保持 dictionary character 的 string

(注释内联)

def is_it_correct(inp):
    if(len(inp) != 5):
        return False
    # store characters in a dictionary; key=ASCII of a character, value=it's frequency in the input string
    ind = 0
    dictionary = dict()
    while(ind < len(inp)):
        cur_char = inp[ind]
        cur_int = ord(cur_char)
        if(dictionary.get(cur_int) == None):
            dictionary[cur_int] = 1
        else:
            dictionary[cur_int] = dictionary[cur_int]+1
        ind = ind+1
    # Make sure there's only one digit (0-9) i.e ASCII b/w 48 & 57
    # Also, make sure that there are exactly 4 lower case alphabets (a-z) i.e ASCII b/w 97 & 122
    digits = 0
    alphabets = 0
    for key, val in dictionary.items():
        if(key >= 48 and key <= 57):
            digits = digits+val
        if(key >= 97 and key <= 122):
            alphabets = alphabets+val
    if(digits != 1 or alphabets != 4):
        return False
    # you should have 4 distinct ASCII values as your dictionary keys (only one ASCII is repeating in 5-length string)
    if(len(dictionary) != 4):
        return False
    return True

ret = is_it_correct("b3ksl")
print(ret)

试试下面的方法,你不需要正则表达式:

>>> import random,string
>>> l=random.sample(string.ascii_lowercase,4)
>>> l.append(str(random.randint(0,10)))
>>> random.shuffle(l)
>>> ''.join(l)
'iNx1k'
>>> 

或者想要更多:

import random,string
lst=[]
for i in range(3):
    l=random.sample(string.ascii_lowercase,4)
    l.append(str(random.randint(0,10)))
    random.shuffle(l)
    lst.append(''.join(l))

import random,string
lst=[]
for i in range(3):
    l=random.sample(string.ascii_lowercase,4)
    l[-1]=l[0]
    l.append(str(random.randint(0,10)))
    random.shuffle(l)
    lst.append(''.join(l))
print(lst)

回答你的问题:

import re
def check(val):
    return len(set(val))!=len(val) and len(re.sub('\d+','',val)) and val.islower() and len(val)==5
a = check("a1abc")
print(a)