python list comprehension to produce two values in one iteration

Question

I want to generate a list in python as follows -

[1, 1, 2, 4, 3, 9, 4, 16, 5, 25 .....]

You would have figured out, it is nothing but n,

Answer 1

You can create a list of lists then use reduce to join them.

print [[n,n*n] for n in range (10)]

[[0, 0], [1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81]]

print reduce(lambda x1,x2:x1+x2,[[n,n*n] for n in range (10)])

[0, 0, 1, 1, 2, 4, 3, 9, 4, 16, 5, 25, 6, 36, 7, 49, 8, 64, 9, 81]

 print reduce(lambda x1,x2:x1+x2,[[n**e for e in range(1,4)]\
 for n in range (1,10)])

[1, 1, 1, 2, 4, 8, 3, 9, 27, 4, 16, 64, 5, 25, 125, 6, 36, 216, 7, 49, 343, 8, 64, 512, 9, 81, 729]

Reduce takes a callable expression that takes two arguments and processes a sequence by starting with the first two items. The result of the last expression is then used as the first item in subsequent calls. In this case each list is added one after another to the first list in the list of lists and then that list is returned as a result.

List comprehensions implicitly call map with a lambda expression using the variable and sequence defined by the "for var in sequence" expression. The following is the same sort of thing.

map(lambda n:[n,n*n],range(1,10))

[[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81]]

I am unaware of a more natural python expression for reduce.

Answer 2

List comprehensions generate one element at a time. Your options are, instead, to change your loop to only generate one value at a time:

[(i//2)**2 if i % 2 else i//2 for i in range(2, 20)]

or to produce tuples then flatten the list using itertools.chain.from_iterable():

from itertools import chain

list(chain.from_iterable((i, i*i) for i in range(1, 10)))

Output:

>>> [(i//2)**2 if i % 2 else i//2 for i in range(2, 20)]
[1, 1, 2, 4, 3, 9, 4, 16, 5, 25, 6, 36, 7, 49, 8, 64, 9, 81]
>>> list(chain.from_iterable((i, i*i) for i in range(1, 10)))
[1, 1, 2, 4, 3, 9, 4, 16, 5, 25, 6, 36, 7, 49, 8, 64, 9, 81]

Answer 3

>>> lst_gen = [[i, i*i] for i in range(1, 10)]
>>> 
>>> lst_gen
[[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81]]
>>> 
>>> [num for elem in lst_gen for num in elem]
[1, 1, 2, 4, 3, 9, 4, 16, 5, 25, 6, 36, 7, 49, 8, 64, 9, 81]

Here is my reference http://docs.python.org/2/tutorial/datastructures.html

Answer 4

As mentioned, itertools is the way to go. Here's how I would do it, I find it more clear:

[i if turn else i*i for i,turn in itertools.product(range(1,10), [True, False])]

Answer 5

lst_gen = sum([(i, i*i) for i in range(1, 10)],())

oh I should mention the sum probably breaks the one iteration rule :(

Answer 6

The question is old, but just for the curious reader, i propose another possibility: As stated on first post, you can easily make a couple (i, i**2) from a list of numbers. Then you want to flatten this couple. So just add the flatten operation in your comprehension.

[x for i in range(1, 10) for x in (i,i**2)]