How to append five random elements from one 5-element list to another list without adding any duplic-CodePudding

def f1():
    a = ['one', 'two', 'three', 'four', 'five']
    def f2():
        b = []
       
        for i in range(5):
            if random.choice(a) not in b:
                b.append(random.choice(a))
            else:
                return f2()
        print(b)
    f2()
f1()

I know that random.choice(a) will change every time it is ran but I need to check a random.choice(a) and append that same random.choice(a) to a list if it is not already present.

CodePudding user response：

You can use random sample to create a new shuffled list:

def f1():
    a = ['one', 'two', 'three', 'four', 'five']
    def f2():
        b = random.sample(a, len(a))
                
        print(b)
    f2()
f1()

CodePudding user response：

import random
def f(l):
  l1=[]
  for i in range(1,6):
       if random.choice(l) not in l1:
             l1.append(random.choice(l)
       else:
             return f(l)
l=["one","two","three","four","five"]
f(l)