如何在 Python 中生成唯一的 ID？

也许 uuid.uuid4()可以完成这项工作。请参阅 Uuid了解更多信息。

您可能需要 Python 的 UUID 函数:

Ref = “ http://docs.python.org/library/UUID.html”rel = “ noReferrer”> 21.15. UUID ー UUID 对象根据 RFC 4122

例如:

import uuid
print uuid.uuid4()

7d529dd4-548b-4258-aa8e-23e34dc8d43d

也许是 Uuid模块？

独特和随机是相互排斥的，也许你想要这个？

import random
def uniqueid():
seed = random.getrandbits(32)
while True:
yield seed
seed += 1

用法:

unique_sequence = uniqueid()
id1 = next(unique_sequence)
id2 = next(unique_sequence)
id3 = next(unique_sequence)
ids = list(itertools.islice(unique_sequence, 1000))

没有两个返回的 id 是相同的(唯一) ，这是基于随机种子值

import time
import random
import socket
import hashlib


def guid( *args ):
"""
Generates a universally unique ID.
Any arguments only create more randomness.
"""
t = long( time.time() * 1000 )
r = long( random.random()*100000000000000000L )
try:
a = socket.gethostbyname( socket.gethostname() )
except:
# if we can't get a network address, just imagine one
a = random.random()*100000000000000000L
data = str(t)+' '+str(r)+' '+str(a)+' '+str(args)
data = hashlib.md5(data).hexdigest()


return data

在这里你可以找到一个实现:

def __uniqueid__():
"""
generate unique id with length 17 to 21.
ensure uniqueness even with daylight savings events (clocks adjusted one-hour backward).


if you generate 1 million ids per second during 100 years, you will generate
2*25 (approx sec per year) * 10**6 (1 million id per sec) * 100 (years) = 5 * 10**9 unique ids.


with 17 digits (radix 16) id, you can represent 16**17 = 295147905179352825856 ids (around 2.9 * 10**20).
In fact, as we need far less than that, we agree that the format used to represent id (seed + timestamp reversed)
do not cover all numbers that could be represented with 35 digits (radix 16).


if you generate 1 million id per second with this algorithm, it will increase the seed by less than 2**12 per hour
so if a DST occurs and backward one hour, we need to ensure to generate unique id for twice times for the same period.
the seed must be at least 1 to 2**13 range. if we want to ensure uniqueness for two hours (100% contingency), we need
a seed for 1 to 2**14 range. that's what we have with this algorithm. You have to increment seed_range_bits if you
move your machine by airplane to another time zone or if you have a glucky wallet and use a computer that can generate
more than 1 million ids per second.


one word about predictability : This algorithm is absolutely NOT designed to generate unpredictable unique id.
you can add a sha-1 or sha-256 digest step at the end of this algorithm but you will loose uniqueness and enter to collision probability world.
hash algorithms ensure that for same id generated here, you will have the same hash but for two differents id (a pair of ids), it is
possible to have the same hash with a very little probability. You would certainly take an option on a bijective function that maps
35 digits (or more) number to 35 digits (or more) number based on cipher block and secret key. read paper on breaking PRNG algorithms
in order to be convinced that problems could occur as soon as you use random library :)


1 million id per second ?... on a Intel(R) Core(TM)2 CPU 6400 @ 2.13GHz, you get :


>>> timeit.timeit(uniqueid,number=40000)
1.0114529132843018


an average of 40000 id/second
"""
mynow=datetime.now
sft=datetime.strftime
# store old datetime each time in order to check if we generate during same microsecond (glucky wallet !)
# or if daylight savings event occurs (when clocks are adjusted backward) [rarely detected at this level]
old_time=mynow() # fake init - on very speed machine it could increase your seed to seed + 1... but we have our contingency :)
# manage seed
seed_range_bits=14 # max range for seed
seed_max_value=2**seed_range_bits - 1 # seed could not exceed 2**nbbits - 1
# get random seed
seed=random.getrandbits(seed_range_bits)
current_seed=str(seed)
# producing new ids
while True:
# get current time
current_time=mynow()
if current_time <= old_time:
# previous id generated in the same microsecond or Daylight saving time event occurs (when clocks are adjusted backward)
seed = max(1,(seed + 1) % seed_max_value)
current_seed=str(seed)
# generate new id (concatenate seed and timestamp as numbers)
#newid=hex(int(''.join([sft(current_time,'%f%S%M%H%d%m%Y'),current_seed])))[2:-1]
newid=int(''.join([sft(current_time,'%f%S%M%H%d%m%Y'),current_seed]))
# save current time
old_time=current_time
# return a new id
yield newid


""" you get a new id for each call of uniqueid() """
uniqueid=__uniqueid__().next


import unittest
class UniqueIdTest(unittest.TestCase):
def testGen(self):
for _ in range(3):
m=[uniqueid() for _ in range(10)]
self.assertEqual(len(m),len(set(m)),"duplicates found !")

希望能有帮助！

也许这对你有用

str(uuid.uuid4().fields[-1])[:5]

这将非常快速地工作，但不会生成随机值，而是单调地增加值(对于给定的线程)。

import threading


_uid = threading.local()
def genuid():
if getattr(_uid, "uid", None) is None:
_uid.tid = threading.current_thread().ident
_uid.uid = 0
_uid.uid += 1
return (_uid.tid, _uid.uid)

它是线程安全的，使用元组可能比使用字符串(如果有的话，更短)有好处。如果你不需要线程安全随意删除线程位(代替 穿线，本地的，使用 Object ()和删除 Tid完全)。

希望能帮上忙。