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In a given array I want to replace the values by the index of this value in an other array (which doesn't contain duplicates). Here is a simple example of I'm trying to do:

import numpy as np
from copy import deepcopy

a = np.array([[0, 1, 2], [2, 1, 3], [0, 1, 3]])
chg = np.array([3, 0, 2, 1])

b = deepcopy(a)
for new, old  in enumerate(chg):
   b[a == old] = new

print b
# [[1 3 2] [2 3 0] [1 3 0]]

But I need to do that on large arrays so having an explicit loop is not acceptable in terms of execution time.

I can't figure out how to do that using fancy numpy functions...

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  • 1
    I dont understand what you are trying to do... Commented Oct 26, 2015 at 15:08
  • 1
    @JoranBeasley: To me it looks like he's taking the values in a, finding their index in chg and creating a new structure b that holds the indexes. Commented Oct 26, 2015 at 15:11
  • Also don't understand this. I don't think the code is doing what the OP expects. By the way you can just do b = a.copy()... Commented Oct 26, 2015 at 15:13
  • @YXD: Regarding using a.copy, I believe modifying the subarrays would modify the original array without deepcopy. Commented Oct 26, 2015 at 15:14
  • 1
    @StevenRumbalski it doesn't: a = np.arange(5); b = a.copy(); b[0] = 7; print(a) - a is not modified Commented Oct 26, 2015 at 15:16

4 Answers 4

Reset to default
2

take is your friend.

a = np.array([[0, 1, 2], [2, 1, 3], [0, 1, 3]])
chg = np.array([3, 0, 2, 1])
inverse_chg=chg.take(chg)
print(inverse_chg.take(a))

gives :

[[1 3 2]
 [2 3 0]
 [1 3 0]]

or more directly with fancy indexing: chg[chg][a], but inverse_chg.take(a) is three times faster.

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1 Comment

Cool use of take (+1) but this might break if chg = np.array([4, 0, 2, 1]), for example.
2

You can convert chg to a 3D array by adding two new axes at the end of it and then perform the matching comparison with a, which would bring in NumPy's broadcasting to give us a 3D mask. Next up, get the argmax on the mask along the first axis to simulate "b[a == old] = new". Finally, replace the ones that had no matches along that axis with the corresponding values in a. The implementation would look something like this -

mask = a == chg[:,None,None]
out = mask.argmax(0)
invalid_pos = ~mask.max(0)
out[invalid_pos] = a[invalid_pos]
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2

This type of replacement operation can be tricky to do in full generality with NumPy, although you could use searchsorted:

>>> s = np.argsort(chg)
>>> s[np.searchsorted(chg, a.ravel(), sorter=s).reshape(a.shape)]
array([[1, 3, 2],
       [2, 3, 0],
       [1, 3, 0]])

(Note: searchsorted doesn't just replace exact matches, so be careful if you have values in a that aren't in chg...)

pandas has a variety of tools which can make these operations on NumPy arrays much easier and potentially a lot quicker / more memory efficient for larger arrays. For this specific problem, pd.match could be used:

>>> pd.match(a.ravel(), chg).reshape(a.shape)
array([[1, 3, 2],
       [2, 3, 0],
       [1, 3, 0]])

This function also allows you to specify what value should be filled if a value is missing from chg.

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3 Comments

Good to see searchsorted! Guess not for cases like chg = np.array([4, 0, 2, 1])? But since OP accepted .take that assumes the same, maybe this would work for OP too.
@Divakar: thanks - this should actually work for the cases like chg = np.array([4, 0, 2, 1]), although I should probably caveat the answer by saying that searchsorted doesn't look for exact matches, so it might replace values in a that don't exist in chg.
Yeah well, if chg has all the values that exist in a, this might just be the fastest.
-2

Check this out:

a = np.array([3,4,1,2,0])
b = np.array([[0,0],[0,1],[0,2],[0,3],[0,4]])

c = b[a]
print(c)

It gives me back:

[[0 3]
 [0 4]
 [0 1]
 [0 2]
 [0 0]]

If you're working with numpy arrays you could do this.

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1 Comment

I'm not sure you understood what I'm trying to do... cause this is just a reordering of your array that you are doing.

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