generate random digits with length of 5

[sotirac]

Wondering if there is a better way to generate string of numbers with
a length of 5 which also can have a 0 in the front of the number.


<pre>
random_number = random.sample([0,1,2,3,4,5,6,7 ,8,9], 5) # choose 5
elements
code = 'this is a string' + str(random_numb er[0]) +
str(random_numb er[1]) + str(random_numb er[2]) + str(random_numb er[3])
+ str(random_numb er[4])
</pre>

[Chris Rebert]

Re: generate random digits with length of 5

On Sun, Sep 28, 2008 at 12:59 PM, sotirac <sotirac@gmail. comwrote:

Wondering if there is a better way to generate string of numbers with
a length of 5 which also can have a 0 in the front of the number.
>
>
<pre>
random_number = random.sample([0,1,2,3,4,5,6,7 ,8,9], 5) # choose 5
elements
code = 'this is a string' + str(random_numb er[0]) +
str(random_numb er[1]) + str(random_numb er[2]) + str(random_numb er[3])
+ str(random_numb er[4])

code = ''.join(str(dig it) for digit in random_number)

Regards,
Chris

</pre>
>
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http://mail.python.org/mailman/listinfo/python-list

>

--
Follow the path of the Iguana...

http://rebertia.com

[Aaron \Castironpi\ Brady]

Re: generate random digits with length of 5

On Sep 28, 2:59 pm, sotirac <soti...@gmail. comwrote:

Wondering if there is a better way to generate string of numbers with
a length of 5 which also can have a 0 in the front of the number.
>
<pre>
 random_number = random.sample([0,1,2,3,4,5,6,7 ,8,9], 5) # choose 5
elements
 code = 'this is a string' + str(random_numb er[0]) +
str(random_numb er[1]) + str(random_numb er[2]) + str(random_numb er[3])
+ str(random_numb er[4])
</pre>

'%05i'%random.r andint(0,99999)

[Gary M. Josack]

Re: generate random digits with length of 5

Chris Rebert wrote:

On Sun, Sep 28, 2008 at 12:59 PM, sotirac <sotirac@gmail. comwrote:
>

>Wondering if there is a better way to generate string of numbers with
>a length of 5 which also can have a 0 in the front of the number.
>>
>>
><pre>
> random_number = random.sample([0,1,2,3,4,5,6,7 ,8,9], 5) # choose 5
>elements
> code = 'this is a string' + str(random_numb er[0]) +
>str(random_num ber[1]) + str(random_numb er[2]) + str(random_numb er[3])
>+ str(random_numb er[4])
>>

>
code = ''.join(str(dig it) for digit in random_number)
>
Regards,
Chris
>
>

></pre>
>>
>--
>http://mail.python.org/mailman/listinfo/python-list
>>
>>

will random.randint( 10000,99999) work for you?

[Gary M. Josack]

Re: generate random digits with length of 5

Aaron "Castironpi " Brady wrote:

On Sep 28, 2:59 pm, sotirac <soti...@gmail. comwrote:
>

>Wondering if there is a better way to generate string of numbers with
>a length of 5 which also can have a 0 in the front of the number.
>>
><pre>
> random_number = random.sample([0,1,2,3,4,5,6,7 ,8,9], 5) # choose 5
>elements
> code = 'this is a string' + str(random_numb er[0]) +
>str(random_num ber[1]) + str(random_numb er[2]) + str(random_numb er[3])
>+ str(random_numb er[4])
></pre>
>>

>
'%05i'%random.r andint(0,99999)
--

http://mail.python.org/mailman/listinfo/python-list

>

This produces numbers other than 5 digit numbers. making the start
number 10000 should be fine.

