How to use numpy empty_like

Question

I've got the following array:

rxn_probability = [1. 0. 0.]

And I want to create another array, num_rxn, that has the same shape and size of rxn_probability, which contains a number of the reactions, so in this case num_rxn would be: [1, 2, 3]. Starting at 1 and increasing until it reaches the same size and shape of rxn_probability, so that if I change the size of rxn_probability the size and shape of num_rxn will automatically be changed.

So far I've tried:

num_rxn = np.array(range(len(rxn_probability + 1)))

(also tried using arange in a similar way)

But this outputs:

[0 1 2]

which isn't what I want because it doesn't start at 1 or end at 3.

I've been reading about numpy.empty_like but I'm not sure if that would be the best or right solution. Any ideas?

Cheers

How are you going to change the size of rxn_probability without creating a new object? — a_guest
– a_guest, Commented May 4, 2020 at 12:52

a_guest · Accepted Answer · 2020-05-04 12:53:39Z

1

You can use np.arange and then reshape to fit the shape of rxn_probability:

num_rxn = np.arange(1, rxn_probability.size + 1).reshape(rxn_probability.shape)

answered May 4, 2020 at 12:53

a_guest

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