Working with 2 arrays with different lengths Numpy Python

Question

Is there a way I could modify the function down below so that it could compute arrays with different length sizes. the length of Numbers array is 7 and the length of the Formating is 5. The code down below compares if any number in Formating is between two values and if it the case then it sums the values that are in between. So for the first calculation since no element in Numbers is between 0, 2 the result will be 0. Link to code was derived from: issue.

Code:

Numbers = np.array([3, 4, 5, 7, 8, 10,20])
Formating = np.array([0, 2 , 5, 12, 15])
x = np.sort(Numbers);
l = np.searchsorted(x, Formating, side='left')
mask=(Formating[:-1,None]<=Numbers)&(Numbers<Formating[1:,None])
N=Numbers[:,None].repeat(5,1).T
result= np.ma.masked_array(N,~mask)
result = result.filled(0)
result = np.sum(result, axis=1)

Expected output:

[ 0  7 30  0]

@hpaulj it doesn't run :-). Should have been N=Numbers[:,None].repeat(4,1).T. That said x and l are redudant, and wrong! — Quang Hoang
– Quang Hoang, Commented Mar 12, 2021 at 2:51
and all that result stuff seems overkilled. (mask * Numbers[None, :]).sum(axis=1) works fine. That said, this series of questions from OP look like an XY-problem. — Quang Hoang
– Quang Hoang, Commented Mar 12, 2021 at 2:55
x and l are from a different solution, mask is what I proposed. The use of 5 in N is left over from when Formatting had 6 terms; obviously (?) if Formating is changed this value should as well. I suggested the np.ma as a way of visualizing the use of mask. np.ma.sum would have taken care of the 0 fill. — hpaulj
– hpaulj, Commented Mar 12, 2021 at 3:27

Quang Hoang · Accepted Answer · 2021-03-12 02:07:47Z

1

Here's an approach with bincounts. Note that you have your x and l messed-up, and I recalled that you could/should use digitize:

# Formating goes here
x = np.sort(Formating);

# digitize
l = np.digitize(Numbers, x)

# output:
np.bincount(l, weights=Numbers)

Out:

array([ 0.,  0.,  7., 30.,  0., 20.])

answered Mar 12, 2021 at 2:07

Quang Hoang

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1 Comment

ege Selcuk Over a year ago

Is there a way i could modify the output to be between the set boundaries and result in [ 0 7 30 0]

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