[bearophileHUGS@lycos.com]

Re: generate random digits with length of 5

sotirac:

random_number = random.sample([0,1,2,3,4,5,6,7 ,8,9], 5) # choose 5 elements

But note that's without replacement. So if you want really random
numbers you can do this:

>>from string import digits
>>from random import choice
>>"".join(choic e(digits) for d in xrange(5))

'93898'

If you need more speed you can invent other solutions, like (but I
don't know if it's faster):

>>from random import shuffle
>>ldigits = list(digits)
>>ldigits

['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']

>>shuffle(ldigi ts)
>>ldigits

['3', '8', '6', '4', '9', '7', '5', '2', '0', '1']

>>"".join(ldigi ts[:5])

'38649'

But this may be the most efficient way:

>>from random import randrange
>>str(randrange (100000)).zfill (5)

'37802'

Bye,
bearophile

[Gary M. Josack]

Re: generate random digits with length of 5

Gary M. Josack wrote:

Aaron "Castironpi " Brady wrote:

>On Sep 28, 2:59 pm, sotirac <soti...@gmail. comwrote:
>>

>>Wondering if there is a better way to generate string of numbers with
>>a length of 5 which also can have a 0 in the front of the number.
>>>
>><pre>
>> random_number = random.sample([0,1,2,3,4,5,6,7 ,8,9], 5) # choose 5
>>elements
>> code = 'this is a string' + str(random_numb er[0]) +
>>str(random_nu mber[1]) + str(random_numb er[2]) + str(random_numb er[3])
>>+ str(random_numb er[4])
>></pre>
>>>

>>
>'%05i'%random. randint(0,99999 )
>--
>http://mail.python.org/mailman/listinfo/python-list
>>

This produces numbers other than 5 digit numbers. making the start
number 10000 should be fine.
--
http://mail.python.org/mailman/listinfo/python-list

nevermind. my brain is tired tonight. this is the best solution.

[Mensanator]

Re: generate random digits with length of 5

On Sep 28, 3:11�pm, "Gary M. Josack" <g...@byoteki.c omwrote:

Chris Rebert wrote:

On Sun, Sep 28, 2008 at 12:59 PM, sotirac <soti...@gmail. comwrote:

>

Wondering if there is a better way to generate string of numbers with
a length of 5 which also can have a 0 in the front of the number.

>

<pre>
�random_numbe r = random.sample([0,1,2,3,4,5,6,7 ,8,9], 5) # choose 5
elements
�code = 'this is a string' + str(random_numb er[0]) +
str(random_numb er[1]) + str(random_numb er[2]) + str(random_numb er[3])
+ str(random_numb er[4])

>

code = ''.join(str(dig it) for digit in random_number)

>

Regards,
Chris

>

</pre>

>

--
>http://mail.python.org/mailman/listinfo/python-list

>
will random.randint( 10000,99999) work for you?

It doesn't meet the OP's requirement that the number
can start with 0. Also, the method the OP asks about
returns a list of unique numbers, so no number can
be duplicated. He can get 02468 but not 13345.

Now, IF it's ok to have an arbitrary number of leading
0s, he can do this:

>>str(random.ra ndint(0,99999)) .zfill(5)

'00089'

>>str(random.ra ndint(0,99999)) .zfill(5)

'63782'

>>str(random.ra ndint(0,99999)) .zfill(5)

'63613'

>>str(random.ra ndint(0,99999)) .zfill(5)

'22315'

[Aaron \Castironpi\ Brady]

Re: generate random digits with length of 5

On Sep 28, 3:44 pm, Mensanator <mensana...@aol .comwrote:

On Sep 28, 3:11 pm, "Gary M. Josack" <g...@byoteki.c omwrote:
>
>
>

Chris Rebert wrote:

On Sun, Sep 28, 2008 at 12:59 PM, sotirac <soti...@gmail. comwrote:

>

>Wondering if there is a better way to generate string of numbers with
>a length of 5 which also can have a 0 in the front of the number.

>

><pre>
>random_numbe r = random.sample([0,1,2,3,4,5,6,7 ,8,9], 5) # choose 5
>elements
>code = 'this is a string' + str(random_numb er[0]) +
>str(random_num ber[1]) + str(random_numb er[2]) + str(random_numb er[3])
>+ str(random_numb er[4])

>

code = ''.join(str(dig it) for digit in random_number)

>

Regards,
Chris

>

></pre>

>

>--
>>http://mail.python.org/mailman/listinfo/python-list

>

will random.randint( 10000,99999) work for you?

