CODEWITHSUNDEEP
pythondjangoredisdjango-cacheredis-py
alonisser
alonisser

Reputation: 12088

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

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:*") and delete them.

I know I can use the cli to do that:

redis-cli KEYS "prefix:*" | xargs redis-cli DEL

But I need to do this from within the app. So I need to use the python binding (I'm using py-redis). I have tried feeding a list into delete, but it fails:

from common.redis_client import get_redis_client
cache = get_redis_client()
x = cache.keys('prefix:*') 

x == ['prefix:key1','prefix:key2'] # True

# And now

cache.delete(x)

# returns 0 . nothing is deleted

I know I can iterate over x:

for key in x:
   cache.delete(key)

But that would be losing redis awesome speed and misusing its capabilities. Is there a pythonic solution with py-redis, without iteration and/or the cli?

Thanks!

Upvotes: 73

Views: 161777

Answers (10)

Pankaj Saini
Pankaj Saini

Reputation: 800

The answer suggested by @radtek is not working for me, since the keys are getting deleted while iterating, which leads to unexpected behavior. Here's an example:

from redis import StrictRedis
cache = StrictRedis()

for i in range(0, 10000):
    cache.set(f'test_{i}', 1)


cursor = '0'
SCAN_BATCH_SIZE = 5000
while cursor != 0:
    cursor, keys = self._redis.scan(cursor=cursor, match='test_*', count=SCAN_BATCH_SIZE)
    if keys:
        cache.delete(*keys)


## Iteration 1
# cursor=5000, keys=['test_0', .... , 'test_4999']
# keys will get deleted

## Iteration 2
# cursor=0, keys=[]
# No remaining keys are found reason being, there are just the 
# 5000 entries left post deletion and the cursor position is already 
# at 5000. Hence, no keys are returned.

You can use redis pipeline in order to solve this issue as mentioned below:

from redis import StrictRedis
cache = StrictRedis()

for i in range(0, 10000):
    cache.set(f'test_{i}', 1)


pipe = cache.pipeline()
cursor = None
SCAN_BATCH_SIZE = 5000
while cursor != 0:
    cursor, keys = self._redis.scan(cursor=cursor or 0, match='test_*', count=SCAN_BATCH_SIZE)
    if keys:
        pipe.delete(*keys)
pipe.execute()

Upvotes: 0

radtek
radtek

Reputation: 36360

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

Upvotes: 46

carton.swing
carton.swing

Reputation: 1727

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.

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

Upvotes: 4

Lynn Han
Lynn Han

Reputation: 503

You can use a specific pattern to match all keys and delete them:

import redis
client = redis.Redis(host='192.168.1.106', port=6379,
                password='pass', decode_responses=True)
for key in client.keys('prefix:*'):
    client.delete(key)

Upvotes: 8

Jijo
Jijo

Reputation: 157

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

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

Upvotes: 0

Blackeagle52
Blackeagle52

Reputation: 1986

cache.delete(*keys) solution of Dirk works fine, but make sure keys isn't empty to avoid a redis.exceptions.ResponseError: wrong number of arguments for 'del' command.

If you are sure that you will always get a result: cache.delete(*cache.keys('prefix:*') )

Upvotes: 7

James
James

Reputation: 1691

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')

Upvotes: 12

Alex Toderita
Alex Toderita

Reputation: 661

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

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

Upvotes: 66

Gleb
Gleb

Reputation: 761

Btw, for the django-redis you can use the following (from https://niwinz.github.io/django-redis/latest/):

from django.core.cache import cache
cache.delete_pattern("foo_*")

Upvotes: 10

Dirk is no longer here
Dirk is no longer here

Reputation: 368509

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.

Upvotes: 28

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