Python list comprehension with lambda

Question

Can someone please explain to me

Why a is a list of True/False, while b is a list of lambdas?
Why does the rule does not apply to c and d?

Codes:

foo = (lambda a, b: a >= b) if False else (lambda a, b: a <= b)
a = [foo(x, x+1) for x in xrange(10)]

foo = lambda a, b: a >= b if False else lambda a, b: a <= b
b = [foo(x, x+1) for x in xrange(10)]

bar = (lambda a, b: a*b*10) if False else (lambda a, b: a*b*100)
c = [bar(x, x+1) for x in xrange(10)]

bar = lambda a, b: a*b*10 if False else lambda a, b: a*b*100
d = [bar(x, x+1) for x in xrange(10)]

Thank you in advance.

In this expression, foo = lambda a, b: a >= b if descending else lambda a, b: a <= b there is no delimeter indicating the end of the lambda, so the if-else is inside the lambda. If descending is false, the else runs, and the outer lambda returns lambda a, b: a <= b. — Parthian Shot
– Parthian Shot, Commented Aug 13, 2014 at 21:16
Made a mistake, all keywords descending has been replaced with True. Thanks @BlackBear, @Joran Beasley — Alan Dong
– Alan Dong, Commented Aug 13, 2014 at 21:17
(b) is no longer a list of lambdas if you replace it with True — Joran Beasley
– Joran Beasley, Commented Aug 13, 2014 at 21:20
Thank you guys, now I understand why. As a question, I now changed all True to False for other people's benefits. — Alan Dong
– Alan Dong, Commented Aug 13, 2014 at 21:23

BlackBear · Accepted Answer · 2014-08-13 21:15:58Z

4

It is just a matter of operator precedence. Let's put some brackets to show how the statements are parsed:

(a) foo = (lambda a, b: a >= b) if True else (lambda a, b: a <= b)
(b) foo = lambda a, b: (a >= b if descending else lambda a, b: a <= b)

When evaluating (b) descending happens to be false, so all the elements become lambda a, b: a <= b

answered Aug 13, 2014 at 21:15

BlackBear

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Joran Beasley · Accepted Answer · 2014-08-13 21:15:17Z

1

foo = lambda a, b: a >= b if descending else lambda a, b: a <= b
b = [foo(x, x+1) for x in xrange(10)]

can be rewritten as

foo = lambda a, b:  (a >= b if descending else lambda a, b: a <= b)
b = [foo(x, x+1) for x in xrange(10)]

which if it is descending is fine as LHS is evaluated to

foo = lambda a,b: a >= b  #does what you would expet

but if its not you get

foo = lambda a,b: lambda a,b:a<=b

which clearly returns a lambda not a value

you could change it to

foo = lambda a, b: a >= b if descending else  a <= b
b = [foo(x, x+1) for x in xrange(10)]

and it should work as you expect

answered Aug 13, 2014 at 21:15

Joran Beasley

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