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

Another option:

reduce(lambda x,y: x + [y, y*y], range(1,10), [])

Answer 2

A little-known trick: list comprehensions can have multiple for clauses.

For example:

>>> [10*x+y for x in range(4) for y in range(3)]
[0, 1, 2, 10, 11, 12, 20, 21, 22, 30, 31, 32]

In your particular case, you could do:

>>> [x*x if y else x for x in range(5) for y in range(2)]
[0, 0, 1, 1, 2, 4, 3, 9, 4, 16]

Answer 3

Try this two liner

lst = [[i, i*i] for i in range(10)]
[lst.extend(i) for i in lst]

Change math as necessary.

EVEN BETTER

#Change my_range to be the number you want range() function of
start = 1
my_range = 10
lst = [i/2 if i % 2 == 0 else ((i-1)/2)**2 for i in range(start *2, my_range*2 - 1)]

Answer 4

Lots of tricks in this thread. Here is another using a one liner generator without imports

x = (lamdba : [[(yield i), (yield i**2)] for i in range(10)])()

EDIT: This will raise DeprecatedWarning in Python 3.7 and SyntaxError in Python 3.8: https://docs.python.org/dev/whatsnew/3.7.html#deprecated-python-behavior

Answer 5

Another option, might seem perverse to some

>>> from itertools import izip, tee
>>> g = xrange(1, 11)
>>> x, y = tee(g)
>>> y = (i**2 for i in y)
>>> z = izip(x, y)
>>> output = []
>>> for k in z:
...     output.extend(k)
... 
>>> print output
[1, 1, 2, 4, 3, 9, 4, 16, 5, 25, 6, 36, 7, 49, 8, 64, 9, 81, 10, 100]

Answer 6

Use itertools.chain.from_iterable:

>>> from itertools import chain
>>> list(chain.from_iterable((i, i**2) for i in xrange(1, 6)))
[1, 1, 2, 4, 3, 9, 4, 16, 5, 25]

Or you can also use a generator function:

>>> def solve(n):
...     for i in xrange(1,n+1):
...         yield i
...         yield i**2

>>> list(solve(5))
[1, 1, 2, 4, 3, 9, 4, 16, 5, 25]