Nick - 1 year ago 65

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!

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()
```