How to create a numpy array of arbitrary length strings?

Question

I\'m a complete rookie to Python, but it seems like a given string is able to be (effectively) arbitrary length. i.e. you can take a string str and keeping add

Answer 1

You could use the object data type:

>>> import numpy
>>> s = numpy.array(['a', 'b', 'dude'], dtype='object')
>>> s[0] += 'bcdef'
>>> s
array([abcdef, b, dude], dtype=object)

Answer 2

You can do so by creating an array of dtype=object. If you try to assign a long string to a normal numpy array, it truncates the string:

>>> a = numpy.array(['apples', 'foobar', 'cowboy'])
>>> a[2] = 'bananas'
>>> a
array(['apples', 'foobar', 'banana'], 
      dtype='|S6')

But when you use dtype=object, you get an array of python object references. So you can have all the behaviors of python strings:

>>> a = numpy.array(['apples', 'foobar', 'cowboy'], dtype=object)
>>> a
array([apples, foobar, cowboy], dtype=object)
>>> a[2] = 'bananas'
>>> a
array([apples, foobar, bananas], dtype=object)

Indeed, because it's an array of objects, you can assign any kind of python object to the array:

>>> a[2] = {1:2, 3:4}
>>> a
array([apples, foobar, {1: 2, 3: 4}], dtype=object)

However, this undoes a lot of the benefits of using numpy, which is so fast because it works on large contiguous blocks of raw memory. Working with python objects adds a lot of overhead. A simple example:

>>> a = numpy.array(['abba' for _ in range(10000)])
>>> b = numpy.array(['abba' for _ in range(10000)], dtype=object)
>>> %timeit a.copy()
100000 loops, best of 3: 2.51 us per loop
>>> %timeit b.copy()
10000 loops, best of 3: 48.4 us per loop