How to avoid circular imports in Python?

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I know the issue of circular imports in python has come up many times before and I have read these discussions. The comment that is made repeatedly in these discussions is that a circular import is a sign of a bad design and the code should be reorganised to avoid the circular import.

Could someone tell me how to avoid a circular import in this situation?: I have two classes and I want each class to have a constructor (method) which takes an instance of the other class and returns an instance of the class.

More specifically, one class is mutable and one is immutable. The immutable class is needed for hashing, comparing and so on. The mutable class is needed to do things too. This is similar to sets and frozensets or to lists and tuples.

I could put both class definitions in the same module. Are there any other suggestions?

A toy example would be class A which has an attribute which is a list and class B which has an attribute which is a tuple. Then class A has a method which takes an instance of class B and returns an instance of class A (by converting the tuple to a list) and similarly class B has a method which takes an instance of class A and returns an instance of class B (by converting the list to a tuple).

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How to avoid circular imports in Python? repeat: Questions

Create list of single item repeated N times

5 answers

I want to create a series of lists, all of varying lengths. Each list will contain the same element e, repeated n times (where n = length of the list).

How do I create the lists, without using a list comprehension [e for number in xrange(n)] for each list?

618

Answer #1

You can also write:

[e] * n

You should note that if e is for example an empty list you get a list with n references to the same list, not n independent empty lists.

Performance testing

At first glance it seems that repeat is the fastest way to create a list with n identical elements:

>>> timeit.timeit("itertools.repeat(0, 10)", "import itertools", number = 1000000)
0.37095273281943264
>>> timeit.timeit("[0] * 10", "import itertools", number = 1000000)
0.5577236771712819

But wait - it"s not a fair test...

>>> itertools.repeat(0, 10)
repeat(0, 10)  # Not a list!!!

The function itertools.repeat doesn"t actually create the list, it just creates an object that can be used to create a list if you wish! Let"s try that again, but converting to a list:

>>> timeit.timeit("list(itertools.repeat(0, 10))", "import itertools", number = 1000000)
1.7508119747063233

So if you want a list, use [e] * n. If you want to generate the elements lazily, use repeat.

How to avoid circular imports in Python? repeat: Questions

What is the best way to repeatedly execute a function every x seconds?

5 answers

DavidM By DavidM

I want to repeatedly execute a function in Python every 60 seconds forever (just like an NSTimer in Objective C). This code will run as a daemon and is effectively like calling the python script every minute using a cron, but without requiring that to be set up by the user.

In this question about a cron implemented in Python, the solution appears to effectively just sleep() for x seconds. I don"t need such advanced functionality so perhaps something like this would work

while True:
    # Code executed here
    time.sleep(60)

Are there any foreseeable problems with this code?

391

Answer #1

If your program doesn"t have a event loop already, use the sched module, which implements a general purpose event scheduler.

import sched, time
s = sched.scheduler(time.time, time.sleep)
def do_something(sc): 
    print("Doing stuff...")
    # do your stuff
    s.enter(60, 1, do_something, (sc,))

s.enter(60, 1, do_something, (s,))
s.run()

If you"re already using an event loop library like asyncio, trio, tkinter, PyQt5, gobject, kivy, and many others - just schedule the task using your existing event loop library"s methods, instead.

391

Answer #2

Lock your time loop to the system clock like this:

import time
starttime = time.time()
while True:
    print "tick"
    time.sleep(60.0 - ((time.time() - starttime) % 60.0))

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