For Python 3.3, how can I make a list of functions using one lambda statement?

Question

Here is what I tried:

quadrant = [lambda x, y: (sign[0] * x, sign[1] * y)
    for sign in ((1, -1), (1, 1), (-1, 1), (-1, -1))]

What I expected to get from this line was a list of functions that each return the input values with some set of signs applied. For example, quadrant[1](x, y) -> (x, y), quadrant[2](x, y) -> (-x, y), and so on.

What I actually got is a list of four identical functions, all the last function that I put into the list. For example, quadrant[1](x, y) -> (-x, -y), quadrant[2](x, y) -> (-x, -y), and so on.

What am I misunderstanding here? Why does each new function added to the list replace all the previously added functions?

Hyperboreus · Accepted Answer · 2013-10-03 19:21:19Z

1

You must break out of the closure:

quadrant = [ (lambda sign: lambda x, y: (sign[0] * x, sign[1] * y) ) (sign)
    for sign in ((1, -1), (1, 1), (-1, 1), (-1, -1))]

Or with named functions as Blender pointed out:

def wrapper (sign):
    def makeQuadrant (x, y):
        return (sign [0] * x, sign [1] * y)
    return makeQuadrant

quadrant = [wrapper (sign) for sign in ((1, -1), (1, 1), (-1, 1), (-1, -1))]

Or with a default named parameter as chepner pointed out:

quadrant = [lambda x, y, sign = sign: (sign[0] * x, sign[1] * y)
    for sign in ((1, -1), (1, 1), (-1, 1), (-1, -1))]

And now to answer your question:

Why does each new function added to the list replace all the previously added functions?

No function is replaced, your list contains four distinct functions as you can see here:

quadrant = [lambda x, y: (sign[0] * x, sign[1] * y, id (sign) )
    for sign in ((1, -1), (1, 1), (-1, 1), (-1, -1))]

for q in quadrant:
    print (id (q) )

What happens is that the value of sign is looked up when the function is called. The current value of sign is the last one assigned during the list comprehension, which is (-1, -1). Which can be seen here:

quadrant = [lambda x, y: (sign[0] * x, sign[1] * y, id (sign) )
    for sign in ((1, -1), (1, 1), (-1, 1), (-1, -1))]

for q in quadrant:
    print (q (2, 3) )

You might want to look a bit into "scoping" and "closures" for further reading.

Your example basically boils down to this:

fs = [lambda: x for x in range (10) ]
for f in fs: print (f () )

edited Oct 3, 2013 at 19:21

answered Oct 3, 2013 at 18:55

Hyperboreus

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6 Comments

Blender Over a year ago

I think at this point a named helper function would make the code a little more readable.

chepner Over a year ago

Or simply make sign an unused parameter with a default value: lambda x, y, sign=sign: ...

Hyperboreus Over a year ago

@chepner Although without doubt your approach is the neatest, I fear that it is the most difficult to understand.

Duncan Over a year ago

@Hyperboreus a name is useful in the stacktrace if it ever throws an exception.

Victor Over a year ago

@Hyperboreus, Thanks. I also appreciate chepner's fix, esp. with a small clarification: ...sign=dflt_sign: (sign[0]...for dflt_sign in .... The scope structure is clear in your explication of Blender's fix, but if straightforward readability is the goal, one may as well just write out quadrant = [lambda x, y: (1, -1), lambda x, y: (1, 1), ...]. I'm still curious how the list [lambda: x for x in range(10)] has 10 functions in it, which seems to mean that the for-loop was applied to each function, but only 1 value for x, which seems to mean that the for-loop also applied to x alone. No?

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