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For example, if I have the following arrays:

arraya = ['z', 'q']
arrayb = ['t', 'f']

I'd like to use the items from each list to create 'pairs' and get the following output:

['zt', 'zf', 'qt', 'qf']

I know I can figure out a jury-rigged way to write a function that produces this result, but a prebuilt function would be better for my purposes.

3
  • Multiplication is mathematically not defined in vector operations, which makes me doubt there is a pre-built method for this. It looks like you are going to have to brute force this method due to it's rather specific needs. Commented Jun 8, 2017 at 22:01
  • What application are you trying to use this for that makes you think of this operation as "multiplication"? Are you trying to implement polynomial multiplication or something? Commented Jun 8, 2017 at 22:05
  • Sorry, wrong word choice. Bad on me. I more correctly meant something along the lines of array item permutations, using the items of two different lists. I`ll update the question to reflect this. Commented Jun 8, 2017 at 22:11

2 Answers 2

1

This is as close to a built-in as you'll be able to get.

>>> import itertools
>>> [''.join(x) for x in itertools.product(arraya, arrayb)]
['zt', 'zf', 'qt', 'qf']

Find the cartesian product of the 2 lists/arrays and then concatenate the product.

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0

There is no builtin function for this, but you could use senderle's cartesian_product or pv's cartesian function. Which is faster may depend on your use case. Then the following yields the desired result:

In [40]: cartesian_product([['z', 'q'], ['t', 'f']]).ravel().view('<U2')
Out[40]: 
array(['zt', 'zf', 'qt', 'qf'], 
      dtype='<U2')

These functions can be faster than using itertools.product. For example,

In [181]: x, y = np.arange(500), np.arange(500)

In [185]: %timeit cartesian_product([x, y])
1000 loops, best of 3: 797 µs per loop

In [184]: %timeit cartesian_product2([x, y])
1000 loops, best of 3: 1.44 ms per loop

In [186]: %timeit cartesian([x, y])
100 loops, best of 3: 4.71 ms per loop

In [100]: %timeit np.array(list(IT.product(x, y)))
10 loops, best of 3: 112 ms per loop

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