Python - optimizing code ( looping through arrays )

Question

I have two arrays, A and B. Based on a function, I create a subset C from A and a subset D from B. Now, I want to join them together ( I want to keep those items that are in C and D with the same index in A and B).

A = np.array([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20])
B = np.array([20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1])
C = np.array([3,7,13])
D = np.array([18,8])

fvl1 = list()
fvl2 = list()


for i in C:
    index = np.where(A == i)
    if B[index] in D:
        fvl1.append(A[index])
        fvl2.append(B[index])

print(fvl1)
print(fvl2)

Output:
[array([3]), array([13])]
[array([18]), array([8])]

This is what I've done until now, but it's really slow. Is there any way to optimize this for it to go faster?

LE: You can see that the element 3 in A has the same index as element 18 in B(index = 3), same with element 13 from A and 8 from B.

Have you tried using sets instead of lists?

Jakob Lovern
– Jakob Lovern

2017-11-27 22:06:41 +00:00
Commented Nov 27, 2017 at 22:06 — Jakob Lovern
– Jakob Lovern, Commented Nov 27, 2017 at 22:06
MCVE please....

timgeb
– timgeb

2017-11-27 22:08:07 +00:00
Commented Nov 27, 2017 at 22:08 — timgeb
– timgeb, Commented Nov 27, 2017 at 22:08
edited example now

Barkz
– Barkz

2017-11-27 22:17:32 +00:00
Commented Nov 27, 2017 at 22:17 — Barkz
– Barkz, Commented Nov 27, 2017 at 22:17
Is A always sorted?

Divakar
– Divakar

2017-11-27 22:25:15 +00:00
Commented Nov 27, 2017 at 22:25 — Divakar
– Divakar, Commented Nov 27, 2017 at 22:25

Eric Duminil · Accepted Answer · 2017-11-27 22:24:13Z

3

You could use np.ind1d or np.isin:

Returns a boolean array of the same shape as element that is True where an element of element is in test_elements and False otherwise.

>>> import numpy as np
>>> A = np.array([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20])
>>> B = np.array([20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1])
>>> C = np.array([3,7,13])
>>> D = np.array([18,8])
>>> np.in1d(A, C) & np.in1d(B, D)
array([False, False,  True, False, False, False, False, False, False,
       False, False, False,  True, False, False, False, False, False,
       False, False], dtype=bool)
>>> indices = np.isin(A, C) & np.isin(B, D)
>>> A[indices]
array([ 3, 13])
>>> B[indices]
array([18,  8])

It's not clear if C and D are converted to sets, but I suppose it will be faster than your current code.

answered Nov 27, 2017 at 22:24

Eric Duminil

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Comments

Alex · Accepted Answer · 2017-11-27 23:40:14Z

I'm not 100% sure I interpreted your question correctly, but a combination of zip and list comprehension might be what you're after.

fv = [(a, b) for (a, b) in zip(A, B)
      if <condition for first list>
      and <condition for second list>]

fv1 = [t[0] for t in fv]
fv2 = [t[1] for t in fv]

zip joins your lists together into a list of tuples.

list(zip([1, 2, 3],['a','b','c'])) == [(1,'a'),(2,'b'),(3,'c')]

You can then filter which outputs you want with conditions in your list comprehension. For example:

fv = [(a, b) for (a, b) in zip(range(10),'abcdefghijklmnop') 
      if a%2 
      and b>'d']
[(5, 'f'), (7, 'h'), (9, 'j')]

Your case can actually be done as a one liner:

fv1, fv2 = zip(*((a, b) for (a, b) in zip(A, B) if a in C and b in D))
fv1 #(3, 13)
fv2 #(18, 8)

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