Unpack tuples in list comprehension with a condition

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I'd like to build one of these lists of tuples:

  • (a, 0), (-a, 0) (b, 0), (-b, 0)
  • (0, a), (0, -a) (0, b), (0, -b)

from scalars a and b.

based on a condition:

  • c = a > b

This is my attempt:

a = 5
b = 2
c = a > b

# Try build two tuples per element, e.g. (5, 0), (-5, 0) (2, 0), (-2, 0)

# This syntax is illegal
#f2 = [(m,0), (-m,0) if c else (0,m), (-0,-m) for m in (a,b)]

# This syntax works but creates tuples of tuples
f2 = [tuple(((m,0), (-m,0))) if c else tuple(((0,m), (-0,-m))) for m in (a,b)]
print(*f2) # ((5, 0), (-5, 0)) ((2, 0), (-2, 0))

# This syntax is illegal
#f3 = [*tuple(((m,0), (-m,0))) if c else *tuple(((0,m), (-0,-m))) for m in (a,b)]
#print(*f3)

f2 builds a list of two tuples of two tuples: ((5, 0), (-5, 0)) ((2, 0), (-2, 0)).
Using * operator in f3 to unpack the outer tuples triggers a syntax error.

What is the correct syntax?

Also I don't understand why f2 is ((5, 0), (-5, 0)) ((2, 0), (-2, 0)), where the outer tuples are not separated by a ,?

list-comprehension tuples python
2 Answers

You can do it with lambda functions.

x1 = lambda x : (x,0)
y1 = lambda x : (-x,0)
x2 = lambda x : (0,x)
y2 = lambda x : (-0,-x)
f2 = [f1(m) if c else f2(m) for m in (a,b) for f1,f2 in zip((x1,y1),(x2,y2))]

Output:
[(5, 0), (-5, 0), (2, 0), (-2, 0)]
A little bit over-engineered but makes sense.

EDIT
Also you can use chain from itertools to put 2 items in a single list comprehension.

from itertools import chain
f2 = list(
    chain.from_iterable(
        ((m,0), (-m,0)) if c else ((0,m), (-0,-m)) for m in (a,b)
    ))

SEE

Revuimar0

In this case you don't really need any loops, you can use condition:

a = 5
b = 2
c = a > b

f2 = [(a, 0), (-a, 0), (b, 0), (-b, 0)] if c else [(0, a), (0, -a), (0, b), (0, -b)]

If (for some reason) you want to solve it with list comprehension, you can use next:

f2 = [(j, 0) if c else (0, j) for i in (a, b) for j in (i, -i)]

You can help my country, check my profile info.

Olvin Roght0
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