How to generate random number with the specific length in python

Question

Let say I need a 3 digit number  so it would be something like     gt  gt  gt  random 3  563  or   gt  gt  gt  random 5  26748  gt  gt  random 2  56

User · Answer

You could write yourself a little function to do what you want:

import random
def randomDigits(digits):
    lower = 10**(digits-1)
    upper = 10**digits - 1
    return random.randint(lower, upper)

Basically, 10**(digits-1) gives you the smallest {digit}-digit number, and 10**digits - 1 gives you the largest {digit}-digit number (which happens to be the smallest {digit+1}-digit number minus 1!). Then we just take a random integer from that range.

User · Answer

To get a random 3-digit number   from random import randint randint 100  999     randint is inclusive at both ends    assuming you really meant three digits  rather than  up to three digits     To use an arbitrary number of digits   from random import randint  def random with N digits n       range start   10   n-1      range end    10  n -1     return randint range start  range end   print random with N digits 2  print random with N digits 3  print random with N digits 4    Output   33 124 5127

User · Answer

From the official documentation  does it not seem that the sample   method is appropriate for this purpose   import random  def random digits n       num   range 0  10      lst   random sample num  n      print str lst  strip         Output    gt  gt  gt random digits 5  2  5  1  0  4

User · Answer

If you need a 3 digit number and want 001-099 to be valid numbers you should still use randrange randint as it is quicker than alternatives  Just add the neccessary preceding zeros when converting to a string   import random num   random randrange 1  10  3    using format num with zeros      03   format num    using string s zfill num with zeros   str num  zfill 3    Alternatively if you don t want to save the random number as an int you can just do it as a oneliner      03   format random randrange 1  10  3     python 3 6  only oneliner   f  random randrange 1  10  3  03     Example outputs of the above are     026   255    512    Implemented as a function   import random  def n len rand len   floor 1       top   10  len      if floor  gt  top          raise ValueError f Floor  floor  must be less than requested top  top        return f  random randrange floor  top  0 len

User · Answer

I really liked the answer of RichieHindle  however I liked the question as an exercise  Here s a brute force implementation using strings    import random first   random randint 1 9  first   str first  n   5  nrs    str random randrange 10   for i in range n-1   for i in range len nrs            first    str nrs i    print str first

User · Answer

Does 0 count as a possible first digit  If so  then you need random randint 0 10  n-1   If not  random randint 10   n-1  10  n-1   And if zero is never allowed  then you ll have to explicitly reject numbers with a zero in them  or draw n random randint 1 9  numbers   Aside  it is interesting that randint a b  uses somewhat non-pythonic  indexing  to get a random number a  lt   n  lt   b  One might have expected it to work like range  and produce a random number a  lt   n  lt  b   Note the closed upper interval    Given the responses in the comments about randrange  note that these can be replaced with the cleaner random randrange 0 10  n   random randrange 10   n-1  10  n  and random randrange 1 10

User · Answer

If you want it as a string  for example  a 10-digit phone number  you can use this   n   10    join       format randint 0  9   for num in range 0  n

User · Answer

You could create a function who consumes an list of int  transforms in string to concatenate and cast do int again  something like this   import random  def generate random number length       return int    join  str random randint 0 10   for   in range length

[python] How to generate random number with the specific length in python

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