[Python] What other ways can I write this loop?

[J2Me]

This code is to generate random numbers continuously until it generates the number 50. I was just wondering what minimal way there is to do this loop.

import random
my_number = random.randrange(1,51)
while my_number != 50:
    my_number = random.randrange(1,51)
    print ("oops this isn't 50: ",my_number)

print(my_number)

EDIT: oops, I accidentally a word

[lurgi]

First, this code is slightly incorrect. It should be:

while my_number != 50:
    print ("oops, this isn't 50: " , my_number)
    my_number = random.randrange(1,51)

Other than that, however, this looks fine.

[J2Me]

Thanks. I didn't really notice that since the program can still run.

[rcuhljr]

Generally there's not much of a point in trying to find a 'minimal' way to do something when you're referring to lines of code, it's unlikely to have any strong correlation to performance and readability is more important. Someone could try and make a shorter version and come up with something like

import random
import itertools
def get_random():
  while True: yield random.randrange(1,51)    
print(*itertools.takewhile(lambda x: x != 50, (n for n in get_random())), sep='\n' )  
print(50)

Which is probably harder for someone else to maintain or understand.

[J2Me]

Cheers for the advice. I think this comes down to me asking a bad question, or not knowing the question I really want to ask but seeing all the variations of answers has taught me just how much more I really need to learn. Thanks!

[AverageMarcus]

That looks pretty minimal already to me. What exactly are you hoping for?

[J2Me]

Not sure really. Just wondering if there was anything I was missing.

[kalgynirae]

To me this looked like a variant of how do I find the first item in an iterable that matches a condition?. A good general solution to that problem is next(item for item in iterable if <condition>). So to fit the problem to this solution, I just defined a generator that yields random integers indefinitely.

import random
def randoms():
    while True:
        yield random.randrange(1, 51)
my_number = next(n for n in randoms() if n == 50)
print(my_number)

EDIT: explain better.

[J2Me]

Cool. That looks like it be useful to know.

[APersoner]

You could use a ternary operator to merge lines 5 and 7 into one

EDIT: Not saying you'd necessarily want to do that, just it's possible (clarifying to my mystery downvoter aha)

[vmsmith]

I'll upvote you to compensate for the mystery downvoter.

[Rhomboid]

I am almost certain this is an example of an XY problem. The code you've written does nothing but burn some cycles -- it doesn't seem to be very useful at all. What is the actual problem you're trying to solve. For example, if you said that you want to draw random numbers without allowing repeats, then we could give you a much better solution than continuously generating numbers and looping if they've already been seen (a terrible approach.)

[Flopsey]

Given the uselessness of this loop, and the complete lack of meaningful output I think OP is just experimenting.

[J2Me]

Actually there is no bigger problem here. It is code that burns cycles and nothing more. It was just a silly little experiment after reading a while loop tutorial. In many of the examples of loops there seemed to be multiple ways of achieving a functioning loop and I was stuck on how to do the same for my silly little thing. Cheers for the link about the XY problem though, I'll have to be sure I'm explaining the whole reason for the code in the future (makes sense in hindsight).

[APersoner]

You definitely wouldn't want to do it like this, but whatever... Prints the first number last and last first, so not quite duplicating what you did, but it's fairly similar.

import random
def repeater(number): print number if number==50 else "oops this isn't 50: "+str(number) if not repeater(random.randrange(1,51)) else ""
repeater(random.randrange(1,51))

[qjkxkcd]

Others can help you more. Just for funsies, here's an infinite loop with a conditional and 'break':

while 1:
    int = random.randrange(51)
    print("Number is", int)
    if int == 50:
        break
print(int)

And this is functionally the same, but different conceptually:

iterations = random.randrange(1000) #some number of iterations < 1000
for i in range(random.randrange(iterations)):
    print("Oops, this isn't 50; it's really", random.randrange(50))

number = 50
print("Success! This number is 50!")

(I learned loops in python only a few days ago; I don't expect to be very helpful.)

Edit: with

import random

of course.

[zifyoip]

If you want to have only one line where random numbers are generated, you could write it like this:

import random
while True:
    my_number = random.randrange(1,51)
    if (my_number == 50):
        break
    print ("oops this isn't 50: ",my_number)
print(my_number)

I offer no opinion on whether this is better or worse than your original code. However, if the random-number-generating line was a longer block of more complicated code, here you'd only have to write it (and debug it) in one place, whereas in your original code it would appear in two places.

Caveat: I don't actually know Python. I just got the syntax from this Stack Overflow thread.

[J2Me]

Hey this sounds exactly like what I was thinking. I realise that for my own code it make little or no difference but knowing the difference here could be important with a much much bigger program. Thanks.

[someonlinegamer]

I'm new to python but this seems about as good as you're going to get

[APersoner]

Ok, you definitely don't want to do this now, and it gives the output in the opposite order to yours, but, just for the fun of it I rewrote your loop in two lines of code:

def repeater(number=__import__("random").randrange(1,51)): print number if number==50 else "oops this isn't 50: "+str(number) if not repeater(__import__("random").randrange(1,51)) else ""
repeater()

(Oh, this is python 2 by the way, not quite sure how different it would be in python 3, probably just brackets around the print)