统一社会信用代码校验位python实现

18位统一代码最后一位计算方法如下：

```

def unioncode_checknum(code):

    alpha_to_num_map={'A':10,'B':11,'C':12,'D':13,'E':14,'F':15,'G':16,'H':17,'J':18,'K':19,'L':20,'M':21,'N':22,'P':23,'Q':24,'R':25,'T':26,'U':27,'W':28,'X':29,'Y':30}

    num_to_alpha_map={}

    for k,v in alpha_to_num_map.items():

        num_to_alpha_map[v]=k

    print(num_to_alpha_map)

    weights = [1,3,9,27,19,26,16,17,20,29,25,13,8,24,10,30,28]

    total = 0

    for i in range(len(code)):

        c = code[i]

        if not c.isdigit():

            total+=alpha_to_num_map[c]*weights[i]

        else:

            total+=int(c)*weights[i]

    pos = 31-(total%31)

    if pos==31:

        return 0

    if pos<10:

        return pos

    else:

        return num_to_alpha_map[pos]

```

统一社会信用代码定义：

http://www.bjdm.org.cn/userfiles/file/1513058211657.pdf