cpython/Lib/test/test_setcomps.py
Serhiy Storchaka 8c579b1cc8
bpo-32856: Optimize the assignment idiom in comprehensions. (GH-16814)
Now `for y in [expr]` in comprehensions is as fast as a simple
assignment `y = expr`.
2020-02-12 12:18:59 +02:00

168 lines
4.1 KiB
Python

doctests = """
########### Tests mostly copied from test_listcomps.py ############
Test simple loop with conditional
>>> sum({i*i for i in range(100) if i&1 == 1})
166650
Test simple case
>>> {2*y + x + 1 for x in (0,) for y in (1,)}
{3}
Test simple nesting
>>> list(sorted({(i,j) for i in range(3) for j in range(4)}))
[(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3), (2, 0), (2, 1), (2, 2), (2, 3)]
Test nesting with the inner expression dependent on the outer
>>> list(sorted({(i,j) for i in range(4) for j in range(i)}))
[(1, 0), (2, 0), (2, 1), (3, 0), (3, 1), (3, 2)]
Test the idiom for temporary variable assignment in comprehensions.
>>> sorted({j*j for i in range(4) for j in [i+1]})
[1, 4, 9, 16]
>>> sorted({j*k for i in range(4) for j in [i+1] for k in [j+1]})
[2, 6, 12, 20]
>>> sorted({j*k for i in range(4) for j, k in [(i+1, i+2)]})
[2, 6, 12, 20]
Not assignment
>>> sorted({i*i for i in [*range(4)]})
[0, 1, 4, 9]
>>> sorted({i*i for i in (*range(4),)})
[0, 1, 4, 9]
Make sure the induction variable is not exposed
>>> i = 20
>>> sum({i*i for i in range(100)})
328350
>>> i
20
Verify that syntax error's are raised for setcomps used as lvalues
>>> {y for y in (1,2)} = 10 # doctest: +IGNORE_EXCEPTION_DETAIL
Traceback (most recent call last):
...
SyntaxError: ...
>>> {y for y in (1,2)} += 10 # doctest: +IGNORE_EXCEPTION_DETAIL
Traceback (most recent call last):
...
SyntaxError: ...
Make a nested set comprehension that acts like set(range())
>>> def srange(n):
... return {i for i in range(n)}
>>> list(sorted(srange(10)))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Same again, only as a lambda expression instead of a function definition
>>> lrange = lambda n: {i for i in range(n)}
>>> list(sorted(lrange(10)))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Generators can call other generators:
>>> def grange(n):
... for x in {i for i in range(n)}:
... yield x
>>> list(sorted(grange(5)))
[0, 1, 2, 3, 4]
Make sure that None is a valid return value
>>> {None for i in range(10)}
{None}
########### Tests for various scoping corner cases ############
Return lambdas that use the iteration variable as a default argument
>>> items = {(lambda i=i: i) for i in range(5)}
>>> {x() for x in items} == set(range(5))
True
Same again, only this time as a closure variable
>>> items = {(lambda: i) for i in range(5)}
>>> {x() for x in items}
{4}
Another way to test that the iteration variable is local to the list comp
>>> items = {(lambda: i) for i in range(5)}
>>> i = 20
>>> {x() for x in items}
{4}
And confirm that a closure can jump over the list comp scope
>>> items = {(lambda: y) for i in range(5)}
>>> y = 2
>>> {x() for x in items}
{2}
We also repeat each of the above scoping tests inside a function
>>> def test_func():
... items = {(lambda i=i: i) for i in range(5)}
... return {x() for x in items}
>>> test_func() == set(range(5))
True
>>> def test_func():
... items = {(lambda: i) for i in range(5)}
... return {x() for x in items}
>>> test_func()
{4}
>>> def test_func():
... items = {(lambda: i) for i in range(5)}
... i = 20
... return {x() for x in items}
>>> test_func()
{4}
>>> def test_func():
... items = {(lambda: y) for i in range(5)}
... y = 2
... return {x() for x in items}
>>> test_func()
{2}
"""
__test__ = {'doctests' : doctests}
def test_main(verbose=None):
import sys
from test import support
from test import test_setcomps
support.run_doctest(test_setcomps, verbose)
# verify reference counting
if verbose and hasattr(sys, "gettotalrefcount"):
import gc
counts = [None] * 5
for i in range(len(counts)):
support.run_doctest(test_setcomps, verbose)
gc.collect()
counts[i] = sys.gettotalrefcount()
print(counts)
if __name__ == "__main__":
test_main(verbose=True)