本人找了很多文章，对于数字转中文大写的都没有很好的支持，尤其是单位不同的时候，于是找了一组代码稍加修改，运行流畅，输出准确。放上来算是给社区一点小贡献。转载记得加上这个链接哦。有问题欢迎留言和私信。有时间都乐意解答。

后面的try，只是我懒得写判断语句，加上判断类型就可以知道是整数还是小数。原则能用就行，反正很简单…

特意说下，用try的速度会比较慢哦。
def shuzi(num):
    dic_num = {"1": "壹", "2": "贰", "3": "叁", "4": "肆", "5": "伍", "6": "陆", "7": "柒", "8": "捌", "9": "玖", }
    dic_unit = {1: "", 2: "拾", 3: "佰", 4: "仟", 5: "万"}
    while 1:
        fs = []
        fs1 = []
        daxie = ""
        xiaoshu = ''
        bs = num.split('.')
        if len(bs) > 1:
            yuan = bs[0]
            lennum = len(bs[0])
            if lennum >= 1:
                for item in bs[0]:
                    if int(item) > 0:
                        fs.append(dic_num[item])
                        fs.append(dic_unit[lennum])
                    elif int(item) == 0:
                        fs.append("零")
                    lennum -= 1
            daxie = "".join(fs)
            lennum1 = len(bs[1])
            if lennum1 >= 1:
                for item in bs[1]:
                    if int(item) > 0:
                        fs1.append(dic_num[item])
                        fs1.append(dic_unit[lennum1])
                    elif int(item) == 0:
                        fs1.append("零")
                    lennum1 -= 1
            while fs1[-1] == "零":
                fs1.pop()

            xiaoshu = "".join(fs1)
            daxie = daxie + "点" + xiaoshu
        else:
            lennum = len(num)
            if lennum >= 1:
                for item in num:
                    if int(item) > 0:
                        fs.append(dic_num[item])
                        fs.append(dic_unit[lennum])
                    elif int(item) == 0:
                        fs.append("零")
                    lennum -= 1
            while fs[-1] == "零":
                fs.pop()
            daxie = "".join(fs)
        daxie += "万圆"#这里修改后缀
        try:
            if int(num) % 10 == 0:
                daxie = daxie + "整"
        except:
            pass
        return daxie

关键在于如何解决10100中0的描述，通过判断上一位的值是否为‘’零‘’，具体看代码：
list1 = ['零', '壹', '贰', '叁', '肆', '伍', '陆', '柒', '捌', '玖', '拾']
list2 = ['圆', '拾', '佰', '仟', '萬']
money = input("请输入金额:")
money2 = ''
for i in range(0, len(money)):
    if int(money[i]) != 0:
        money2 += list1[int(money[i])]+list2[len(money)-i-1]
    else:
        if money2[-1] != "零":
            money2 += "零"
if money2[-1] == "零":
    print(money2[0:len(money2)-1]+"圆整")
else:
    print(money2+"整")

今天有人在群里问到Python怎么把小写数字转换成中文的大写字符，觉得这个问题很有趣，于是想了一下，就开始写代码了。
以下是代码，感觉不足的地方就是如果是1001元，那么就会转换成 壹仟零佰零拾壹圆，多了个零。这个代码python2和python3都可以运行。
# -*- coding:utf-8 -*-
# date: 2018-02-28 11:48
def digital_to_chinese(digital):
str_digital = str(digital)
chinese = {"1": "壹", "2": "贰", "3": "叁", "4": "肆", "5": "伍", "6": "陆", "7": "柒", "8": "捌", "9": "玖", "0": "零"}
chinese2 = ["拾", "佰", "仟", "万", "厘", "分", "角"]
jiao = ""
bs = str_digital.split(".")
yuan = bs[0]
if len(bs) > 1:
jiao = bs[1]
r_yuan = [i for i in reversed(yuan)]
count = 0
for i in range(len(yuan)):
if i == 0:
r_yuan[i] += "圆"
continue
r_yuan[i] += chinese2[count]
count += 1
if count == 4:
count = 0
chinese2[3] = "亿"
s_jiao = [i for i in jiao][:3]  # 去掉小于厘之后的
j_count = -1
for i in range(len(s_jiao)):
s_jiao[i] += chinese2[j_count]
j_count -= 1
last = [i for i in reversed(r_yuan)] + s_jiao
last_str = "".join(last)
print(str_digital)
print(last_str)
last_str = last_str.replace("0百", "0").replace("0十", "0").replace("000", "0").replace("00", "0").replace("0圆", "圆")
for i in range(len(last_str)):
digital = last_str[i]
if digital in chinese:
last_str = last_str.replace(digital, chinese[digital])
print(last_str)
return last_str
if __name__ == "__main__":
digital_to_chinese(123456789.456)
这是输出
123456789.456
1亿2仟3佰4拾5万6仟7佰8拾9圆4角5分6厘
壹亿贰仟叁佰肆拾伍万陆仟柒佰捌拾玖圆肆角伍分陆厘

python数字转换成中文大写

代码



#-*- coding:utf-8 -*-
count_map = {1:1, 2:10, 3:100, 4:1000, 5:10000}
#用来统计传入数值的级别
def set_to_chinese(money, s):
    end_map = {10:u'拾',100:u'佰',1000:u'仟',10000:u'万'}
    trans_map = {0:u'零',1:u'壹',2:u'贰',3:u'叁',4:u'肆',
                 5:u'伍',6:u'陆',7:u'柒',8:u'捌',9:u'玖'}
    if s == 1:
        return trans_map[money]
    if money%s == 0:
        return trans_map[money/s]+end_map[s]
    else:
        mid = trans_map[0] if len(str(money))\
              -len(str(money%s))>1 else u''
        return trans_map[money/s]+end_map[s]+ mid +\
               set_to_chinese(money%s, 
money = 3840
print set_to_chinese(money, count_map[len(str(money))])
#叁仟捌佰肆拾