Python | Avoid previous value from random selection from list -

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basically want code print out 5 wrong() func's without there being 2 in row of same text. , of course don't want luck. :)

though don't worry part of printing out 5 wrong()s, want make sure if use function @ least twice 100% sure previous value won't same next.

for example, want avoid:

wrong! wrong!

though still fine:

wrong! incorrect! wrong!

my code:

import random  def wrong():     wrong_stats=["\n wrong!","\n tough luck!","\n better luck next time!","\n not there yet!","\n incorrect!"]     rand = random.choice(wrong_stats)     rand3 = random.choice(wrong_stats)     norep(rand,rand3,wrong_stats)  def norep(rand,rand3,wrong_stats):     if rand == rand3:         same = true         while same:             rand = random.choice(wrong_stats)             if rand != rand3:                 print(rand)                 break      elif rand != rand3:         print(rand)  wrong() wrong() wrong() wrong() wrong()

you'll need keep track of last value returned; could

  • use module global (usually messy in practice),
  • or turn class (kind of verbose),
  • or keep track externally , pass in each time (clunky , tedious),

but imo nicest way turn wrong function generator instead: way can keep track of last returned value in generator execution state, , avoid next time around, without having worry in external code anywhere.

def wrong():     wrong_stats = ["wrong!","tough luck!","better luck next time!","not there yet!","incorrect!"]     previous_value = none     while true:         value = random.choice(wrong_stats)         if value != previous_value:             yield value             previous_value = value

and usage:

w = wrong() in range(5):     print(next(w))  # tough luck! # incorrect! # not there yet! # tough luck! # better luck next time!

you can keep calling next generator , produce infinite number of strings without ever repeating previous value.


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