CODEWITHSUNDEEP
pythonpermutation
AziMez
AziMez

Reputation: 2072

Python Permutations unexpected behavior?

Given the sample below:

from itertools import permutations
p=permutations("abc", 2)

def func():
  for i in p:
      print("values=", i)
        

print("First use case:") 
func()

print("Second use case:") 
func()

print("The End!")

Output:

First use case:
values= ('a', 'b')
values= ('a', 'c')
values= ('b', 'a')
values= ('b', 'c')
values= ('c', 'a')
values= ('c', 'b')
Second use case:
The End!

The question is about the second function CALL, why doesn't print the values?!

Upvotes: 0

Views: 31

Answers (1)

mujjiga
mujjiga

Reputation: 16856

p is and iterator which means once you finish iterating through it there are no more elements in it. You will have to regenerate the iterator.

Like:

def func():
  p=permutations("abc", 2)
  for i in p:
      print("values=", i)

If you have situation where the iterator is generated randomly and you want to use the same random elements then you will have to move them to something like a list.

Example:

from numpy.random import permutation
import numpy as np
p =  list(iter(np.random.randint(0,10,10)))

def func():
  for i in p:
      print("values=", i)

Upvotes: 1

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