# Help converting a transformation matrix

**URL:** <https://discourse.julialang.org/t/help-converting-a-transformation-matrix/46095>\
**Category:** General Usage\
**Tags:** question, matlab\
**Created:** [September 5, 2020, 9:46am UTC](https://discourse.julialang.org/t/help-converting-a-transformation-matrix/46095 "2020-09-05T09:46:00Z")\
**Posts on this page:** 6\
**Page:** 1

<div class="post-metadata">

**Author:** ![yakir12](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/yakir12/32/297_2.png) [@yakir12](https://discourse.julialang.org/u/yakir12)\
**Post date:** [September 5, 2020, 9:46am UTC](https://discourse.julialang.org/t/help-converting-a-transformation-matrix/46095/1 "2020-09-05T09:46:01Z")

</div>

For lack of a better alternative, I’m using Matlab’s image calibration toolbox. One of the things I’m after is the transformation matrix Matlab’s `fitgeotrans` outputs:

```matlab
tform = fitgeotrans(imagePoints, worldPoints, 'projective');

```

In `tform` is a 3x3 matirx, `tform.T`, which I think is a transformation matrix. **I’d like to use this `T` in Julia**.

I “found out” that applying Matlab’s `transformPointsForward` with `tform`:

```matlab
[worldPoints_X, worldPoints_Y] = transformPointsForward(tform, imagePoints_X, imagePoints_Y);

```

has the exact same results when using `T` in Julia:

```julia
worldPoint_XY = [imagePoint_X imagePoint_Y 1.0]*T

```

where `imagePoint_X` and `imagePoint_Y` is the image coordinate and `worldPoint_XY` now contains the world coordinate.

My question is: **How can I convert this `tform` to a transformation in `CoordinateTransformations.jl`?**

One of my goals is to use the `CoordinateTransformations.Transformation` with `ImageTransformations.warp` as well as transforming image coordinates to world coordinates.

Thanks in advance!

---

<div class="post-metadata">

**Author:** ![touste](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/touste/32/10758_2.png) [@touste](https://discourse.julialang.org/u/touste)\
**Post date:** [September 7, 2020, 8:06am UTC](https://discourse.julialang.org/t/help-converting-a-transformation-matrix/46095/2 "2020-09-07T08:06:56Z")

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Hi, looking at Matlab’s documentation:

Affine2D:  
Forward 2-D affine transformation, specified as a nonsingular 3-by-3 numeric matrix.  
The matrix `T` uses the convention:  
`[x y 1] = [u v 1] * T`  
where `T` has the form:  
`[a b 0; c d 0; e f 1];`

`T` is composed of a 2D rotation:  
`rot = LinearMap(T[1:2,1:2]')`  
and a translation:  
`trans = Translation(T[3,1:2])`  
such as:  
`tform = trans ∘ rot`

Note: `T[1:2,1:2]'` is used because CoordinateTransformations uses the other convention:  
`[x,y] = T*[u,v]`

`tform` can now be applied to imagePoints:  
`worldPoints_XY = [tform(p) for p in eachrow(imagePoints)]`

Note: for better performance, it is recommended to use StaticArrays for representing these small vectors.

---

<div class="post-metadata">

**Author:** ![yakir12](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/yakir12/32/297_2.png) [@yakir12](https://discourse.julialang.org/u/yakir12)\
**Post date:** [September 7, 2020, 8:19am UTC](https://discourse.julialang.org/t/help-converting-a-transformation-matrix/46095/3 "2020-09-07T08:19:59Z")

</div>

Awesome! I’ll try it out! Thank you.

---

<div class="post-metadata">

**Author:** ![yakir12](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/yakir12/32/297_2.png) [@yakir12](https://discourse.julialang.org/u/yakir12)\
**Post date:** [September 10, 2020, 12:23pm UTC](https://discourse.julialang.org/t/help-converting-a-transformation-matrix/46095/4 "2020-09-10T12:23:13Z")

</div>

I’m running into problems because the third column of `T` is not exactly `0,0,1`:

```julia
julia> tform
3×3 Array{Float64,2}:
    0.371599 -0.0328841 0.000191534
    0.0463963 0.368681 0.000134089
 -285.262 -96.424 1.0

```

This is admittedly odd and “out of reach” because it’s coming from Matlab’s `fitgeotrans`, but I thought it’s worth asking… Any idea what I can do about it?