>
It doesn't meet the OP's requirement that the number
can start with 0. Also, the method the OP asks about
returns a list of unique numbers, so no number can
be duplicated. He can get 02468 but not 13345.
>
Now, IF it's ok to have an arbitrary number of leading
0s, he can do this:
>

>str(random.ran dint(0,99999)). zfill(5)

'00089'

>str(random.ran dint(0,99999)). zfill(5)

'63782'

>str(random.ran dint(0,99999)). zfill(5)

'63613'

>str(random.ran dint(0,99999)). zfill(5)

>
'22315'

Is a while loop until there are 5 distinct digits best otherwise?

while 1:
a= '%05i'% random.randint( 0, 99999 )
if len( set( a ) )== 5: break

[Tim Chase]

Re: generate random digits with length of 5

Wondering if there is a better way to generate string of numbers with

a length of 5 which also can have a 0 in the front of the number.

If you want to resample the same digit multiple times, either of these
two will do:

>>from random import choice
>>''.join(str(c hoice(range(10) )) for _ in range(5))

'06082'

>>from string import digits
>>''.join(choic e(digits) for _ in range(5))

'09355'


If you need to prevent the digits from being reused

>>d = list(digits)
>>random.shuffl e(digit)
>>''.join(d[:5])

'03195'

I suspect that the zfill responses don't have the property of equally
distributed "randomness ", as the first digit may more likely be a zero.
The methods here should give equal probabilities for each choice in
each place.

-tkc

[=?ISO-8859-1?Q?Michael_Str=F6der?=]

Re: generate random digits with length of 5

Gary M. Josack wrote:

Aaron "Castironpi " Brady wrote:

>On Sep 28, 2:59 pm, sotirac <soti...@gmail. comwrote:
>>

>>Wondering if there is a better way to generate string of numbers with
>>a length of 5 which also can have a 0 in the front of the number.
>>>
>><pre>
>> random_number = random.sample([0,1,2,3,4,5,6,7 ,8,9], 5) # choose 5
>>elements
>> code = 'this is a string' + str(random_numb er[0]) +
>>str(random_nu mber[1]) + str(random_numb er[2]) + str(random_numb er[3])
>>+ str(random_numb er[4])
>></pre>
>>>

>>
>'%05i'%random. randint(0,99999 )
>--
>http://mail.python.org/mailman/listinfo/python-list
>>

This produces numbers other than 5 digit numbers. making the start
number 10000 should be fine.

Why do you think it's wrong?

>>import random
>>'%05i'%random .randint(0,9999 9)

'09449'

>>>

IMO it's exactly what was required.

Ciao, Michael.

[Aaron \Castironpi\ Brady]

Re: generate random digits with length of 5

On Sep 28, 4:08 pm, Michael Ströder <mich...@stroed er.comwrote:

Gary M. Josack wrote:

Aaron "Castironpi " Brady wrote:

On Sep 28, 2:59 pm, sotirac <soti...@gmail. comwrote:

>

>Wondering if there is a better way to generate string of numbers with
>a length of 5 which also can have a 0 in the front of the number.

>

><pre>
> random_number = random.sample([0,1,2,3,4,5,6,7 ,8,9], 5) # choose5
>elements
> code = 'this is a string' + str(random_numb er[0]) +
>str(random_num ber[1]) + str(random_numb er[2]) + str(random_numb er[3])
>+ str(random_numb er[4])
></pre>

>

'%05i'%random.r andint(0,99999)
--
>http://mail.python.org/mailman/listinfo/python-list

>

This produces numbers other than 5 digit numbers. making the start
number 10000 should be fine.

>
Why do you think it's wrong?
>
>
>

>import random
>'%05i'%random. randint(0,99999 )

'09449'
>
IMO it's exactly what was required.
>
Ciao, Michael.

As you read, there isn't agreement on whether the OP wanted
replacement. His original code didn't; his spec seemed to.

