如何在python中增加字母数字？

Question

我正在创建一个由16个字母数字字符组成的密钥，其中我将生成从A001、A002、A003到A999的4位数字。直到A999，字母表将自动增加到B，数字将再次从001开始。同样的过程将持续到Z999。一旦A-Z系列结束，它将以AA01开始，依此类推。如何在python中实现这一点？因为我是python新手，所以我自己尝试过，也尝试过一些示例，但我无法增加字符数。 任何想法或想法都将不胜感激

非常感谢

rec=0
new_list16 = []
def autoIncrement():
    global rec
    first = 'A'
    i = chr(ord(first))
    new_list16.append(i)

    while True:
        pStart = 1 #adjust start value, if req'd 
        pInterval = 1 #adjust interval value, if req'd
        if (rec == 0):
            rec += pStart
        else:
            rec = rec + pInterval
        return str(rec).zfill(3)
#print(autoIncrement())
new_list16.append(autoIncrement())

print(*new_list16, sep = '')

Answer 1

def auto_increment(number):
    if number == 'ZZZZ':
        return 'ZZZZ'

    digits = "".join([i for i in number if i.isdigit()])
    chars = "".join([i for i in number if not i.isdigit()])
    if int(digits) == int('9' * len(digits)):
        digits = "000"
        new_char = [ord(i) for i in chars]
        if new_char[-1] % ord('Z') == 0:
            new_char = "".join([chr(i) for i in new_char]) + 'A'
        else:
            new_char[-1] = new_char[-1] + 1
            new_char = "".join([chr(i) for i in new_char])
    else:
        new_char = chars

    new_digit = int(digits) + 1
    l = len(new_char) 
    ll = len(str(new_digit))
    new_digit = (("0" * (3-ll)) + str(new_digit))[(l-1):]
    return f"{new_char}{new_digit}"

此函数用于返回给定任何数字的下一个数字。 例如：A999将返回AB01

现在可以在循环中使用此函数

Answer 2

从A999到B001而不是B000确实会把事情弄得有点乱，但是你仍然可以对a-Z部分使用this，对数字使用简单的模运算

def excel_format(num):
    # see https://stackoverflow.com/a/182924/1639625
    res = ""
    while num:
        mod = (num - 1) % 26
        res = chr(65 + mod) + res
        num = (num - mod) // 26
    return res

def full_format(num, d=3):
    chars = num // (10**d-1) + 1 # this becomes   A..ZZZ
    digit = num %  (10**d-1) + 1 # this becomes 001..999
    return excel_format(chars) + "{:0{}d}".format(digit, d)

for i in range(10000):
    print(i, full_format(i, d=2))

数字部分的位数由可选的d参数控制。我将使用2进行演示，但是3也同样有效

0 A01
...
98 A99
99 B01
...
2573 Z99
2574 AA01
...
9998 CW99
9999 CX01

Answer 3

这可能需要更多的测试和重构，但这里有一个开始：

def leadingZeros(number, digits):
    numberString = str(number)
    for digit in range(1, digits):
        if number < 10**digit:
            numberString = '0' + numberString
    return numberString


def autoIncrement(oldNumber):
    order = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ!'
    lastDigitOrder = order.find(oldNumber[3])
    newNumber = ''
    if order.find(oldNumber[1]) <= 9:
        # 3 digit number
        number = int(oldNumber[1:]) + 1
        letter = oldNumber[0]
        if 1000 == number:
            letterOrder = order.find(oldNumber[0])
            letter = order[letterOrder + 1]
        newNumber = letter + leadingZeros(number % 1000, 3)
    elif order.find(oldNumber[2]) <= 9:
        # 2 digit number
        number = int(oldNumber[2:]) + 1
        letters = oldNumber[0:2]
        if 100 == number:
            letterOrder = order.find(oldNumber[1])
            letter = order[letterOrder + 1]
            letters = oldNumber[0] + letter
        newNumber = letters + leadingZeros(number % 100, 2)
    elif order.find(oldNumber[3]) <= 9:
        # 1 digit number
        number = int(oldNumber[3]) + 1
        letters = oldNumber[0:3]
        if 10 == number:
            letterOrder = order.find(oldNumber[2])
            letter = order[letterOrder + 1]
            letters = oldNumber[0:2] + letter
        newNumber = letters + leadingZeros(number % 10, 1)
    else:
        # just letters
        print(oldNumber)
        letterOrder = order.find(oldNumber[3])
        letter = order[letterOrder + 1]
        newNumber = oldNumber[0:3] + letter

    # if one of the digits has gone past Z then we need to update the letters
    if '!' == newNumber[3]:
        # past Z in 4th digit
        letterOrder = order.find(oldNumber[2])
        newNumber = newNumber[0:2] + order[letterOrder + 1] + 'A'
    if '!' == newNumber[2]:
        # past Z in 3rd digit
        letterOrder = order.find(oldNumber[1])
        newNumber = newNumber[0:1] + order[letterOrder + 1] + 'A' + newNumber[3]
    if '!' == newNumber[1]:
        # past Z in 2nd digit
        letterOrder = order.find(oldNumber[0])
        newNumber = order[letterOrder + 1] + 'A' + newNumber[2:]

    return newNumber


print(autoIncrement('A999'))
print(autoIncrement('AA99'))
print(autoIncrement('AAA9'))
print(autoIncrement('AAAA'))
print(autoIncrement('AZZ9'))