在python中从两个数组查找的最快方法

Question

I have three arrays of same size (can be very long upto 5000). I have to look up for pair of values (this will always be unique) in the first two arrays, for ex,(2,3) and accordingly fetch the value from third array at the same index. What is the fastest way to do this or any simple python inbuilt library present for this? The simplest solution for the problem is:

a = [1,1,1,2,2,2,3,3,3,4,4,4]
b = [2,3,4,3,4,5,7,3,2,1,8,9]
c = [4,5,6,13,4,8,80,4,2,3,7,11]
for i in range(0, len(a)):
  if a[i] == 2 and b[i] == 3:
    fetch = c[i]

Answer 1

Find the index and use it:

>>> c[zip(a, b).index((2, 3))]
13

Or prepare a dict to look the pair up:

>>> dict(zip(zip(a, b), c))[2, 3]
13

This would be faster if you wanted to look up many pairs, not just one. And you could use its get in case it's possible that the pair doesn't exist.

Answer 2

You could use a generator expression with next() and zip() over all the lists together:

>>> next((z for x, y, z in zip(a, b, c) if (x, y) == (2, 3)), 'None found')
13

Answer 3

Benchmarking different solutions, using lists of length 10000 similar to the shown ones and searching all existing pairs, using Python 2.7.14:

 2.380 seconds rajat      # The original
 1.712 seconds rajat2     #   Precomputed range
 1.843 seconds rajat3     #   xrange instead of range
 5.243 seconds stefan1    # zip(a, b).index
 0.954 seconds stefan1b   #   Precomputed zip(a, b).index
16.108 seconds stefan2    # dict
 0.002 seconds stefan2b   #   Precomputed dict
10.782 seconds chris      # next(generator)
 6.728 seconds chris2     #   bit optimized
 1.754 seconds chris3     #   Precomputed zip(a, b, c)

The code:

from timeit import timeit

b = range(100) * 100
a = sorted(b)
c = range(10000)

#-------------------------------------------------------------------------

def rajat(A, B):
    'The original'
    for i in range(0, len(a)):
        if a[i] == A and b[i] == B:
            return c[i]

def rajat2(A, B, r=range(0, len(a))):
    '  Precomputed range'
    for i in r:
        if a[i] == A and b[i] == B:
            return c[i]

def rajat3(A, B):
    '  xrange instead of range'
    for i in xrange(0, len(a)):
        if a[i] == A and b[i] == B:
            return c[i]

def stefan1(A, B):
    'zip(a, b).index'
    return c[zip(a, b).index((A, B))]

def stefan1b(A, B, index=zip(a, b).index):
    '  Precomputed zip(a, b).index'
    return c[index((A, B))]

def stefan2(A, B):
    'dict'
    return dict(zip(zip(a, b), c))[A, B]

def stefan2b(A, B, d=dict(zip(zip(a, b), c))):
    '  Precomputed dict'
    return d[A, B]

def chris(A, B):
    'next(generator)'
    return next((z for x, y, z in zip(a, b, c) if (x, y) == (A, B)), 'None found')

def chris2(A, B):
    '  bit optimized'
    return next((z for x, y, z in zip(a, b, c) if x == A and y == B), 'None found')

def chris3(A, B, abc=zip(a, b, c)):
    '  Precomputed zip(a, b, c)'
    return next((z for x, y, z in abc if x == A and y == B), 'None found')

#-------------------------------------------------------------------------

ab = zip(a, b)
def test():
    for A, B in ab:
        func(A, B)

for func in rajat, rajat2, rajat3, stefan1, stefan1b, stefan2, stefan2b, chris, chris2, chris3:
    t = timeit(test, number=1)
    print '%6.3f seconds %-10s # %s' % (t, func.__name__, func.__doc__)