First, every item in your zset should have a score of 0. I'm going to
make a temporary copy of the zset so that we don't have any race
conditions, but with work, you can address that. Below is the code to
solve your problem using a zset...

def data_range(conn, data, prefix):
    if ord(prefix[-1]) == 255:
        # we don't want to deal with having to carry
        end = prefix + 20*'\xff'
    else:
        end = prefix[:-1] + chr(ord(prefix[-1]) + 1)
    conn.zinterstore('temporary', data)
    r1 = conn.zadd('temporary', 0, prefix)
    r2 = conn.zadd('temporary', 0, end)
    b = conn.zrank('temporary', prefix)
    e = conn.zrank('temporary', end)
    # if you just want the count...
    # count = e + 1 - b - r1 - r2
    results = conn.zrange('temporary', b + r1, e - r2)
    conn.delete('temporary')
    return len(results), results

Redis scripting can offer a similar solution without copying and
without worrying about race conditions. You can also do a similar
thing just using the scores to store the prefixes of the values, which
would allow you to just use zrangebyscore without needing to
add/remove items or copy the zset, but it is limited to 53 bits
naively, and almost 64 bits if you know what you are doing.

Regards,
 - Josiah

On Wed, Jan 4, 2012 at 6:37 AM, Salimane Adjao Moustapha

In terms of using the score and searching that way... your original
post suggested that you wanted to do prefix lookups. If you still want
to do prefix lookups and are having problems figuring out how to do
it, reply to this thread and I'll point you in the right direction.

Regards,
 - Josiah

On Wed, Jan 4, 2012 at 6:57 PM, Salimane Adjao Moustapha

To find the ids with a given prefix, you need to calculate a start/end
score for performing ZRANGEBYSCORE.1. Calculate the "start" as the
score method above.2. Calculate "end" by replacing all trailing '0'
digits with hex 'f' (technically 'b' is sufficient, and the only '0'
digits should be at the right side of the score).
Looking at an example, if we wanted to find ids with prefixes of 726,
we would generate a "start" score of 0x8370000000. We would also
generate an "end" score of 0x837fffffff. Our example id from before
has a prefix of 726, but also has a score of 0x8372670000, which falls
inside the range that we generated.
In Python:
def score(id):    # step 1    id = map(int, str(id))    # step 2
id.extend((10 - len(id)) * [-1])    # step 3 and 4    return
int(''.join(i+1 for i in id), 16)
def end_score(score):    # get the hex digits back    id =
hex(score)[2:].rstrip('L')    # replace the trailing 0 with 'f'
return int(id.replace('0', 'f'), 16)
Regards,
 - Josiah

On Wed, Jan 4, 2012 at 9:16 PM, Salimane Adjao Moustapha

1. Take your id as a sequence of integers...
726156 -> [7, 2, 6, 1, 5, 6]

2. Append -1 values to the right until you have 10 digits (technically
enough for ids of up to 33 bits)...
[7, 2, 6, 1, 5, 6, -1, -1, -1, -1]

3. Add 1 to all of your digits...
[8, 3, 7, 2, 6, 7, 0, 0, 0, 0]

4. Consider each of these digits as a 4-bit portion of a 40-bit score...
score = 0x8372670000

To find the ids with a given prefix, you need to calculate a start/end
score for performing ZRANGEBYSCORE.

1. Calculate the "start" as the score method above.
2. Calculate "end" by replacing all trailing '0' digits with hex 'f'
(technically 'b' is sufficient, and the only '0' digits should be at
the right side of the score).

Looking at an example, if we wanted to find ids with prefixes of 726,
we would generate a "start" score of 0x8370000000. We would also
generate an "end" score of 0x837fffffff. Our example id from before
has a prefix of 726, but also has a score of 0x8372670000, which falls
inside the range that we generated.

In Python:
def score(id):
    # step 1
    id = map(int, str(id))
    # step 2
    id.extend((10 - len(id)) * [-1])
    # step 3 and 4
    return int(''.join(i+1 for i in id), 16)

def end_score(score):
    # get the hex digits back
    id = hex(score)[2:].rstrip('L')
    # replace the trailing 0 with 'f'
    return int(id.replace('0', 'f'), 16)

On Wed, Jan 4, 2012 at 11:11 PM, Josiah Carlson

1. Take your id as a sequence of integers...
726156 -> [7, 2, 6, 1, 5, 6]

2. Append -1 values to the right until you have 10 digits (technically
enough for ids of up to 33 bits)...
[7, 2, 6, 1, 5, 6, -1, -1, -1, -1]

3. Add 1 to all of your digits...
[8, 3, 7, 2, 6, 7, 0, 0, 0, 0]

4. Left-shift your digits incrementally until you find a '0' digit at
the front, and remember how many times you rotated it for each:
0, [8, 3, 7, 2, 6, 7, 0, 0, 0, 0]
1, [3, 7, 2, 6, 7, 0, 0, 0, 0, 8]
2, [7, 2, 6, 7, 0, 0, 0, 0, 8, 3]
3, [2, 6, 7, 0, 0, 0, 0, 8, 3, 7]
4, [6, 7, 0, 0, 0, 0, 8, 3, 7, 2]
5, [7, 0, 0, 0, 0, 8, 3, 7, 2, 6]
6, [0, 0, 0, 0, 8, 3, 7, 2, 6, 7] <- stop and ignore this one

5. Again consider those as 4 bit portions of a 40 bit score, but
create your members with information about rotations a little
differently...

726156_0 -> 0x8372670000
726156_1 -> 0x3726700008
726156_2 -> 0x7267000083
726156_3 -> 0x2670000837
726156_4 -> 0x6700008372
726156_5 -> 0x7000083726

To search for a '615' anywhere in the string, we generate the same
start/stop endpoints:
0x72600000 and 0x726fffff. Performing the range scan, we get 726156_2
-> 0x7267000083, which is the 2nd rotation of id 726156.

Unfortunately, you lose 1 hex digit in the process, because you *need*
at least 1 trailing 0 in the zero-rotation version to only get exactly
what you want when you perform your scan. On the other hand, if you
also do a quick 'is this string in that string' check for everything
returned by the zrangebyscore call, you can get that last digit back
and can use all 40 bits for information.

This does potentially increase the amount of space used by a factor of
10x, but that all depends on your data.

Regards,
 - Josiah

On Thu, Feb 16, 2012 at 3:05 AM, Salimane Adjao Moustapha

0, [8, 3, 7, 2, 6, 7, 0, 0, 0, 0]
1, [3, 7, 2, 6, 7, 0, 0, 0, 0, 0]
2, [7, 2, 6, 7, 0, 0, 0, 0, 0, 0]
3, [2, 6, 7, 0, 0, 0, 0, 0, 0, 0]
4, [6, 7, 0, 0, 0, 0, 0, 0, 0, 0]
5, [7, 0, 0, 0, 0, 0, 0, 0, 0, 0]
6, [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] <- stop and ignore this one

And you stop when you run out of positive digits on the left. Then you
always get what you want with your zrangebyscore query.

Regards,
 - Josiah

On Thu, Feb 16, 2012 at 8:45 AM, Josiah Carlson