R function rep() in Python (replicates elements of a list/vector)

Question

The R function rep() replicates each element of a vector:

> rep(c(\"A\",\"B\"), times=2)
[1] \"A\" \"B\" \"A\" \"B\"

This is like the list m

Answer 1

l = ['A','B']
n = [2, 4]

Your example uses strings which are already iterables. You can produce a result string which is similar to a list.

''.join([e * m for e, m in zip(l, n)])
'AABBBB'

Update: the list comprehension is not required here:

''.join(e * m for e, m in zip(l, n))
'AABBBB'

Answer 2

The numpy.repeat has been mentioned, and that's clearly the equivalent to what you want. But for completenes' sake, there's also repeat from the itertools standard library. However, this is intended for iterables in general, so it doesn't allow repetions by index (because iterables in general do not have an index defined).

We can use the code given there as a rough equivalent

def repeat(object, times=None):
    # repeat(10, 3) --> 10 10 10
    if times is None:
        while True:
            yield object
    else:
        for i in xrange(times):
            yield object

to define our own generalised repeat:

def repeat_generalised(object, times=None):
    # repeat(10, 3) --> 10 10 10
    if times is None:
        while True:
            yield object
    else:
        for reps, elem in zip(times, object):
            for i in xrange(reps): 
                yield elem

The problem of course is that there's a lot of possible edge cases you have to define (What should happen if object and times have a different number of elements?), and that would depend on you individual use case.

Answer 3

The following might work for you:

>>>[['a','b'],['A','B']]*5


[['a', 'b'], ['A', 'B'], ['a', 'b'], ['A', 'B'], ['a', 'b'], ['A', 'B'], ['a', 'b'], ['A', 'B'], ['a', 'b'], ['A', 'B']]

Answer 4

Since you say "array" and mention R. You may want to use numpy arrays anyways, and then use:

import numpy as np
np.repeat(np.array([1,2]), [2,3])

EDIT: Since you mention you want to repeat rows as well, I think you should use numpy. np.repeat has an axis argument to do this.

Other then that, maybe:

from itertools import izip, chain, repeat
list(chain(*(repeat(a,b) for a, b in izip([1,2], [2,3]))))

As it doesn't make the assumption you have a list or string to multiply. Though I admit, passing everything as argument into chain is maybe not perfect, so writing your own iterator may be better.

Answer 5

Not sure if there's a built-in available for this, but you can try something like this:

>>> lis = ["A", "B"]
>>> times = (2, 3)
>>> sum(([x]*y for x,y in zip(lis, times)),[])
['A', 'A', 'B', 'B', 'B']

Note that sum() runs in quadratic time. So, it's not the recommended way.

>>> from itertools import chain, izip, starmap
>>> from operator import mul
>>> list(chain.from_iterable(starmap(mul, izip(lis, times))))
['A', 'A', 'B', 'B', 'B']

Timing comparions:

>>> lis = ["A", "B"] * 1000
>>> times = (2, 3) * 1000
>>> %timeit list(chain.from_iterable(starmap(mul, izip(lis, times))))
1000 loops, best of 3: 713 µs per loop
>>> %timeit sum(([x]*y for x,y in zip(lis, times)),[])
100 loops, best of 3: 15.4 ms per loop

Answer 6

Use numpy arrays and the numpy.repeat function:

import numpy as np

x = np.array(["A", "B"])
print np.repeat(x, [2, 3], axis=0)

['A' 'A' 'B' 'B' 'B']