Redis Python - how to delete all keys according to a specific pattern In python, without python iterating

Question

I\'m writing a django management command to handle some of our redis caching. Basically, I need to choose all keys, that confirm to a certain pattern (for example: \"prefix:

Answer 1

From the Documentation

delete(*names)
    Delete one or more keys specified by names

This just wants an argument per key to delete and then it will tell you how many of them were found and deleted.

In the case of your code above I believe you can just do:

    redis.delete(*x)

But I will admit I am new to python and I just do:

    deleted_count = redis.delete('key1', 'key2')

Answer 2

I think the

 for key in x: cache.delete(key)

is pretty good and concise. delete really wants one key at a time, so you have to loop.

Otherwise, this previous question and answer points you to a lua-based solution.

Answer 3

Use SCAN iterators: https://pypi.python.org/pypi/redis

for key in r.scan_iter("prefix:*"):
    r.delete(key)

Answer 4

Here is a full working example using py-redis:

from redis import StrictRedis
cache = StrictRedis()

def clear_ns(ns):
    """
    Clears a namespace
    :param ns: str, namespace i.e your:prefix
    :return: int, cleared keys
    """
    count = 0
    ns_keys = ns + '*'
    for key in cache.scan_iter(ns_keys):
        cache.delete(key)
        count += 1
    return count

You can also do scan_iter to get all the keys into memory, and then pass all the keys to delete for a bulk delete but may take a good chunk of memory for larger namespaces. So probably best to run a delete for each key.

Cheers!

UPDATE:

Since writing the answer, I started using pipelining feature of redis to send all commands in one request and avoid network latency:

from redis import StrictRedis
cache = StrictRedis()

def clear_cache_ns(ns):
    """
    Clears a namespace in redis cache.
    This may be very time consuming.
    :param ns: str, namespace i.e your:prefix*
    :return: int, num cleared keys
    """
    count = 0
    pipe = cache.pipeline()
    for key in cache.scan_iter(ns):
        pipe.delete(key)
        count += 1
    pipe.execute()
    return count

UPDATE2 (Best Performing):

If you use scan instead of scan_iter, you can control the chunk size and iterate through the cursor using your own logic. This also seems to be a lot faster, especially when dealing with many keys. If you add pipelining to this you will get a bit of a performance boost, 10-25% depending on chunk size, at the cost of memory usage since you will not send the execute command to Redis until everything is generated. So I stuck with scan:

from redis import StrictRedis
cache = StrictRedis()
CHUNK_SIZE = 5000

def clear_ns(ns):
    """
    Clears a namespace
    :param ns: str, namespace i.e your:prefix
    :return: int, cleared keys
    """
    cursor = '0'
    ns_keys = ns + '*'
    while cursor != 0:
        cursor, keys = cache.scan(cursor=cursor, match=ns_keys, count=CHUNK_SIZE)
        if keys:
            cache.delete(*keys)

    return True

Here are some benchmarks:

5k chunks using a busy Redis cluster:

Done removing using scan in 4.49929285049
Done removing using scan_iter in 98.4856731892
Done removing using scan_iter & pipe in 66.8833789825
Done removing using scan & pipe in 3.20298910141

5k chunks and a small idle dev redis (localhost):

Done removing using scan in 1.26654982567
Done removing using scan_iter in 13.5976779461
Done removing using scan_iter & pipe in 4.66061878204
Done removing using scan & pipe in 1.13942599297

Answer 5

Use delete_pattern: https://niwinz.github.io/django-redis/latest/

from django.core.cache import cache
cache.delete_pattern("prefix:*")

Answer 6

According to my test, it will costs too much time if I use scan_iter solution (as Alex Toderita wrote).

Therefore, I prefer to use:

from redis.connection import ResponseError

try:
    redis_obj.eval('''return redis.call('del', unpack(redis.call('keys', ARGV[1])))''', 0, 'prefix:*')
except ResponseError:
    pass

The prefix:* is the pattern.

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refers to: https://stackoverflow.com/a/16974060