Python 数字转财务大写数字（二）

今天把功能补全了，可以实现完全的转换功能，包括小数部分。
主要就是增加了用float函数判断输入是否合法，用preTreatment函数将数字分为整数和小数部分，然后用convert函数来对两部分分别进行处理。

import re
digits = list('0123456789')
accDigits = list('零壹贰叁肆伍陆柒捌玖')
units = ['', '拾', '佰', '仟']
sep = list('分角元万亿兆')
acc_conv_dict = dict(zip(digits, accDigits))


def preTreatment(num):
    num = '{:.2f}'.format(float(num))
    partition = num.partition('.')
    iPart, fPart = partition[0], partition[2]
    return iPart, fPart


def convert(num):
    iPart, fPart = preTreatment(num)
    blocks = four_digits_block(iPart)
    iPartStr, fPartStr = '', ''
    for blockIndex in range(len(blocks)):
        block = blocks[blockIndex]
        for index in range(4):
            if not (blockIndex == 0 and index == 2 and block[index] == '1'):
                iPartStr += acc_conv_dict[block[index]]
            if block[index] != '0':
                iPartStr += units[3 - index]
        iPartStr = iPartStr.strip('零')
        iPartStr += sep[len(blocks) - blockIndex + 1]

    iPartStr = re.sub('零+', '零', iPartStr)

    for index in range(len(fPart)):
        fPartStr += acc_conv_dict[fPart[index]]
        if fPart[index] != '0':
            fPartStr += sep[1 - index]
    fPartStr = fPartStr.strip('零')

    if fPartStr == '':
        result = iPartStr + '整'
    else:
        result = iPartStr + fPartStr

    return result


def four_digits_block(iPart):
    fill_counter = 4 - (len(iPart) % 4) if len(iPart) % 4 != 0 else 0
    zero_fill = iPart.zfill(fill_counter + len(iPart))
    blocks = [zero_fill[i * 4:(i + 1) * 4] for i in range(len(zero_fill) // 4)]
    return blocks


while True:
    num = input("请输入数字：")
    if num == '':
        print('欢迎您再次使用，再见！')
        break
    try:
        float(num)
    except:
        print('输入数字不合法!')
        pass

    result = convert(num)
    print(result)