在Python中生成一个随机字母

Question

有没有办法在Python中生成随机字母（如random.randint但字母）？ random.randint的范围功能会很好但是有一个只输出随机字母的生成器会比没有好。

Answer 1

简单：

>>> import string
>>> string.ascii_letters
'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
>>> import random
>>> random.choice(string.ascii_letters)
'j'

string.ascii_letters根据当前语言环境返回包含小写字母和大写字母的字符串。

random.choice从序列中返回单个随机元素。

Answer 2

import random
def Random_Alpha():
    l = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z']
    return l[random.randint(0,25)]

print(Random_Alpha())

Answer 3

import string
import random

KEY_LEN = 20

def base_str():
    return (string.letters+string.digits)   
def key_gen():
    keylist = [random.choice(base_str()) for i in range(KEY_LEN)]
    return ("".join(keylist))

你可以得到这样的随机字符串：

g9CtUljUWD9wtk1z07iF
ndPbI1DDn6UvHSQoDMtd
klMFY3pTYNVWsNJ6cs34
Qgr7OEalfhXllcFDGh2l

Answer 4

def create_key(key_len):
    key = ''
    valid_characters_list = string.letters + string.digits
    for i in range(key_len):
        character = choice(valid_characters_list)
        key = key + character
    return key

def create_key_list(key_num):
    keys = []
    for i in range(key_num):
        key = create_key(key_len)
        if key not in keys:
            keys.append(key)
    return keys

Answer 5

所有以前的答案都是正确的，如果你正在寻找各种类型的随机字符（即字母数字和特殊字符），那么这里是我创建的脚本，用于演示各种类型的创建随机函数，它有三个函数，一个用于数字，alpha-字符和特殊字符。该脚本只是生成密码，只是演示生成随机字符的各种方法的示例。

import string
import random
import sys

#make sure it's 3.7 or above
print(sys.version)

def create_str(str_length):
    return random.sample(string.ascii_letters, str_length)

def create_num(num_length):
    digits = []
    for i in range(num_length):
        digits.append(str(random.randint(1, 100)))

    return digits

def create_special_chars(special_length):
    stringSpecial = []
    for i in range(special_length):
        stringSpecial.append(random.choice('!$%&()*+,-.:;<=>?@[]^_`{|}~'))

    return stringSpecial

print("how many characters would you like to use ? (DO NOT USE LESS THAN 8)")
str_cnt = input()
print("how many digits would you like to use ? (DO NOT USE LESS THAN 2)")
num_cnt = input()
print("how many special characters would you like to use ? (DO NOT USE LESS THAN 1)")
s_chars_cnt = input()
password_values = create_str(int(str_cnt)) +create_num(int(num_cnt)) + create_special_chars(int(s_chars_cnt))

#shuffle/mix the values
random.shuffle(password_values)

print("generated password is: ")
print(''.join(password_values))

结果：

Answer 6

您可以使用

map(lambda a : chr(a),  np.random.randint(low=65, high=90, size=4))

Answer 7

好吧，这是我的答案！它运作良好。只需将所需的随机字母数放入“数字”......（Python 3）

import random

def key_gen():
    keylist = random.choice('abcdefghijklmnopqrstuvwxyz')
    return keylist

number = 0
list_item = ''
while number < 20:
    number = number + 1
    list_item = list_item + key_gen()

print(list_item)

Answer 8

import string
import random

def random_char(y):
    return ''.join(random.choice(string.ascii_letters+string.digits+li) for x in range(y))
no=int(input("Enter the number of character for your password=  "))
li = random.choice('!@#$%^*&( )_+}{')
print(random_char(no)+li)

Answer 9

我过于复杂的代码：

import random

letter = (random.randint(1,26))
if letter == 1:
   print ('a')
elif letter == 2:
    print ('b')
elif letter == 3:
    print ('c')
elif letter == 4:
    print ('d')
elif letter == 5:
    print ('e')
elif letter == 6:
    print ('f')
elif letter == 7:
    print ('g')
elif letter == 8:
    print ('h')
elif letter == 9:
    print ('i')
elif letter == 10:
    print ('j')
elif letter == 11:
    print ('k')
elif letter == 12:
    print ('l')
elif letter == 13:
    print ('m')
elif letter == 14:
    print ('n')
elif letter == 15:
    print ('o')
elif letter == 16:
    print ('p')
elif letter == 17:
    print ('q')
elif letter == 18:
    print ('r')
elif letter == 19:
    print ('s')
elif letter == 20:
    print ('t')
elif letter == 21:
    print ('u')
elif letter == 22:
    print ('v')
elif letter == 23:
    print ('w')
elif letter == 24:
    print ('x')
elif letter == 25:
    print ('y')
elif letter == 26:
    print ('z')

它基本上从26中生成一个随机数，然后转换成相应的字母。这可能会有所改善，但我只是一个初学者，我为这段代码感到自豪。

Answer 10

也许这可以帮助你：

import random
for a in range(64,90):
    h = random.randint(64, a)
    e += chr(h)
print e

Answer 11

在键盘上放一个蟒蛇让他滚过字母，直到你找到你喜欢的随机组合开玩笑！

import string #This was a design above but failed to print. I remodled it.
import random
irandom = random.choice(string.ascii_letters) 
print irandom

Answer 12

>>> import random
>>> import string
>>> random.choice(string.ascii_letters)
'g'

Answer 13

>>>def random_char(y):
       return ''.join(random.choice(string.ascii_letters) for x in range(y))

>>>print (random_char(5))
>>>fxkea

生成y个随机字符

Answer 14

>>> import random
>>> import string    
>>> random.choice(string.ascii_lowercase)
'b'

Answer 15

另一种方式，为了完整性：

>>> chr(random.randrange(97, 97 + 26))

使用ascii'a'为97的事实，字母表中有26个字母。

在确定random.randrange()函数调用的上限和下限时，请记住random.randrange()在其上限是唯一的，这意味着它只会生成比提供的值少1个单位的整数。

Answer 16

def randchar(a, b):
    return chr(random.randint(ord(a), ord(b)))

Answer 17

import random
def guess_letter():
    return random.choice('abcdefghijklmnopqrstuvwxyz')

Answer 18

你可以列一个清单：

import random
list1=['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
b=random.randint(0,7)
print(list1[b])

Answer 19

您可以使用它来获得一个或多个随机字母

import random
import string
random.seed(10)
letters = string.ascii_lowercase
rand_letters = random.choices(letters,k=5) # where k is the number of required rand_letters

print(rand_letters)

['o', 'l', 'p', 'f', 'v']