CODEWITHSUNDEEP
pythonarraysnumpyindexing
Giggi
Giggi

Reputation: 720

Multiple slice in list indexing for numpy array

Numpy array admits a list of indices, for example

a = np.arange(1000)
l = list([1,44,66,33,90,345])
a[l] = 22

But this method don't work if we want to use a multiple slice indexing or indices plus a slice, for example.

a = np.arange(1000)
l = list([1,44,66,33,90, slice(200,300) , slice(500,600) ])
a[l] = 22

This code returns an error message:

IndexError: too many indices

My question is very simple: do you know if in numpy or scipy there exist an efficient method for using this kind of indexing?

Or what's a good and efficient way for using an indexing method like this?

Don't forget that the usage of slices produce a very fast code; and my problem is to have as faster as possible code.

Upvotes: 5

Views: 8697

Answers (2)

Henry Gomersall
Henry Gomersall

Reputation: 8692

You can use fancy indexing to build an index list.

l = numpy.array([1,44,66,33,90]+range(200,300)+range(500,600))
a[l] = 22

But as @Lev pointed out, this may not be any faster (though it almost certainly will be if you can precompute the index list).

However, fancy indexing applies per-axis. So you can fancy index on one axis, and slice the others, if that helps at all:

a = numpy.random.randn(4, 5, 6)
l = numpy.array([1, 2])
a[l, slice(None), slice(2, 4)] = 10

Upvotes: 0

Lev Levitsky
Lev Levitsky

Reputation: 65781

What comes to my mind:

a = np.arange(1000)
l = np.hstack(([1, 44, 66, 33, 90], np.arange(200, 300), np.arange(500, 600)))
a[l] = 22

I'm not sure if it's the simplest way, but it works.

Edit: you're right that this is slower than using slices; but you cannot create a slice object with arbitrary values. Maybe you should just do several assignments then:

%timeit a[np.hstack(([1, 44, 66, 33, 90], np.arange(200, 300), np.arange(500, 600)))] = 22
10000 loops, best of 3: 39.5 us per loop

%timeit a[[1, 44, 66, 33, 90]] = 22; a[200:300] = 22; a[500:600] = 22
100000 loops, best of 3: 18.4 us per loop

Upvotes: 4

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