Nick - 1 year ago 88

Python Question

I tried to translate a piece of code from Matlab to Python and I'm running into some errors:

Matlab:

`function [beta] = linear_regression_train(traindata)`

y = traindata(:,1); %output

ind2 = find(y == 2);

ind3 = find(y == 3);

y(ind2) = -1;

y(ind3) = 1;

X = traindata(:,2:257); %X matrix,with size of 1389x256

beta = inv(X'*X)*X'*y;

Python:

`def linear_regression_train(traindata):`

y = traindata[:,0] # This is the output

ind2 = (labels==2).nonzero()

ind3 = (labels==3).nonzero()

y[ind2] = -1

y[ind3] = 1

X = traindata[ : , 1:256]

X_T = numpy.transpose(X)

beta = inv(X_T*X)*X_T*y

return beta

I am receiving an error: operands could not be broadcast together with shapes (257,0,1389) (1389,0,257) on the line where beta is calculated.

Any help is appreciated!

Thanks!

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Answer Source

The problem is that you are working with numpy arrays, not matrices as in MATLAB. Matrices, by default, do matrix mathematical operations. So `X*Y`

does a matrix multiplication of `X`

and `Y`

. With arrays, however, the default is to use element-by-element operations. So `X*Y`

multiplies each corresponding element of `X`

and `Y`

. This is the equivalent of MATLAB's `.*`

operation.

But just like how MATLAB's matrices can do element-by-element operations, Numpy's arrays can do matrix multiplication. So what you need to do is use numpy's matrix multiplication instead of its element-by-element multiplication. For Python 3.5 or higher (which is the version you should be using for this sort of work), that is just the `@`

operator. So your line becomes:

```
beta = inv(X_T @ X) @ X_T @ y
```

Or, better yet, you can use the simpler `.T`

transpose, which is the same as `np.transpose`

but much more concise (you can get rid of the `np.transpose line entirely):

```
beta = inv(X.T @ X) @ X.T @ y
```

For Python 3.4 or earlier, you will need to use `np.dot`

since those versions of python don't have the `@`

matrix multiplication operator:

```
beta = np.dot(np.dot(inv(np.dot(X.T, X)), X.T), y)
```

Numpy has a matrix object that uses matrix operations by default like the MATLAB matrix. *Do not use it!* It is slow, poorly-supported, and almost never what you really want. The Python community has standardized around arrays, so use those.

There may also be some issues with the dimensions of `traindata`

. For this to work properly then `traindata.ndim`

should be equal to `3`

. In order for `y`

and `X`

to be 2D, `traindata`

should be `3D`

.

This could be an issue if `traindata`

is 2D and you want `y`

to be MATLAB-style "vector" (what MATLAB calls "vectors" aren't really vectors). In numpy, using a single index like `traindata[:, 0]`

reduces the number of dimensions, while taking a slice like `traindata[:, :1]`

doesn't. So to keep `y`

2D when `traindata`

is 2D, just do a length-1 slice, `traindata[:, :1]`

. This is exactly the same values, but this keeps the same number of dimensions as `traindata`

.

**Notes**: Your code can be significantly simplified using logical indexing:

```
def linear_regression_train(traindata):
y = traindata[:, 0] # This is the output
y[labels == 2] = -1
y[labels == 3] = 1
X = traindata[:, 1:257]
return inv(X.T @ X) @ X.T @ y
return beta
```

Also, your slice is wrong when defining `X`

. Python slicing excludes the last value, so to get a 256 long slice you need to do `1:257`

, as I did above.

Finally, please keep in mind that modifications to arrays inside functions carry over outside the functions, and indexing does not make a copy. So your changes to `y`

(setting some values to `1`

and others to `-1`

), will affect `traindata`

outside of your function. If you want to avoid that, you need to make a copy before you make your changes:

```
y = traindata[:, 0].copy()
```

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