How to make a function that returns a random string based on different characters that have certain-CodePudding

For my specific assignment, I should create a function that takes in a positive integer width and a Boolean value. The function should then return a string based on the given global variables:

      ARROWS = ‘<>^v’
      PERIOD = ‘.’
      ADDITION = ‘ ’
      MULTIPLICATION = ‘X’
      RANDOM_THINGS = ‘*|’

The width integer should determine the length of the string and the Boolean value should determine whether or not the global variable RANDOM_THINGS can be used (if set to false, only use the first four variables).

So far my code accounts for this part. It is as follows:

  Def function(n, true_or_false)
     arrows = random.choice(ARROWS) #to separate the characters in the string
     list1 = (arrows, ADDITION, MULTIPLICATION, PERIOD)
     x = random.choices(list1, k =n)
     print(“”.join(x))

This is where the probability comes in: Each character of the string should have a 5/6 chance of being from the variables ARROWS, ADDITION or MULTIPLICATION and a 1/6 chance of being from the variable PERIOD. If the Boolean value is set to True, there should be a 50% chance that one (and only one) character in the string is replaced by a character in the RANDOM_THINGS variable.

CodePudding user response：

from random import choice, choices

ARROWS = '<>^v'
PERIOD = '.'
ADDITION = ' '
MULTIPLICATION = 'X'
RANDOM_THINGS = '*|'

def function(n, maybe_replace):
    
    list1 = [None, ADDITION, MULTIPLICATION, PERIOD]
    x = []
    
    # Loop to vary the chosen arrow
    for i in range(n):
        
        # update chosen arrow and first element in list1
        list1[0] = choice(ARROWS)
       
        # 5/6 is 5 times greater than 1/6
        # so use 5 and 1 at probability weights
        x.append(*choices(list1, weights=[5, 5, 5, 1], k=1))
    
    if maybe_replace:
        
        position = choice(range(n))
        others = [x[position], choice(RANDOM_THINGS)]
        
        # others has two elements, so 50% prob for each...
        x[position] = choice(others)
        
    print("".join(x))

CodePudding user response：

Do you mean this?

def function(n, includeothers = False):
    alphabet = ARROWS PERIOD ADDITION MULTIPLICATION
    if includeothers:
        alphabet  = RANDOM_THINGS
    x = random.choices(alphabet, k =n)
    print(“”.join(x))

Did you want all of the ARROWS in the list of options? What you had would only include ONE if the ARROWS and ignore the rest.