CODEWITHSUNDEEP
pythonpython-3.xlistnumpy
Burak
Burak

Reputation: 11

Python: Create an array using values and indices of a given array

Given the following list (or numpy array):

x = [4, 3, 1, 2]

I want to generate another list (or numpy array) with 1+4+3+2=10 elements such as:

y = [1, 1, 1, 1, 2, 2, 2, 3, 4, 4]

Where y will have x[i] successive elements with a value of i.

Other examples:

x = [0,3,1]
y = [2,2,2,3]

x = [2,0,2]
y = [1,1,3,3]

x = [1,1,1,1,1]
y = [1,2,3,4,5]

How can this be done efficiently? Thanks a lot.

Upvotes: 1

Views: 67

Answers (4)

hpaulj
hpaulj

Reputation: 231665

In [52]: alist, reps = [1,2,3,4], [4,3,1,2]

np.repeat does this kind of repetition nicely - but it makes things arrays (which takes time):

In [53]: np.repeat(alist,reps)
Out[53]: array([1, 1, 1, 1, 2, 2, 2, 3, 4, 4])

List repeat can be used with:

In [54]: [[i]*j for i,j in zip([1,2,3,4],[4,3,1,2])]
Out[54]: [[1, 1, 1, 1], [2, 2, 2], [3], [4, 4]]

and the lists of lists can be flattened with:

In [55]: [k for l in ([i]*j for i,j in zip(alist, reps)) for k in l]
Out[55]: [1, 1, 1, 1, 2, 2, 2, 3, 4, 4]
In [56]: list(itertools.chain(*([i]*j for i,j in zip(alist, reps))))
Out[56]: [1, 1, 1, 1, 2, 2, 2, 3, 4, 4]

Some timings:

In [57]: timeit np.repeat(alist,reps)
10.9 µs ± 398 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

If we start with arrays, the result is much faster:

In [58]: %%timeit a,b = np.array(alist), np.array(reps)
    ...:  np.repeat(a,b)
    ...: 
2.97 µs ± 103 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

But for this small sample, the list methods are faster:

In [59]: timeit [k for l in ([i]*j for i,j in zip(alist, reps)) for k in l]
2.33 µs ± 70.4 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [60]: timeit list(itertools.chain(*([i]*j for i,j in zip(alist, reps))))
2.46 µs ± 76.7 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

For larger cases, I expect the [58] times to scale the best. I'm not sure about the others.

Upvotes: 0

s3dev
s3dev

Reputation: 9721

If you wanted to use numpy:

x = np.array([4, 3, 1, 2])
a = np.arange(1, x.size+1)

np.repeat(a, x)

Output:

array([1, 1, 1, 1, 2, 2, 2, 3, 4, 4])

Upvotes: 1

Alex Mandelias
Alex Mandelias

Reputation: 517

An unnecessarily complicated way to do this is the following, utilizing list comprehensions and the reduce function:

from functools import reduce

def f(ls):
    return reduce(lambda x, y: x+y, [[i+1]*ls[i] for i in range(len(ls))])

>>> f([0, 3, 1])
[2, 2, 2, 3]
>>> f([2, 0, 2])
[1, 1, 3, 3]
>>> f([1, 1, 1, 1, 1])
[1, 2, 3, 4, 5]

Upvotes: 0

lorenzozane
lorenzozane

Reputation: 1279

This do the work:

x = [4,3,1,2]

y = []
for index, num in enumerate(x):
    for i in range(num):
        y.append(index + 1)
        
print(y)

or if you prefer with list comprehension in one line:

x = [4,3,1,2]

y = [index + 1 for index, num in enumerate(x) for i in range(num)]
        
print(y)

Output:

[1, 1, 1, 1, 2, 2, 2, 3, 4, 4]

Upvotes: 1

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