如何在python中的几个可用函数中随机重复10次？

Question

For several functions: ftn_1, ftn_2, ... , ftnk I want to repeat 10 times randomly among above. (The whole time is k^10.)

I find some function which acts like this way but it seems to act just for string : itertools.product

def ftn1():
   ...
def ftn2():
   ...
x=0;
itertools.product(x=ftn1(x), x=ftn2(x), .., x=ftnk(x), [repeat=10])

I want to do similar to above and find the smallest value for all cases.

Answer 1

You just need to create a list (or a tuple) of all your functions, and then use random on it, for instance:

import random

funcs = [ftn1, ftn2, ftn3, ..., ftnN]
my_func = random.choice(funcs)

Then you can call my_func with any argument of your choice.

Answer 2

If you mean you have k functions and want to call all of them 10 times in total, but in random order, you could shuffle() a list of them.
For sanity purposes this example has 3 functions and calls them 2 times. An argument is supplied too for all of them, it is just written out:

import random

def f1(x):
  print("f1:"+str(x))
def f2(x):
  print("f2:"+str(x))
def f3(x):
  print("f3:"+str(x))
# here you could have more functions...

repeats=[f1,f2,f3]*2 # ... and they could be listed here, and 2 could be 10
print("Default order:")
for i in range(len(repeats)):
  repeats[i](i) # repeats[i] is a function here, which is invoked with i

random.shuffle(repeats)
print("Shuffled order:")
for i in range(len(repeats)):
  repeats[i](i)

Example output (of course it varies because of randomness)

 Default order: f1:0 f2:1 f3:2 f1:3 f2:4 f3:5 Shuffled order: f3:0 f1:1 f1:2 f2:3 f2:4 f3:5 

Though the number of them is not 3^2, but 3*2.