def some_generator(some_args):
    Import = namedtuple("Import", ["module", "name", "alias"]) # "module", "name" and "alias" can be listed
...
    for n in node.names:
        yield Import(module, n.name.split('.'), n.asname)
...
files = list()
requirements = list()
...
for member in members:
    files.append(member)
...
for requirement in self.some_generator(some_args):
    requirements.append(requirement)
...

for file in files:
    if file in requirements:
        print(f"{file} is in requirements")

(Oct-25-2022, 05:06 PM)ibreeden Wrote: Do you mean this?

for file in files:
    if file in requirements:
        print(f"{file} is in requirements")

yes, but i have namedtuple (module, name, alias and they can be listed)

(Oct-25-2022, 06:40 PM)deanhystad Wrote: This is vague:

yes, but i have namedtuple (module, name, alias and they can be listed)

Do you want to check if file matches any of the fields in the tuple or a particular field?

If there is a match, what do you want to happen?

from random import randint
from secrets import randbelow

Thing = namedtuple("Thing", ("A", "B", "C"))

things = [Thing(randint(1, 10), randint(1, 10), randint(1, 10)) for _ in range(10)]

for thing in things:
    if 5 in thing:
        print(thing)

Output:
Thing(A=8, B=6, C=5)
Thing(A=4, B=5, C=8)
Thing(A=5, B=2, C=7)

from collections import namedtuple
from random import randint
from itertools import chain

Import = namedtuple("Import", ("module", "name", "alias"))
requirements = [Import(randint(1, 1000), randint(1, 1000), randint(1, 1000)) for _ in range(10)]
all_requirements = set(chain(*requirements))  # This will contain every value for module, name and alias that appears in any requirement

files = list(range(100, 200))
for file in files:
    if file in all_requirements:
        print(file)

Output:
126
133

(Oct-25-2022, 08:15 PM)deanhystad Wrote: iberdeen's code does that. Is the problem that it does it for 1 tuple? Here I search multiple tuples (but I have only one file).

from random import randint
from secrets import randbelow

Thing = namedtuple("Thing", ("A", "B", "C"))

things = [Thing(randint(1, 10), randint(1, 10), randint(1, 10)) for _ in range(10)]

for thing in things:
    if 5 in thing:
        print(thing)

Output:
Thing(A=8, B=6, C=5)
Thing(A=4, B=5, C=8)
Thing(A=5, B=2, C=7)

A namedtuple is still a tuple. The names don't make any difference to "in".

If you have many of these tuples and you want to check many filenames, and you don't really care which tuple is matched (that is a lot of if's), consider making a set that contains all requirements.

from collections import namedtuple
from random import randint
from itertools import chain

Import = namedtuple("Import", ("module", "name", "alias"))
requirements = [Import(randint(1, 1000), randint(1, 1000), randint(1, 1000)) for _ in range(10)]
all_requirements = set(chain(*requirements))  # This will contain every value for module, name and alias that appears in any requirement

files = list(range(100, 200))
for file in files:
    if file in all_requirements:
        print(file)

Output:
126
133

        for file in files:
            for requirement in requirements:
                if file in requirement:
                    requirements.remove(requirement)

(Oct-26-2022, 10:32 AM)elnk Wrote: so if i want to delete element from "requirements" wich conrain in "files" is it code right and it work?

for file in files:
    for requirement in requirements:
        if file in requirement:
            requirements.remove(requirement)

from collections import namedtuple
from random import randint

Import = namedtuple("Import", ("module", "name", "alias"))

files = set(range(100, 200))

# All the requirements
requirements = [Import(randint(1, 1000), randint(1, 1000), randint(1, 1000)) for _ in range(10)]

# Requirements that have no match with any file.
requirements = [req for req in requirements if set(req).isdisjoint(files)]
print(*requirements, sep="\n")