I know there are some similar questions,but all of them just return a list,I just need a number,each time random generate a unique single number:
Now I'm using Python to build a loop:
for _ in range(9):
random_number=random.randint(1,11)
print(random_number)
My question is how to make sure each time the random_number is unique,not repeating in all 9 times. I don't need a list,just the random_number
You can use random.sample over the target range of numbers to pick a desired number of unique numbers from them:
import random
for random_number in random.sample(range(1, 12), 9):
print(random_number)
If you want each number to be unique from all the others, you don't want a truly random number each time (it's in fact very improbable for random numbers within a small set to be unique). I'd do it like this:
numbers = list(range(1, 12))
random.shuffle(numbers)
for random_number in numbers[:9]:
print(random_number)
Essentially it's like taking a deck of cards, shuffling it, and then dealing the top nine cards. You'll get nine "random" cards but they will never repeat since they're coming from the same deck and you aren't reshuffling in between deals (consequently they aren't truly random, which is why card-counting is a thing).
random.sample is definitely the way to go, but another idea is to initialize all numbers in a list and then randomly pop one at a time:
import random
numbers = list(range(1, 12))
for _ in range(9):
random_index = random.randint(0, len(numbers)-1)
selected_rn = numbers.pop(random_index)
This has a nice interpretation: a predefined sequence of numbers that you point to a number of times and grab one number at random.
This produces a random number each time the codes are run, using numpy
import numpy as np
np.random.sample(1)[0]
Other applications can be added easily to any context. for example when range is defined:
import numpy as np
each_time_new_random_number = np.random.sample(1)[0]
start = 5
stop = 16
start + (stop-start)*each_time_new_random_number
Good wind to your sails.
