Repeat Each Item In A List A Number Of Times Specified In Another List

I have two lists, x and y: >>> x = [2, 3, 4] >>> y = [1, 2, 3] I want to use these to create a new list. The new list will have each element in x repeated the n

Solution 1:

numpy's repeat function gets the job done:

>>>import numpy as np>>>x = [2, 3, 4]>>>y = [1, 2, 3]>>>np.repeat(x, y)
array([2, 3, 3, 4, 4, 4])

Solution 2:

1. You can use list comprehension, like this

   >>>x = [2, 3, 4]>>>y = [1, 2, 3]>>>[item for item, count inzip(x, y) for i inrange(count)]
   [2, 3, 3, 4, 4, 4]
   

   Here, we zip the x and y so that the element from x and its corresponding count from y are grouped as a single tuple. Then, we iterate count number of items to produce the same item.

2. If your objects in x are immutables, then you can create count copies of the same and put them together in a list, like this

   >>>[i for item, count inzip(x, y) for i in [item] * count]
   [2, 3, 3, 4, 4, 4]
   

3. You can do the same lazily, with itertools.repeat, like this

   >>>from itertools import chain, repeat>>>chain.from_iterable((repeat(item, count) for item, count inzip(x,y)))
   <itertools.chain object at 0x7fabe40b5320>
   >>>list(chain.from_iterable((repeat(item, cnt) for item, cnt inzip(x,y))))
   [2, 3, 3, 4, 4, 4]
   

   Please note that the chain returns an iterable, not a list. So, if you don't want all the elements at once, you can get the items one by one from it. This will be highly memory efficient if the count is going to be a very big number, as we don't create the entire list in the memory immediately. We generate the values on-demand.

4. Thanks ShadowRanger. You can actually apply repeat over x and y and get the result like this

   >>>list(chain.from_iterable(map(repeat, x, y)))
   [2, 3, 3, 4, 4, 4]
   

   here, map function will apply the values from x and y to repeat one by one. So, the result of map will be

   >>> list(map(repeat, x, y))
   [repeat(2, 1), repeat(3, 2), repeat(4, 3)]
   

   Now, we use chain.from_iterable to consume values from each and every iterable from the iterable returned by map.

Solution 3:

Simple using for loop.

>>>x = [2, 3, 4]>>>y = [1, 2, 3]>>>final = []>>>for index, item inenumerate(y):
        final.extend([x[index]]*item)

Solution 4:

One way to achieve this is via using .elements() function of collections.Counter() along with zip. For example:

>>> from collections import Counter

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

# `.elements()` returns an object of `itertool.chain` type, which is an iterator.# in order to display it's content, here type-casting it to `list` >>> list(Counter(dict(zip(x,y))).elements())
[2, 3, 3, 4, 4, 4]