),(10,20). The output is incorret when using Eudclidean distance method, but correct when using y-coordinates order method.

Output:

ordered by Euclidean distance

[[10 10]

[20 20]

[30 10]

[10 20]]

orderd by y-coordinates

[[10 10]

[30 10]

[20 20]

[10 20]]

For the test code, please go to gist: https://gist.github.com/flashlib/e8261539915426866ae910d55a3f9959

Fantastic as always 🙂

Thanks Arif! 🙂

Hi Adrian ,et. everyone ,

I used the code above as video feed on rasPi inorder to zoom in the image containing

a ‘white’ text embedded on a ‘red’ square, mounted on a drone as part of a project.

The algorithm works perfectly and the zoomed image appears perfectly upright text

even when image is slightly skewed ….. However when the square was turned

completely on the l.h.s or r.h.s …. The text appeared sideways too ….

The same situation was when the text was facing bottom

Unfortunately my text recognition software (pytesseract) couldnt read the text sideways/bottomways

There are other recog. engines that can deal with this but are not free

Is there a way i could modify the code so that my embedded text always upright.

I did give myself a thinking but could’nt go that far becoz i thought that for the case

if the image is completely sideways … i might say that the distance between top-left

and top right corner is less than what was before and hence rotate by 90` but the

thing is the case wont work for both situations … i.e. completley l.h.s and r.h.s

and i am absolutely clueless on how to solve when the text is facing bottom

Oh crisis !! I just realized my first solution method wont work as well because my

target is a red square not a rectangle ….. *facepalms*

is it possible to distinguish between real objects and floor lines ? I am trying to detect objects on a floor that has lines all over and I don’t know how to separate the real objects from rectangles/squares on the floor.

In most cases, yes, this should be possible. Using the Canny edge detector, you can determine lines that run most of the width/height of the image. Furthermore, these lines should also be equally spaced and in many cases intersecting. You can use this information to help prune out the floro lines that you are not interested in.

Useful post as always.

If I am reading this right, another special case that may need to be accounted for is skewed four-sided objects where the order of the x values may not reliably give you lefts and rights. Take for example an object with the points {(0,0), (2,0), (3,4), (5,4)}. The top right x is smaller than the bottom left x and the sort by x’s will result in top left and top right being identified as top left and bottom left respectively.

Hi Adrian,

Question on the circular items in the picture. Why is it that the bounding rectangle is upright (that is, not angled) for the two quarters, but is angled for the nickle?

It’s simply due to how the left-most and right-most coordinates are sorted. You can also see results like these if there is noise due to shadowing, lighting spaces, etc.

It almost seemed that the bounding rectangle detected the rotation angle of the coin (relative to the the head and neck of the President being upright). Just wanted to make sure this was coincidence and not a desired feature of the algorithm.

Yep, that’s a total coincidence.

You write

# now that we have the top-left coordinate, use it as an

# anchor to calculate the Euclidean distance between the

# top-left and right-most points; …

# … the point with the largest distance will be

# our bottom-right point

This may be true for images of quadrilaterals but it’s not generally true. Imagine a square oriented with its edges horizontal and vertical and now deform it by sliding the right-hand vertical image straight upwards so that we get a series of parallelograms. At some point it consists of two equilateral triangles stuck together and now the two right-hand points are equidistant from the top-left corner. From now on, the upper-right corner is further from the anchor than is the lower-right corner.

I discovered this the hard way while trying to find the grid-lines on a go board. There were a lot of false lines from diagonals and they created some very skewed parallelograms.

I think I should have written ‘images of rectangles’ instead of ‘images of quadrilaterals’

Addendum

I’m now using code that finds the top-left and bottom-left points by your method but then calculates the angle bl->tl->r for r in the rightMost points and assigns tr and br accordingly. It seems to work.

Thank you for sharing your insights David, I appreciate it.

This approach is not working for Video Frame object, as object from first frame appearing in second frame , its counting as second object. kindly help…

Hi Varsha — I’m not sure what you mean by “counting as second object”. Can you please elaborate?

I need to solve my programs like this but i need in C++ program.. can you give me a suggestion ? Please.. and thanks..

Hello — I only provide Python on this blog. Best of luck with the project!

