CODEWITHSUNDEEP
pythonpython-3.x
Ylonec
Ylonec

Reputation: 35

Unwanted random iterator in For-Loop

I tried to create a function for generating a set number (numbersToChoose) of values between two other values (startFrom and stopAt) but for some reason the iterator (?) in the second for-loop (line 7), in this case a seems to be randomly generated even though I don't see any reason for that.

def chooseRandomNumbers(numbersToChoose, startFrom, stopAt):

 numbers = [(randrange(startFrom, stopAt))]  # initialize empty list of numbers

 for n in range(numbersToChoose-1):  # choose random number from 1 to 49 a total of 7 times
    x = randrange(startFrom, stopAt+1)
    for a in numbers:  # check for already existing elements in "numbers"
        if x == numbers[a]:  # if new random number already exists in "numbers" discard and run loop again
            n -= 1  # decreases n by 1 so a new (hopefully not already existing) number can be generated
        else:  # generated numbers does not exist in "numbers" yet
            numbers.append(x)  # appends randomly chosen number to list "numbers"

 return numbers  # returns said list "numbers"

Any advice on how to deal with this is greatly appreciated. Also please tell me if possible anything else bad in the code (just started python).

Upvotes: 3

Views: 85

Answers (1)

Pruthvi Raj
Pruthvi Raj

Reputation: 3036

Your code to check if generated number is already in the list is wrong. for a in numbers, in this loop, you are using numbers[a] while a is a member of list but not the index of the member. use in to test if a number is in list:

from random import randrange

def chooseRandomNumbers(numbersToChoose, startFrom, stopAt):
     numbers = []
     for n in range(numbersToChoose):
        x = randrange(startFrom, stopAt+1)
        if not x in numbers:
            numbers.append(x)
        else:
            n -= 1
     return numbers

or simply:

from random import sample
def chooseRandomNumbers(numbersToChoose, startFrom, stopAt):
    return sample(range(startFrom,stopAt+1),numbersToChoose)

Upvotes: 2

Related Questions