Element-wise string concatenation in numpy

Question:

Is this a bug?

import numpy as np
a1=np.array(['a','b'])
a2=np.array(['E','F'])

In [20]: add(a1,a2)
Out[20]: NotImplemented

I am trying to do element-wise string concatenation. I thought Add() was the way to do it in numpy but obviously it is not working as expected.

Asked By: Dave31415

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Answers:

This can (and should) be done in pure Python, as numpy also uses the Python string manipulation functions internally:

>>> a1 = ['a','b']
>>> a2 = ['E','F']
>>> map(''.join, zip(a1, a2))
['aE', 'bF']
Answered By: Niklas B.

This can be done using numpy.core.defchararray.add. Here is an example:

>>> import numpy as np
>>> a1 = np.array(['a', 'b'])
>>> a2 = np.array(['E', 'F'])
>>> np.core.defchararray.add(a1, a2)
array(['aE', 'bF'], 
      dtype='<U2')

There are other useful string operations available for NumPy data types.

Answered By: Mike T

Another solution is to convert string arrays into arrays of python of objects so that str.add is called:

>>> import numpy as np
>>> a = np.array(['a', 'b', 'c', 'd'], dtype=np.object)   
>>> print a+a
array(['aa', 'bb', 'cc', 'dd'], dtype=object)

This is not that slow (less than twice as slow as adding integer arrays).

Answered By: jonathanrocher

You can use the chararray subclass to perform array operations with strings:

a1 = np.char.array(['a', 'b'])
a2 = np.char.array(['E', 'F'])

a1 + a2
#chararray(['aE', 'bF'], dtype='|S2')

another nice example:

b = np.array([2, 4])
a1*b
#chararray(['aa', 'bbbb'], dtype='|S4')
Answered By: Saullo G. P. Castro

One more basic, elegant and fast solution:

In [11]: np.array([x1 + x2 for x1,x2 in zip(a1,a2)])
Out[11]: array(['aE', 'bF'], dtype='<U2')

It is very fast for smaller arrays.

In [12]: %timeit np.array([x1 + x2 for x1,x2 in zip(a1,a2)])
3.67 µs ± 136 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [13]: %timeit np.core.defchararray.add(a1, a2)
6.27 µs ± 28.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [14]: %timeit np.char.array(a1) + np.char.array(a2)
22.1 µs ± 319 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

For larger arrays, time difference is not much.

In [15]: b1 = np.full(10000,'a')    
In [16]: b2 = np.full(10000,'b')    

In [189]: %timeit np.array([x1 + x2 for x1,x2 in zip(b1,b2)])
6.74 ms ± 66.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [188]: %timeit np.core.defchararray.add(b1, b2)
7.03 ms ± 419 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [187]: %timeit np.char.array(b1) + np.char.array(b2)
6.97 ms ± 284 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Answered By: Gaurav Singhal

Adding to Niklas B. answer as in later versions of Python this may have changed because as of Python 3.10 this will result in a map object.

To fix this you need to add the list function..

>>> a1 = ['a','b']
>>> a2 = ['E','F']
>>> list(map(''.join, zip(a1, a2)))  # <--- See here we have added list()
['aE', 'bF']
Answered By: DiegoV