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pythonarraysnumpyreplacemultidimensional-array
Saullo G. P. Castro
Saullo G. P. Castro

Reputation: 58895

Change the values of a NumPy array that are NOT in a list of indices

I have a NumPy array like:

a = np.arange(30)

I know that I can replace the values located at positions indices=[2,3,4] using for instance fancy indexing:

a[indices] = 999

But how to replace the values at the positions that are not in indices? Would be something like below?

a[ not in indices ] = 888

Upvotes: 35

Views: 28143

Answers (4)

Alexey Trofimov
Alexey Trofimov

Reputation: 5007

Just overcome similar situation, solved this way:

a = np.arange(30)
indices=[2,3,4]

a[indices] = 999

not_in_indices = [x for x in range(len(a)) if x not in indices]

a[not_in_indices] = 888

Upvotes: 4

aaren
aaren

Reputation: 5675

Only works for 1d arrays:

a = np.arange(30)
indices = [2, 3, 4]

ia = np.indices(a.shape)

not_indices = np.setxor1d(ia, indices)
a[not_indices] = 888

Upvotes: 6

mgilson
mgilson

Reputation: 309929

I don't know of a clean way to do something like this:

mask = np.ones(a.shape,dtype=bool) #np.ones_like(a,dtype=bool)
mask[indices] = False
a[~mask] = 999
a[mask] = 888

Of course, if you prefer to use the numpy data-type, you could use dtype=np.bool_ -- There won't be any difference in the output. it's just a matter of preference really.

Upvotes: 50

kirelagin
kirelagin

Reputation: 13616

Obviously there is no general not operator for sets. Your choices are:

  1. Subtracting your indices set from a universal set of indices (depends on the shape of a), but that will be a bit difficult to implement and read.
  2. Some kind of iteration (probably the for-loop is your best bet since you definitely want to use the fact that your indices are sorted).

  3. Creating a new array filled with new value, and selectively copying indices from the old one.

    b = np.repeat(888, a.shape)
b[indices] = a[indices]

Upvotes: 4

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