How to merge 2 numpy arrays and concatenate their values?

Question

Is there a way to obtain the same result from this Python-arrays merge:

a = [1,2,3,4]

b = [4,3,2,1]

c = [ int(''.join (map (str, xs))) for xs in zip (a,b) ]

c
Out[4]: [14, 23, 32, 41]

But operating directly on Numpy-arrays:

a
Out[9]: array([1, 2, 3, 4])

b
Out[10]: array([4, 3, 2, 1])

c = Your Answer

c
# desired output: array([14, 23, 32, 41])

My first (and obvious) solution was:

c = np.array([int(''.join (map(str, xs))) for xs in zip(a.tolist(),b.tolist())])

c
Out[12]: array([14, 23, 32, 41])

But I would like to know if it's possible to do that directly with the numpy-arrays, without converting them to python-arrays.

Note: I use 1,2,3,4 values for simplification, I would like to have a solution that work with +double digits on both arrays of size > 10**4.

Updated with timings for different solutions:

a = np.arange(1000000)

b = np.arange(1,1000001)

#: Mi first Solution
%%timeit
c = np.array([int(''.join (map(str, xs))) for xs in zip(a.tolist(),b.tolist())])
1 loop, best of 3: 1.99 s per loop

#: Donkey's Solution (thought to smaller arrays)
%%timeit
c = np.char.add(a.astype(str),b.astype(str)).astype(int)
1 loop, best of 3: 1.8 s per loop

#: My second Solution
%%timeit
c = merge(a,b)
10 loops, best of 3: 128 ms per loop

#: Divakar's Solution
%%timeit
c = a*(10**(np.log10(b).astype(int)+1)) + b
10 loops, best of 3: 117 ms per loop

Verify results:

c1 = np.array([int(''.join (map(str, xs))) for xs in zip(a.tolist(),b.tolist())])

c2 = np.char.add(a.astype(str),b.astype(str)).astype(int)

c3 = merge(a,b)

np.alltrue(np.logical_and(c1==c2,c2==c3))
Out[51]: True


c4 = a*(10**(np.log10(b).astype(int)+1)) + b

np.alltrue(np.logical_and(c1==c2,c2==c4))
Out[58]: True

I've checked the module and found a quickly answer. Thanks Eelco. — David de la Iglesia
– David de la Iglesia, Commented Apr 18, 2016 at 17:54
Thanks for the comment Andrey, but that operation doesn't return the answer that I want when the 2 arrays contain double digits. Probably my fault the fact of not indicate that I wanted an answer apliccable to double digits also. — David de la Iglesia
– David de la Iglesia, Commented Apr 18, 2016 at 18:15

miradulo · Accepted Answer · 2016-04-18 18:05:43Z

4

You can use the dtype parameter to have your numpy arrays be string arrays, on which you can simply use the free function add in numpy.char to concatenate them string-wise, like so

a = numpy.array([1,2,3,4], dtype=numpy.str)
b = numpy.array([4,3,2,1], dtype=numpy.str)

c = numpy.char.add(a, b).astype(int)

Outputs:

[14 23 32 41]

edited Apr 18, 2016 at 18:05

answered Apr 18, 2016 at 17:54

miradulo

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3 Comments

David de la Iglesia Over a year ago

Thanks for the answer, I was just writing the same solution at the same time ^^

miradulo Over a year ago

@DaviddelaIglesia Cheers, glad I could help.

David de la Iglesia Over a year ago

I think I found a quite faster solution ^^

Divakar · Accepted Answer · 2016-04-18 19:23:54Z

2

Here's an approach using NumPy mathematical functions -

a*(10**(np.log10(b).astype(int)+1)) + b

Sample run -

In [32]: a
Out[32]: array([ 16,   2, 399,   4])

In [33]: b
Out[33]: array([  4,  38,   2, 190])

In [34]: a*(10**(np.log10(b).astype(int)+1)) + b
Out[34]: array([ 164,  238, 3992, 4190])

answered Apr 18, 2016 at 19:23

Divakar

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2 Comments

David de la Iglesia Over a year ago

Thanks divakar. I had founded about 10 minutes ago a similar approach. Just like yours but with support to 0s and numbers formed by only 9s.

Divakar Over a year ago

@DaviddelaIglesia Could you add this to your timings test in the question? Thanks!

David de la Iglesia · Accepted Answer · 2016-04-18 19:10:00Z

1

I write a function with a solution that I found after thinking a while:

def merge(a,b):
#: I don't find a better way to create this array
nines = np.array([9,99,999,9999,99999,999999,9999999, 99999999])

#: get number of digits
exp = np.log10(a)+1

#: fix the zeros
exp[exp == -np.inf] = 1

#: fix the nines
exp[np.in1d(a,nines)] += 1

c = a * 10**exp.astype(int) + b
return c

It might appear to be too much overthinking but it's a lot (x10) faster than the other solutions:

%%timeit
c = merge(a,b)
10 loops, best of 3: 128 ms per loop

edited Apr 18, 2016 at 19:10

answered Apr 18, 2016 at 17:53

David de la Iglesia

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3 Comments

miradulo Over a year ago

I don't think it is any faster, actually. Rather twice as slow. I'd double check your use of timeit.

miradulo Over a year ago

Ah, no now I see you've tested for 10 million elements. For a smaller case (like your example in your question) the solution I posted is twice as fast.

David de la Iglesia Over a year ago

Sorry about the misleading example, I was trying to simplificate the situation but I should especify wich what kind of arrays I was working at (big and +double digits). I updated the answer

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