I have a Python list of lists:

l = [[1, 2, 3], [4], [5, 6], [7, 8, 9, 10]]

What I want is to repeat the first element of each list based on the length of the list:

result = [1, 1, 1, 4, 5, 5, 7, 7, 7, 7]

I can achieve this using list comprehension but my list is very long and so the method is slow:

result = [[x[0]]*len(x) for x in l]

[[1, 1, 1], [4], [5, 5], [7, 7, 7, 7]]

Although, this still returns a list of lists rather than a flat list. So, I am trying to figure out the fastest method for creating that flat list based on the criteria mentioned above.

Update: I want the fastest performing method since the list is long

Using itertools.repeat with chain is the most efficient using python 2:

In [13]: l = [choice(l) for _ in xrange(1000000)]
In [14]: timeit list(itertools.chain(*[[i[0]]*len(i) for i in l]))
1 loops, best of 3: 416 ms per loop

In [15]: timeit [i[0] for i in l for _ in xrange(len(i))]
1 loops, best of 3: 245 ms per loop

In [16]: timeit list(itertools.chain.from_iterable(repeat(i[0],len(i)) for i in l))
1 loops, best of 3: 223 ms per loop

In [17]: timeit [i for x in l for i in [x[0]]*len(x)]
1 loops, best of 3: 332 ms per loop

Interestingly using python3, using a list instead of a generator expression is faster:

In [8]: timeit list(chain.from_iterable(repeat(i[0], len(i)) for i in l))
1 loops, best of 3: 372 ms per loop

In [9]: timeit [i[0] for i in l for _ in range(len(i))]
1 loops, best of 3: 433 ms per loop

In [10]: timeit list(chain.from_iterable([repeat(i[0],len(i)) for i in l]))
1 loops, best of 3: 296 ms per loop

In [11]: timeit list(chain(*[[i[0]]*len(i) for i in l]))
1 loops, best of 3: 460 ms per loop

In [12]: timeit [i for x in l for i in [x[0]]*len(x)]
1 loops, best of 3: 348 ms per loop

If you want a compromise between time and space then iterate over the chain object getting an element at a time:

In [18]: %%timeit
for ele in chain.from_iterable([repeat(i[0],len(i)) for i in l]):
        pass
   ....:
1 loops, best of 3: 306 ms per lo

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