How to randomly choose method from a list of methods with different arguments?
python random choice multiple
python random selection from list without repetition
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random function in python
I have object
class b with methods
foo(), boo(), noo(arg). Now I want to randomly choose one of them without extending the
class b. The problem is that if I use technics supposed here: Syntax to call random function from a list , then I have to have the same arguments of my methods.
So I can't use smth like this:
my_list = [a.foo, a.boo, a.noo] random.choice(my_list)()
I case of
noo method I need to call it with
a.noo(arg). So how to do it in pythonic way?
import random funcs = [foo, boo, noo] args = [[1,2,3], [4,5], ] choice = random.randint(len(funcs)) funcs[choice](*args[choice])
This way you have the flexibility to call any function with dynamic arguments, depending on your methods.
Remember to keep the arguments and the functions in the same order
Edit: You can also do as suggested in the comments (which i think it's easier, thank you!):
import random calls = [(foo, [1,2,3]), (boo, [4,5]), (noo, )] choice = random.choice(calls) choice(*choice)
Python Random choices() Method, Method. ❮ Random Methods The choices() method returns a list with the randomly selected element from the specified sequence. You can weigh the possibility of each result with the weights parameter or the cum_weights parameter. The sequence can be a string, a range, a list, a tuple or any other kind of sequence. Example. Return a list with 14 items. The list should contain a randomly selection of the values from a specified list, and there should be 10 times higher possibility to select "apple" than the other two:
This is how I would do it. Although if you want to have functions with more than 1 argument it's probably best to send it as a list and unpack it at the function.
import random def foo(): print('foo') def boo(): print('boo') def noo(arg): print('noo:', arg) my_list = [(foo,), (boo,), (noo, 'hello world')] for i in range(10): choice = random.choice(my_list) if len(choice) == 1: choice() else: choice(choice)
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Something like this was suggested in comment:
import random my_list = [a.foo, a.boo, a.noo] rand = random.choice(my_list) if rand.__name__ == 'noo': rand(arg) else: rand()
Python random.sample() to choose multiple items from any sequence, sample() function has two arguments, and both are required. The population can be any sequence such as list, set from which you want to select Arguments. The random.sample() function has two arguments, and both are required. The population can be any sequence such as list, set from which you want to select a k length number. The k is the number of random items you want to select from the sequence. k must be less than the size of the specified list.
Data Science from Scratch: First Principles with Python, More frequently we'll use the iteritems() method, which lazily yields the 2 arguments and returns an element chosen randomly from the corresponding range(): choose randomly from range(3, 6) = [3, 4, 5] There are a few more methods will probably be different) If you need to randomly pick one element from a list you The Next() Method of System.Random class in C# is used to get a random integer number. This method can be overloaded by passing different parameters to it as follows: Next() Next(Int32) Next(Int32, Int32) Next() Method. This method is used to returns a non-negative random integer.
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- Why don't you use an
ifdo the a.noo case ?
- but is this pythonic way?
- If the solution works for you, it doesn't need to be 'pythonic'
- I mean, it might be shorter. Can't I specify somehow the arg for noo function in a list?
- You need to call a random function, but each call should be made with a particular set of arguments, is that it?