Python: How to use random to get two numbers in different orders-CodePudding

I am looking for a fast way to get 2 integers randomly say 1, 2 or 2, 1.

    var1 = random.randint( 1, 2 )
    # do something
    var2 = random.???       # say if var1 == 1, var2 should be 2.
    # do something else

Is there something that can be done to achieve following (Some other syntactical way)

    var2 = ( 1, 2 ) - var1

CodePudding user response：

Many solutions are possible, but the one which matches your idea most would be to use a compact if else statement:

import random

var1 = random.randint(1, 2)
var2 = 2 if var1 == 1 else 1

Another easy way to go is to create a list of values and shuffle them:

import random

l = list(range(1, 3))
random.shuffle(l)
var1 = l[0]
var2 = l[1]

Or use an in build function of random directly sovling your issue:

import random

var1, var2 = random.sample([1, 2], k=2)

CodePudding user response：

I guess you're looking for random.sample:

Return a k length list of unique elements chosen from the population sequence or set. Used for random sampling without replacement.

import random
first, second = random.sample([1, 2], k=2)
print(first, second)

CodePudding user response：

If you should mix the inputs, you could use

a, b = random.sample([1, 2], k=2) # Where k would be count of elements

If you ok that there could be 2 same outputs

a, b = random.choices([1, 2], k=2)  # a == b possible

CodePudding user response：

You asked for "a fast way", so here's one 10 times faster than the other solutions:

from random import random

var1 = 1 if getrandbits(1) else 2
var2 = 2 if var1 == 1 else 1

Benchmark results including two even faster but longer ways:

  62 ns ± 0 ns  if getrandbits(1):
                    var1 = 1
                    var2 = 2
                else:
                    var1 = 2
                    var2 = 1

  65 ns ± 0 ns  if getrandbits(1):
                    var1, var2 = 1, 2
                else:
                    var1, var2 = 2, 1

  82 ns ± 0 ns  var1 = 1 if getrandbits(1) else 2
                var2 = 2 if var1 == 1 else 1

  88 ns ± 2 ns  var1 = 1 if random_() < 0.5 else 2
                var2 = 2 if var1 == 1 else 1

 841 ns ± 3 ns  var1 = randint(1, 2)
                var2 = 2 if var1 == 1 else 1

 868 ns ± 4 ns  var1 = random.randint(1, 2)
                var2 = 1   2 - var1

 875 ns ± 2 ns  var1 = random.randint(1, 2)
                var2 = 2 if var1 == 1 else 1

1430 ns ± 4 ns  l = list(range(1, 3))
                random.shuffle(l)
                var1 = l[0]
                var2 = l[1]

2400 ns ± 4 ns  var1, var2 = random.sample([1, 2], k=2)

Benchmark code (Try it online!):

from timeit import repeat
from statistics import median, stdev
from random import shuffle

setup = '''
import random
from random import randint, sample, random as random_, getrandbits
'''
codes = '''
var1 = random.randint(1, 2)
var2 = 2 if var1 == 1 else 1

var1 = randint(1, 2)
var2 = 2 if var1 == 1 else 1

l = list(range(1, 3))
random.shuffle(l)
var1 = l[0]
var2 = l[1]

var1, var2 = random.sample([1, 2], k=2)

var1 = 1 if random_() < 0.5 else 2
var2 = 2 if var1 == 1 else 1

var1 = 1 if getrandbits(1) else 2
var2 = 2 if var1 == 1 else 1

if getrandbits(1):
    var1 = 1
    var2 = 2
else:
    var1 = 2
    var2 = 1

if getrandbits(1):
    var1, var2 = 1, 2
else:
    var1, var2 = 2, 1

var1 = random.randint(1, 2)
var2 = 1   2 - var1
'''.strip().split('\n\n')

times = {c: [] for c in codes}
def stats(code):
    ts = [t * 1e9 for t in sorted(times[code])[:5]]
    return 'M ns ±  ns ' % (median(ts), stdev(ts))

number = 10000
for _ in range(50):
  for code in codes:
    t = min(repeat(code, setup, number=number)) / number
    times[code].append(t)
for code in sorted(codes, key=stats):
  print(stats(code), code.replace('\n', '\n' ' '*16))
  print()

CodePudding user response：

One approach specific to two integers that would match your original idea closely would be to subtract the chosen first integer from the sum of the two integers:

var1 = random.randint(1, 2)
var2 = 1   2 - var1