Construct two dimensional numpy array from indices and values of a one dimensional array

Question

Say I've got

Y = np.array([2, 0, 1, 1])

From this I want to obtain a matrix X with shape (len(Y), 3). In this particular case, the first row of X should have a one on the second index and zero otherwhise. The second row of X should have a one on the 0 index and zero otherwise. To be explicit:

X = np.array([[0, 0, 1], [1, 0, 0], [0, 1, 0], [0, 1, 0]])

How do I produce this matrix? I started with

X = np.zeros((Y.shape[0], 3))

but then couldn't figure out how to populate/fill in the ones from the list of indices

As always, thanks for your time!

What is the question?

milancurcic
– milancurcic

2013-09-23 20:55:03 +00:00
Commented Sep 23, 2013 at 20:55 — milancurcic
– milancurcic, Commented Sep 23, 2013 at 20:55
Fair enough: edited to explicitly ask the question

cd98
– cd98

2013-09-23 21:04:15 +00:00
Commented Sep 23, 2013 at 21:04 — cd98
– cd98, Commented Sep 23, 2013 at 21:04

DSM · Accepted Answer · 2013-09-23 21:01:57Z

13

Maybe:

>>> Y = np.array([2, 0, 1, 1])
>>> X = np.zeros((len(Y), 3))
>>> X[np.arange(len(Y)), Y] = 1
>>> X
array([[ 0.,  0.,  1.],
       [ 1.,  0.,  0.],
       [ 0.,  1.,  0.],
       [ 0.,  1.,  0.]])

answered Sep 23, 2013 at 21:01

DSM

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Comments

Jaime · Accepted Answer · 2013-09-23 21:21:40Z

3

To give a one-liner alternative to DSM's perfectly good answer:

>>> Y = np.array([2, 0, 1, 1])
>>> np.arange(3) == Y[:, np.newaxis]
array([[False, False,  True],
       [ True, False, False],
       [False,  True, False],
       [False,  True, False]], dtype=bool)

answered Sep 23, 2013 at 21:21

Jaime

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2 Comments

DSM Over a year ago

I'd throw in a *1 or an .astype(int), but +1.

Jaime Over a year ago

@DSM Your code is already very significantly faster than this without the type conversion...

Joran Beasley · Accepted Answer · 2013-09-23 20:59:30Z

1

Y = np.array([2, 0, 1, 1])
new_array = np.zeros((len(Y),3))
for i in range(len(Y)):
    new_array[i,Y[i]] = 1

I think ... i dont think there is an easier way (but i might be wrong)

answered Sep 23, 2013 at 20:59

Joran Beasley

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3 Comments

Steven Rumbalski Over a year ago

np.zeros((len(Y), 3)). Also may want to specify dtype=np.int as example is showing integers.

Joran Beasley Over a year ago

DSM comes through with a better answer :P

cd98 Over a year ago

Thanks! In case anyone is interested, I timed @JoranBeasley's and DSM's answer with a 10000 by 3 array and the vectorising gives 744 µs per loop over 11.8 ms per loop

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