How Extract Items Of Sublists In A One-line-comprehension In Python?

I am currently learning the concept of list comprehensions in python. However, I have huge problems when the list I am iterating over contains sublists of equal or different length

Solution 1:

Using list-comprehension, you can do something like that:

>>>L1 = [1, 2, 3]>>>L2 = [4, 5, 6]>>>L3 = [7, 8, 9]>>>L = [L1, L2, L3]>>>s=set([x for y in L for x in y])>>>s
set([1, 2, 3, 4, 5, 6, 7, 8, 9])

y is iterating over the sublist, while x iterates over items in y.

Solution 2:

Use an empty set and .union it:

L1 = [1, 2, 3]
L2 = [4, 5, 6]
L3 = [7, 8, 9]

print set().union(L1, L2, L3)

Used in your code as:

L = [L1, L2, L3]

defunion_set(L):
    returnset().union(*L)

Solution 3:

Use * for unpacking and pass the unpacked items to set.union:

>>>L = [L1, L2, L3]>>>set.union(*(set(x) for x in L))
set([1, 2, 3, 4, 5, 6, 7, 8, 9])

Efficient versions using itertools:

>>>from itertools import islice>>>set.union(set(L[0]),*islice(L,1,None))
set([1, 2, 3, 4, 5, 6, 7, 8, 9])

>>>from itertools import chain>>>set(chain.from_iterable(L))
set([1, 2, 3, 4, 5, 6, 7, 8, 9])

Timing comparisons:

>>>L = [L1, L2, L3]*10**5>>>%timeit set.union(*(set(x) for x in L))
1 loops, best of 3: 416 ms per loop

>>>%timeit set(chain.from_iterable(L))               # winner
1 loops, best of 3: 69.4 ms per loop

>>>%timeit set.union(set(L[0]),*islice(L,1,None))
1 loops, best of 3: 78.6 ms per loop

>>>%timeit set().union(*L)
1 loops, best of 3: 105 ms per loop

>>>%timeit set(chain(*L))
1 loops, best of 3: 79.2 ms per loop

>>>%timeit s=set([x for y in L for x in y])
1 loops, best of 3: 151 ms per loop

Solution 4:

You could use itertools.chain like this

>>>L1 = [1, 2, 3]>>>L2 = [4, 5, 6]>>>L3 = [7, 8, 9]>>>L = [L1,L2,L3]>>>set(itertools.chain(*L))
set([1, 2, 3, 4, 5, 6, 7, 8, 9])

* unpacks the list, and chain creates a list out of sublists.