---

<div class="post-metadata">

**Author:** ![yakir12](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/yakir12/32/297_2.png) [@yakir12](https://discourse.julialang.org/u/yakir12)\
**Post date:** [September 10, 2020, 2:42pm UTC](https://discourse.julialang.org/t/help-converting-a-transformation-matrix/46095/5 "2020-09-10T14:42:25Z")

</div>

I’ve tracked down Matlab’s implementation for these transformation matrices. Here it is:

```matlab
imagePoints = rand(10,2);
worldPoints = rand(10,2);
tform = fitgeotrans(imagePoints, worldPoints, 'projective'); % create the transformation matrix

Uorg = [1, 2; 3, 4]; % some coordinates we want to transform
V1 = transformPointsForward(tform, Uorg) % Matlab's own function for transforming these coordinates

%% the following is what's happening in the `transformPointsForward` function:
U = padarray(Uorg,[0 1],1,'post') # add a third column of ones
X = U*tform.T % linear algebra
m = repmat(X(:,3),[1 2]) % these are residuals we need to normalize with..?
X(:,1:2) = X(:,1:2)./m % normalizing? 
V2 = X(:,1:2) % this is equal to `V1`

```

I think I failed to recognize `T` (from before) as a projective transformation matrix (and not an affine one). It is therefore not enough to include just translation and rotation (and within rotation, scale too). We need another term I think? I strongly suspect that the answer is in [https://github.com/JuliaGeometry/CoordinateTransformations.jl#perspective-transformations](https://github.com/JuliaGeometry/CoordinateTransformations.jl#perspective-transformations). I’m just not sure how exactly I can decompose this `T` to a `PerspectiveMap() ∘ inv(AffineMap(...))`.

Appreciate any input you might have!

---

<div class="post-metadata">

**Author:** ![yakir12](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/yakir12/32/297_2.png) [@yakir12](https://discourse.julialang.org/u/yakir12)\
**Post date:** [September 11, 2020, 11:48am UTC](https://discourse.julialang.org/t/help-converting-a-transformation-matrix/46095/6 "2020-09-11T11:48:31Z")

</div>

OK!  
I got it figured out, it’s of course much simpler than I thought. My original goal was to convert matlab’s transformation matrix to a `CoordinateTransformations` one. I can do that without decomposing them into translation, rotation, scale, and shear (or what not). I can simply do:

```julia
M = LinearMap(SMatrix{3,3, Float64}(T'))
tform = PerspectiveMap() ∘ M ∘ push1

```

where `T` is matlab’s transformation matrix (and therefore needs a transpose, `'`), and `push1` is simply:

```julia
push1(x) = CoordinateTransformations.push(x, 1)

```

All of this is from Tim Holy’s answer on SO to a related question ([Perspective warp an image in Julia - Stack Overflow](https://stackoverflow.com/a/45782741/2261957)). Thank you @tim.holy and @touste !

So now, the coordinate transformation in Julia , `tform` from above, generates the same results as Matlab’s `T` when applied to 2D coordinates. Yay! 🎆

But I’m still stuck on one irritating detail: When I apply this projective transformation matrix to an image with `ImageTransformations.warp` I run into problems related to the fact that the transformation `tform` is not invertible. The only way I managed to solve this (I think) is:

```julia
wimg = warp(img, itform, ImageTransformations.autorange(img, tform))

```

where `itform` is simply:

```julia
itform = PerspectiveMap() ∘ inv(M) ∘ push1

```

and to plot it (with Makie):

```julia
image!(ax, ImageTransformations.autorange(img, tform)..., parent(wimg))

```

where `ax` is a axis from `MakieLayout`.

This seems some what convoluted, which is totally fine, but I’m just worried I messed it up in some way…

I’ll also flag for future readers the issue of the convention about where the origin is in an image. In Matlab it’s in the top left corner of the image, so when you detect the checkerboard corners with `detectCheckerboardPoints` you get `xy` coordinates that are plottable (they plot correctly on the image) but cannot be used as indices (you will not find the corner in the `x` column and `y` row, rather `ImageHeight - y`). This complicates things even further… I think @tim.holy solved this ambiguity in Julia with [ImageAxes.jl](https://juliaimages.org/latest/pkgs/axes/#ImageAxes.jl-1).