如何生成大小为N的字符串,由数字和大写英文字母组成,例如:
6个754z4英国U911K4型
如何生成大小为N的字符串,由数字和大写英文字母组成,例如:
6个754z4英国U911K4型
当前回答
我想建议您下一个选择:
import crypt
n = 10
crypt.crypt("any sring").replace('/', '').replace('.', '').upper()[-n:-1]
妄想模式:
import uuid
import crypt
n = 10
crypt.crypt(str(uuid.uuid4())).replace('/', '').replace('.', '').upper()[-n:-1]
其他回答
>>> import random
>>> str = []
>>> chars = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890'
>>> num = int(raw_input('How long do you want the string to be? '))
How long do you want the string to be? 10
>>> for k in range(1, num+1):
... str.append(random.choice(chars))
...
>>> str = "".join(str)
>>> str
'tm2JUQ04CK'
random.choice函数在列表中选择一个随机条目。还可以创建一个列表,以便可以在for语句中追加字符。在结尾str是[t','m','2','J','U','Q','0','4','C','K'],但是str=“”.join(str)会处理这一点,留下'tm2JUQ04CK'。
希望这有帮助!
如果需要随机字符串而不是伪随机字符串,则应使用os.urandom作为源
from os import urandom
from itertools import islice, imap, repeat
import string
def rand_string(length=5):
chars = set(string.ascii_uppercase + string.digits)
char_gen = (c for c in imap(urandom, repeat(1)) if c in chars)
return ''.join(islice(char_gen, None, length))
我使用此方法从a->z生成长度为n的随机字符串
输入随机数s=“”.join(随机选择([chr(i)for i in range(ord('a'),ord('z'))])for _ in range(10))
一种无重复的随机生成器函数,使用一个集合来存储以前生成的值。注意,如果字符串或数量非常大,这将消耗一些内存,而且可能会稍微慢一点。发电机将在给定量或达到最大可能组合时停止。
代码:
#!/usr/bin/env python
from typing import Generator
from random import SystemRandom as RND
from string import ascii_uppercase, digits
def string_generator(size: int = 1, amount: int = 1) -> Generator[str, None, None]:
"""
Return x random strings of a fixed length.
:param size: string length, defaults to 1
:type size: int, optional
:param amount: amount of random strings to generate, defaults to 1
:type amount: int, optional
:yield: Yield composed random string if unique
:rtype: Generator[str, None, None]
"""
CHARS = list(ascii_uppercase + digits)
LIMIT = len(CHARS) ** size
count, check, string = 0, set(), ''
while LIMIT > count < amount:
string = ''.join(RND().choices(CHARS, k=size))
if string not in check:
check.add(string)
yield string
count += 1
for my_count, my_string in enumerate(string_generator(6, 20)):
print(my_count, my_string)
输出:
0 RS9N3P
1 S0GDGR
2 ZNBLFV
3 96FF97
4 38JJZ3
5 Q3214A
6 VLWNK1
7 QMT05E
8 X1ZFP0
9 MZF442
10 10L9AZ
11 GE8HIQ
12 S7PA43
13 MVLXO9
14 YX7Y0G
15 GIIKPF
16 3KCUQA
17 XHIXFV
18 BJQ5VG
19 HQF01Q
如果字符串始终需要包含字母和数字,请使用以下命令:
#!/usr/bin/env python
from typing import Generator
from random import SystemRandom as RND
from string import ascii_uppercase, digits
def string_generator(size: int = 2, amount: int = 1) -> Generator[str, None, None]:
"""
Return x random strings of a fixed length.
:param size: string length, defaults to 1
:type size: int, optional
:param amount: amount of random strings to generate, defaults to 1
:type amount: int, optional
:yield: Yield composed random string if unique
:rtype: Generator[str, None, None]
"""
if size < 2:
print(
'Since the string must contain letters and numbers,',
'its size must be at least two characters',
)
return
CHARS = list(ascii_uppercase + digits)
LIMIT = len(CHARS) ** size
count, check, string = 0, set(), ''
while LIMIT > count < amount:
while 1:
string = ''.join(RND().choices(CHARS, k=size))
if not (string.isalpha() or string.isnumeric()):
break
if string not in check:
check.add(string)
yield string
count += 1
for my_count, my_string in enumerate(string_generator(6, 20)):
print(my_count, my_string)
输出:
0 5JIITL
1 3W7AI1
2 8QMO10
3 ETHAI8
4 BLFPP9
5 50X2CC
6 1LI8WA
7 39CKET
8 R4IM0E
9 Q3KHGS
10 65CBIU
11 XTRHO4
12 N2XIGO
13 06VFCW
14 VE9MJ8
15 A9ADHK
16 Y5ROJ0
17 OH7EJF
18 TQ709S
19 P0DDNJ
我想建议您下一个选择:
import crypt
n = 10
crypt.crypt("any sring").replace('/', '').replace('.', '').upper()[-n:-1]
妄想模式:
import uuid
import crypt
n = 10
crypt.crypt(str(uuid.uuid4())).replace('/', '').replace('.', '').upper()[-n:-1]