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*Question edited/updated to add an example

Hi all! I have this a np.array. Based on the reference values of it, I want to update array b, which is my matrix. The "1st column" of a represents a code and the "2nd column" is my reference value. The matrix is populated with codes and I must replace them. See below the example.

import numpy as np
a = np.asarray([[0, 11], [1, 22], [2, 33]])
b = np.asarray([[0, 14, 12, 2], [1, 1, 7, 0], [0, 0,3,5], [1, 2, 2, 6]])

In other words: I want to replace the 0, 1, 2 values in "b" by 11, 22, 33, respectively.

Which is the best way to do that, considering that my real a array has +- 50 codes and my real b matrices have a shape of (850,850).

Thanks in advance!

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  • Could you add some example of expected output? Commented Dec 15, 2018 at 23:03
  • @RenardKorzeniowski question updated ! Thanks Commented Dec 16, 2018 at 10:10

2 Answers 2

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If I understand the question correctly, this example should show what you're asking for?

Assuming a is the matrix as you've listed above, and b is the list you want to write to

import numpy as np
a = np.asarray([[0, 10], [2, 30], [1, 40]])
b = np.zeros(3)
b[a[:, 0]] = a[:, 1]

where the [:, 0] is the index to be changed, and [:, 1] is what to populate it with

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Thanks! It works when arrays have same dimension, right? In my case the dimension differs so I'm still figuring out how to make it happen. I updated the question with one example. Thank you!
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If codes are not too long integers, You just have to build the correct lookup table :

lut = np.arange(b.max()+1)
k,v = a.T
lut[k] = v

For :

>>> b
[[ 0 14 12  2]
 [ 1  1  7  0]
 [ 0  0  3  5]
 [ 1  2  2  6]]

>>> lut[b]
[[11 14 12 33]
 [22 22  7 11]
 [11 11  3  5]
 [22 33 33  6]]

undefined codes are mapped to themselves,code=value.

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4 Comments

Great! It is working. You've said that the approach can be used for codes that are not too long integers. If they are, why is not a good idea? Is it because processing time or sth? In my case, the code range is 1-10000 but the values have around 12-16 digits
No 10000 << 850², so lut will be little in regard to your data.
One more question! My real data has is populated with a negative value that represents "no data" for my analysis . I've tested your solution with one example, but because of the inverted index my output matrix is not what I was expecting. Is there any way to determine that the negative value remains imutable? My example array: b = np.asarray([[0, 1, 2, 3, -9999], [0, 1, 2, 10000, -9999], [0, 1, 2, 3, -9999], [0, 1, 2, 3, -9999]])
Clearly this -9999 is a bad idea, there is always a day where it will be problematic. work around : lut = np.arange(b.max()+2); b[b==-9999]=b.max()+1; solve your problem;b[b==b.max()+1]=-9999

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