John Jaques - 8 months ago 61

Python Question

What is the best (elegant and efficient) way in Theano to convert a vector of indices to a matrix of zeros and ones, in which every row is the one-of-N representation of an index?

`v = t.ivector() # the vector of indices`

n = t.scalar() # the width of the matrix

convert = <your code here>

f = theano.function(inputs=[v, n], outputs=convert)

Example:

`n_val = 4`

v_val = [1,0,3]

f(v_val, n_val) = [[0,1,0,0],[1,0,0,0],[0,0,0,1]]

Answer

I didn't compare the different option, but you can also do it like this. It don't request extra memory.

```
import numpy as np
import theano
n_val = 4
v_val = np.asarray([1,0,3])
idx = theano.tensor.lvector()
z = theano.tensor.zeros((idx.shape[0], n_val))
one_hot = theano.tensor.set_subtensor(z[theano.tensor.arange(idx.shape[0]), idx], 1)
f = theano.function([idx], one_hot)
print f(v_val)[[ 0. 1. 0. 0.]
[ 1. 0. 0. 0.]
[ 0. 0. 0. 1.]]
```

Source (Stackoverflow)