Generate random integers between 0 and 9

Question

How can I generate random integers between 0 and 9 (inclusive) in Python?

For example, 0, 1, 2, 3, 4

Answer 1

I had better luck with this for Python 3.6

str_Key = ""                                                                                                
str_RandomKey = ""                                                                                          
for int_I in range(128):                                                                                    
      str_Key = random.choice('0123456789')
      str_RandomKey = str_RandomKey + str_Key 

Just add characters like 'ABCD' and 'abcd' or '^!~=-><' to alter the character pool to pull from, change the range to alter the number of characters generated.

Answer 2

Generating random integers between 0 and 9.

import numpy
X = numpy.random.randint(0, 10, size=10)
print(X)

Output:

[4 8 0 4 9 6 9 9 0 7]

Answer 3

In case of continuous numbers randint or randrange are probably the best choices but if you have several distinct values in a sequence (i.e. a list) you could also use choice:

>>> import random
>>> values = list(range(10))
>>> random.choice(values)
5

choice also works for one item from a not-continuous sample:

>>> values = [1, 2, 3, 5, 7, 10]
>>> random.choice(values)
7

If you need it "cryptographically strong" there's also a secrets.choice in python 3.6 and newer:

>>> import secrets
>>> values = list(range(10))
>>> secrets.choice(values)
2

Answer 4

if you want to use numpy then use the following:

import numpy as np
print(np.random.randint(0,10))

Answer 5

The secrets module is new in Python 3.6. This is better than the random module for cryptography or security uses.

To randomly print an integer in the inclusive range 0-9:

from secrets import randbelow
print(randbelow(10))

For details, see PEP 506.

Answer 6

from random import randint

x = [randint(0, 9) for p in range(0, 10)]

This generates 10 pseudorandom integers in range 0 to 9 inclusive.