Posts: 5 Threads: 2 Joined: Oct 2022 Oct-25-2022, 04:26 PM (This post was last modified: Oct-25-2022, 04:27 PM by elnk.) Hello, i have this code: 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) ...and then i need to check if an element from "files" is in "requirements" (module, name or alias) how to implement it? Posts: 581 Threads: 1 Joined: Aug 2019 Do you mean this? for file in files: if file in requirements: print(f"{file} is in requirements") Posts: 5 Threads: 2 Joined: Oct 2022 (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) Posts: 6,920 Threads: 22 Joined: Feb 2020 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? Posts: 5 Threads: 2 Joined: Oct 2022 (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? "file" is it string, i need to check if an element from "files" (string) is in "requirements" (namedtuple) (module, name or alias) Posts: 6,920 Threads: 22 Joined: Feb 2020 Oct-25-2022, 08:15 PM (This post was last modified: Oct-25-2022, 08:15 PM by deanhystad.) 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
Posts: 5 Threads: 2 Joined: Oct 2022 (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
thank you! 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) Posts: 581 Threads: 1 Joined: Aug 2019 (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) No. You must never change a collection while iterating over it. You are iterating over "requirements" and then remove items of it. Instead make a copy of "requirements" and iterate over that copy. Posts: 6,920 Threads: 22 Joined: Feb 2020 Oct-26-2022, 04:03 PM (This post was last modified: Oct-26-2022, 04:03 PM by deanhystad.) Don't remove, append. I would build a new list that only contains the requirements you want to keep, and I would use set operations to determine which requirements to save 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") |
