Repeat each item in a list a number of times specified in another list

问题

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 number of times specified by the corresponding element in y. Hence, the desired output is

>>> new_list
[2, 3, 3, 4, 4, 4]

The order of the elements in new_list doesn't matter to me. It's also not crucial that it be a list -- any sequence type is fine.

What is the fastest, most efficient, most Pythonic way to achieve this?

回答1:

You can use list comprehension, like this
```
>>> x = [2, 3, 4]
>>> y = [1, 2, 3]
>>> [item for item, count in zip(x, y) for i in range(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.
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 in zip(x, y) for i in [item] * count]
[2, 3, 3, 4, 4, 4]
```
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 in zip(x,y)))
<itertools.chain object at 0x7fabe40b5320>
>>> list(chain.from_iterable((repeat(item, cnt) for item, cnt in zip(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.
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.

回答2:

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])

回答3:

Simple using for loop.

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

回答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]

来源：https://stackoverflow.com/questions/33382474/repeat-each-item-in-a-list-a-number-of-times-specified-in-another-list

标签

python

list

repeat