Your article is very good but..

I can’t understand what line number 86, perspective. order_points(pts) do..

In perspective manner..

Please guide me..

Thanks for article.

Hi Umesh — I’m not sure what your question is. What specifically are you trying to understand with the

`order_points`

function?Hi, Adiran！

The blogpost is very useful for me, many thanks for sharing!

There is a little problem, I am processing a short video with this method, when there exists objects in video, it perfectly draws rectangles on objects, but when comes to blank (simple black color) part of the video, it gives me an error like:

in line : (cnts, _) = contours.sort_contours(cnts)

not enough values to unpack (expected 2, got 0)

The error seems like becouse of cannot find any contours in blank part of the video, how to i solve this? … I am a beginner on CV, thank you again!

I would need to see the full traceback of the error to determine the exact issue; however, I would suggest checking

`if len(cnts) == 0`

. If this is the case you can skip the frame since no contours can be found. Since you’re new to OpenCV and image processing I would definitely recommend working through Practical Python and OpenCV where I discuss the fundamentals of OpenCV in detail. By the time you work through the book you’ll be able to work through the majority of tutorials here on PyImageSearch with ease.i got an assignment to measure the length and height of the rice grain and i am new to this language and can’t understand anything but i am reading this site and it is really helpful but can you tell me how to do that?

Hey Waqar — take a look at this blog post where I share more information on measuring object sizes in images.

I find it doesn’t work in some case. The target is not Rect, it is Quadrilateral. For example, the four points are:[[ 96 263] [ 98 380] [100 382] [ 97 263]]. Do you know any solution for this case?

Thanks.

Take a look at my updated blog post.

Adrian thanks for this solution, really! But I was left with a question: Why can’t you reorder the rightmost points in respect to y-axis to get TR and BR just as you did for TL and BL ?

It creates a few edge cases, unfortunately. Refer to the comments section of the previous post.

My appreciation for such a detailed explanation, a great introduction to the CV.

Code works well while detecting a bounders, but where is an issue with a counting of objects for some reason.. Some of the digits get skipped, like #3, #5, #6 and #7 instead of 1, 2, 3, 4.

Any ideas why it could happed?

We’re not performing digit recognition in this tutorial so I’m not sure what you are referring to the digits being skipped. Could you please clarify?

Thank you for your crystal clear solutions.

I am using your new method for a application where my box is narrow and quite long. When this is rotated approx 45° clockwise happen that the most-left points are the two bottom ones: the result of new algorithm will give as tl point the bl one.

For this special condition the old algorithm works better but fails in the cases you already mentioned.

I cant figure out how could we fix it in order to be always reliable.

Any suggestion would be very much appreciated.

Hi,

I have some objects on a wooden textured background, took a photo of it and tried out this algorithm, it breaks at finding the contour. The contour includes the background texture as well in it, making tiny boxes, I tried around tweaking the contourArea size, but some of the objects have similar size of that some of the texture.

How can i completely remove the textures? any ideas?

Hey Hola — do you have any example images of what you’re working with? That would likely help me provide suggestions.

Hey Adrian! When I try to execute the program the following error occurs.

ipykernel_launcher.py: error: the following arguments are required: -i/–image, -w/–width

I am using jupyter notebook.

First make sure you read this tutorial on command line arguments so you can get up to speed on what they are and how they work.

From there you have two options:

1. Execute the script via command line instead of Jupyter notebook.

2. Update the “argparse” code per the recommendations in the argparse tutorial I linked you to.

HI Adrian, how can a measure the objects(height and widht) in an image knowing the distance of picamera to the object????

Make sure you follow this tutorial. Once you have the triangle similarity you can compute the width and height.

I’m not sure the euclidean method, as used, is fool-proof.

After the anchor point, let’s denote the two ordered, right-most points a and b. a will always be closer to the anchor than b as they have been ordered (a_x is smaller than b_x).

For a fixed value of x_y, there are two cases for y_b:

1- when y_b is smaller, (point a) will be both below point b and will be the bottom right-most even though it does not make the longest line from the anchor. Point b will make the longest line from the anchor.

2- when y_b is larger, (point b) will be below a, at the bottom, and will make the longest line from the anchor.

Fawzi Sdudah