Create List With Numbers Getting Greater Each Time Python [closed]

Question

How can I make a function that will create a list, increasing the amount of numbers it contains each time to a specified value?

For example if the max was 4, the list would contain

1, 2, 2, 3, 3, 3, 4, 4, 4, 4

It's difficult to explain what I'm looking for, but from the example I think you'll understand!

Thanks

Answer 1

I'd use itertools.chain:

itertools.chain(*([i] * i for i in range(1, 5)))

or itertools.chain.from_iterable to do it slightly more lazily:

itertools.chain.from_iterable([i] * i for i in range(1, 5))

---

And for the ultimate laziness, pair with itertools.repeat -- (using xrange in you're using python2.x):

import itertools as it
it.chain.from_iterable(it.repeat(i, i) for i in range(1, 5))

As a function:

def lazy_funny_iter(n):
    return it.chain.from_iterable(it.repeat(i, i) for i in range(1, n+1))

def lazy_funny_list(n):
    return list(lazy_funny_iter(n))

Answer 2

A Nested loop. This would be a very basic way to do it. There are much better ways, this should give you the general idea.

>>> def listmaker(num):
    l = []
    for i in xrange(1, num+1):
        for j in xrange(i):
            l.append(i)
    return l

>>> print listmaker(4)
[1, 2, 2, 3, 3, 3, 4, 4, 4, 4]

Here is doing it with list comprehension:

>>> def listmaker2(num):
    return [y for z in [[x]*(x) for x in xrange(1, num+1)] for y in z]

>>> print listmaker2(4)
[1, 2, 2, 3, 3, 3, 4, 4, 4, 4]

Using extend as suggested.

>>> def listmaker3(num):
    l = []
    for i in xrange(1, num+1):
        l.extend([i]*(i))
    return l

>>> print listmaker3(4)
[1, 2, 2, 3, 3, 3, 4, 4, 4, 4]