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This list would be made of the results of multiplying any 2 integers in the range, including 2 of the same integer. The order of the output list doesn't matter, nor does the presence of duplicate integers.

For example, given a range of 1 to 3, the output should be something like `[1, 2, 3, 4, 6, 9]`.
Top Answer
import itertools
[x*y for x, y in itertools.combinations_with_replacement([1,2,3], 2)]
Answer #2
Using list comprehension.

def get_products(minFactor, maxFactor):
  """Assumes first parameter < 2nd parameter."""
  products = [
    for multiplicand in range(minFactor, maxFactor+1)
    for multiplier in range(multiplicand, maxFactor+1)
  return products

Try in []( --- along with other ways to do it.

This will return without duplicates---just so it's easier to verify the correctness of the output.

Otherwise, replace the 2nd for loop's range to `range(minFactor, maxFactor+1)`.

# References

[ --- List Comprehensions](
Answer #3
Using only builtins, you can do something like the following

::: tio lZHNToUwEIX3fYoxLijKT27uzoSFb@DemKaGgr3eDk0Lhj49DgWEG@PC2Uwy3@mZ4WBD/9Hhebq/KwfvyneNpcIvsHHMmDa2cz1I11rpvGIsNgfVz6h4du1gFPYvkfB0lRSyroVcGU/yMfmLBCKsVg0Y@akEDsbzMQvpEwMqb7rrPKONmtRGY4RphFdyOUI5HqAJtMTJQPD1LU6azoEmJTiJreKbdbb7PMJp3bvpL//Qz4XzLfAAl5upbghg189m2123744XF9JahTXH5UOc6ge3v6Of0oAQKI0SAqoKEiGM1ChEsljScZ6uWMOObY7b81@57IHPvBiz2MKisy5muqpTNlHlI5ymPMD5Gw
§§§ python python3
#!/usr/bin/env python

import argparse

parser = argparse.ArgumentParser()

def make_nums(x,y):
    smol_num = int(min(x,y))
    larg_num = int(max(x,y))
    my_array = []
    for i in range(smol_num, larg_num + 1):
        for j in range(smol_num, larg_num + 1):
            n = i * j
            if n not in my_array:
    return my_array

if __name__ == '__main__':
    args = parser.parse_args()
    my_array = make_nums(args.x,args.y)


If the above code exists on your desktop as ``, you might execute it as below...

![Screenshot 2020-06-30 at 16.45.57.png](/image?hash=bae1fa294c3ecc4b8e9a8a301075029e09222ddd6cc7d77bd7ccae90ec0b6f14)

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