Making random phone number

Question

I want to create phone number generator with the following restrictions:

• start with 0.
• Second number must be 1.
• Third number can be on these 2,3,5.
• Number is 11 digits.
• Don't want the same number twice.

This is the code I used:

# import module import random as r

ph_no = []
ph_no2 = []

# the first number should be in the range of 6 to 9
ph_no.append(r.randint(0, 1))

ph_no2.append(r.randint(1, ))
# the for loop is used to append the other 9 numbers.
# the other 9 numbers can be in the range of 0 to 9.
for i in range(1, 10):
    ph_no.append(r.randint(0, 9))
    ph_no2.append(r.randint(0,9))

# printing the number
for i in ph_no:
    print(i, end="")

This is the problem:

ph_no2.append(r.randint(1, ))
TypeError: Random.randint() missing 1 required positional argument: 
'b'

I want help to make this code work.

Edite:

I used this code to print the output into a txt file but it doesn't print all the output:

with open("textfile.txt", "w") as external_file:
    add_text = "This text will be added to the file"
    print(pn(), file=external_file)
    external_file.close()

Answer 1

randint has an argument for start and stop. You're missing one

# 0 is start, 10 is stop
>>> r.randint(0,10)
6

Use ph_no2.append(r.randint(start, stop))

---

Here's a phone number generator with your requirements:

import random as r


def pn():

    # start with 0.
    # Second number must be 1.
    number = '01'

    # Third number can be on these 2,3,5.
    number += r.choice('235')

    # Number is 11 digits.
    for i in range(8):
        number += r.choice('0123456789')

    return number

>>> pn()
'01523693965'

---

Edit:

To get n numbers, and none are the same

# Set does not have duplicates
all_numbers = set()
n = 10

while len(all_numbers) < n:
    # Continue adding until you have `n` numbers
    x = pn()
    print(x)
    all_numbers.add(x)

01561915353
01558729923
01512093981
01589320526
...

set docs

Answer 2

This uses random's random.randint(a, b) and random.choice(seq):

#!/usr/bin/env python
import random


def random_number():
    number = list(
        str(random.randint(11111111111, 99999999999))
    )  # 11 random digits to str
    number[0] = '0'  # start with 0
    number[1] = '1'  # second number must be 1
    number[2] = str(random.choice([2, 3, 5]))  # third number can be 2, 3, or 5

    return ''.join(number)

Usage: random_number() returns a random number

for i in range(0, 9):
    print(random_number())

Output example:

01247829529
01389527108
01513481001
01284462585
01305116767
01219669837
01239182461
01245292538
01284163077

Explanation: Works by casting int -> str -> list -> editing the individual numbers in place -> rejoining to a single str -> int()

Answer 3

import random as r
for x in range(10):
  ph_no = [0,1]   # first 2 numbers
  ph_no.append(*r.sample([2,3,5],1))  # Third number

  #remaining numbers
  while len(ph_no)<11:
    k=r.randint(0, 9)
    if k not in ph_no[1:]: # we are skipping zero as we cannot have 11 unique numbers from 0-9
      ph_no.append(k)
    else:
      continue
  print(''.join(str(x) for x in ph_no))

#output
01327486095
01508746239
01567293084
01346857920
01520693874
01369405827
01384290657
01563082479
01520947386
01532460879