Error: Using functions in list comprehension

[comment comments]

Any time I use a function in a list comprehension, it gives me a " SyntaxError: Unexpected identifier 'fnName' "

For example:
Web VPython 3.2
def asdf():
    pass
[asdf() for a in [1,2,3]]

gives the error "SyntaxError: Unexpected identifier 'asdf'"

[comment comments]

Adding info: it appears that (at least some) glowscript functions work as expected (i.e. [sphere(pos=a) for a in [vector(1,2,3),vector(1,1,1), vector(0,0,0)]] gives three spheres), but (at least some) python functions do not (i.e. [sum(a) for a in [[1,2,3],[1,1,1],[0,0,0]]] gives an error)

[Harlan Gilbert]

Your example works properly through VPython version 2.8, in versions thereafter throws an error.

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Harlan Gilbert, Ph.D.

High School Co-chair and Math, Physics, Computer Science, and Philosophy Teacher

Green Meadow Waldorf High School

Chestnut Ridge, NY 10977

(845) 356-2514 x. 320

[Joe Heafner]

> On May 30, 2023, at 12:47, Harlan Gilbert <hgil...@gmws.org> wrote:
>
> Your example works properly through VPython version 2.8, in versions thereafter throws an error.

RapydScript (https://github.com/kovidgoyal/rapydscript-ng) doesn't support nested list comprehensions, but that doesn't seem to be what's going on here. I wonder if it's a bug elsewhere.


Joe Heafner
Sent from one of my Macs

[Bruce Sherwood]

When I run the following in installed Python

def asdf():
    pass
c = [asdf() for a in [1,2,3]]
print(c)

I get the result [None, None, None]. Which leaves me wondering what is intended. Can you provide a simple test routine that produces something other than a list of Nones when running with installed Python?

[Harlan Gilbert]

This would be a more normal usage of list comprehension

def sqs(x):

    return x**2

c = [sqs(a) for a in [1,2,3]]

print(c)

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[Bruce Sherwood]

Thanks for the help. The problem is that the generated JavaScript code contains inappropriate async and await statements . I suspect that it will be difficult to fix this. The workaround of course is to use a loop.

Bruce