[Mensanator]

Re: generate random digits with length of 5

On Sep 28, 3:54�pm, "Aaron \"Castironpi \" Brady"
<castiro...@gma il.comwrote:

On Sep 28, 3:44�pm, Mensanator <mensana...@aol .comwrote:
>
>
>
>
>

On Sep 28, 3:11 pm, "Gary M. Josack" <g...@byoteki.c omwrote:

>

Chris Rebert wrote:
On Sun, Sep 28, 2008 at 12:59 PM, sotirac <soti...@gmail. comwrote:

>

Wondering if there is a better way to generate string of numbers with
a length of 5 which also can have a 0 in the front of the number.

>

<pre>
random_number = random.sample([0,1,2,3,4,5,6,7 ,8,9], 5) # choose5
elements
code = 'this is a string' + str(random_numb er[0]) +
str(random_numb er[1]) + str(random_numb er[2]) + str(random_numb er[3])
+ str(random_numb er[4])

>

code = ''.join(str(dig it) for digit in random_number)

>

Regards,
Chris

>

</pre>

>

--
>http://mail.python.org/mailman/listinfo/python-list

>

will random.randint( 10000,99999) work for you?

>

It doesn't meet the OP's requirement that the number
can start with 0. Also, the method the OP asks about
returns a list of unique numbers, so no number can
be duplicated. He can get 02468 but not 13345.

>

Now, IF it's ok to have an arbitrary number of leading
0s, he can do this:

>

>>str(random.ra ndint(0,99999)) .zfill(5)

'00089'

>>str(random.ra ndint(0,99999)) .zfill(5)

'63782'

>>str(random.ra ndint(0,99999)) .zfill(5)

'63613'

>>str(random.ra ndint(0,99999)) .zfill(5)

>

'22315'

>
Is a while loop until there are 5 distinct digits best otherwise?

Of course not.

>
while 1:
� a= '%05i'% random.randint( 0, 99999 )
� if len( set( a ) )== 5: break

How is this better than the OP's original code?

[Mensanator]

Re: generate random digits with length of 5

On Sep 28, 4:02�pm, Tim Chase <python.l...@ti m.thechases.com wrote:

Wondering if there is a better way to generate string of numbers with
a length of 5 which also can have a 0 in the front of the number.

>
If you want to resample the same digit multiple times, either of these
two will do:
>
�>>from random import choice
�>>''.join(st r(choice(range( 10))) for _ in range(5))
'06082'
>
�>>from string import digits
�>>''.join(ch oice(digits) for _ in range(5))
'09355'
>
If you need to prevent the digits from being reused
>
�>>d = list(digits)
�>>random.shu ffle(digit)
�>>''.join(d[:5])
'03195'
>
I suspect that the zfill responses don't have the property of equally
distributed "randomness ", as the first digit may more likely be a zero.
� The methods here should give equal probabilities for each choice in
each place.

Does he want equal probabilities of each digit in each place?

>
-tkc

[bearophileHUGS@lycos.com]

Re: generate random digits with length of 5

Tim Chase:

I suspect that the zfill responses don't have the property of equally
distributed "randomness ", as the first digit may more likely be a zero.

This code I have shown before:
str(randrange(1 00000)).zfill(5 )

If the numbers are equally distributed in [0, 99999], then the leading
zeros have the correct distribution.

A little test program for you:

from string import digits
from random import choice, randrange
from collections import defaultdict

def main():
N = 1000000

freqs1 = defaultdict(int )
for i in xrange(N):
n = "".join(choice( digits) for d in xrange(5))
freqs1[n[0]] += 1
print [freqs1[str(i)] for i in xrange(10)]

freqs2 = defaultdict(int )
for i in xrange(N):
n = str(randrange(1 00000)).zfill(5 )
freqs2[n[0]] += 1
print [freqs2[str(i)] for i in xrange(10)]

import psyco; psyco.full()
main()

The output:
[100153, 99561, 99683, 100297, 99938, 100162, 99738, 100379, 100398,
99691]
[99734, 100153, 100091, 100683, 99580, 99676, 99671, 100131, 100102,
100179]

Of course with a bit of math you can also demonstrate it :-)

Bye,
bearophile