I am not to figure out how to form a ndarray from existing numpy arrays. Suppose I have three arrays - index, distA, and distB. Now the I want to form a ndarray of these three arrays

What I have tried -

>>> indices = np.array([[5,7,4], [6,4,8]])
>>> distances = np.array([[0.2, 0.3, 0.4], [0.3, 0.4, 0.5]])
>>> np.column_stack((indices, distances))
array([[5. , 7. , 4. , 0.2, 0.3, 0.4],
   [6. , 4. , 8. , 0.3, 0.4, 0.5]])

But I want to form a table-like structure so that I can retrieve values in distA and distB.

5, 0.2, 0
7, 0.3, 0
4, 0.4, 0.4
6, 0, 0.3
8, 0, 0.5

Is it possible to do this with numpy?

  • 1
    $\begingroup$ I can't see how you aim to obtain a 5x3 matrix from two 2x3 arrays $\endgroup$
    – yatu
    Dec 11, 2018 at 8:55
  • $\begingroup$ @nixon Neither me, I was thinking there would be some to accomplish this. $\endgroup$ Dec 11, 2018 at 9:02
  • 1
    $\begingroup$ Its just that I can't see the logic on how to build this ndarray, can you better explain please? $\endgroup$
    – yatu
    Dec 11, 2018 at 9:15
  • $\begingroup$ There would not be a direct way to do this, but if we could do a union of indices and find the corresponding distance in a and in b, then using pandas dataframe we can build this. Not sure if this a good question to ask, still a noob. $\endgroup$ Dec 11, 2018 at 9:24

1 Answer 1


If your purpose is to deal with some data analysis from your arrays and have a nice view of it, try this using pandas:

import pandas as pd

df = pd.DataFrame({'distA' : np.array([0.2, 0.3, 0.4]),
                   'distB' : np.array([0.3, 0.4, 0.5])})


enter image description here

In the case you want to make computation with vectors, matrixes, tensors, etc .. NumPy is probably a good option. I didn't understand exactly what you've asked, though. Can you explain me what you're asking?

  • $\begingroup$ Yes, this is what I required, but it neither analysis nor computation on vectors, matrices, or tensors. It about the computation specific to distance calculation an efficient manner, and the algorithm I wrote, this formation was an integral part. Thanks. $\endgroup$ Dec 24, 2018 at 4:40

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