最高支持九千万亿,还可以扩展,但是不支持零,不支持小数,不支持负数
由于一万亿等于一兆,但是兆这个单位不常用,我这里用的是万亿
如果发现bug欢迎及时反馈
def fun(number, recursive_depth=0): str_number = str(number) if len(str_number) > 4: str_number = str_number[-4:] bits = "零 一 二 三 四 五 六 七 八 九".split(" ") units = " 十 百 千".split(" ") large_unit = ' 万 亿 万'.split(" ") # 可扩展,以万为单位 number_len = len(str_number) result = "" for i in range(number_len): result += bits[int(str_number[i])] if str_number[i] != "0": result += units[number_len - i - 1] # 去除连续的零 while "零零" in result: result = result.replace("零零", "零") # 去除尾部的零 if result[-1] == "零": result = result[:-1] # 调整10~20之间的数 if result[:2] == "一十": result = result[1:] # 字符串连接上大单位 result += large_unit[recursive_depth] # 判断是否递归 if len(str(number)) > 4: recursive_depth += 1 return fun(str(number)[:-4], recursive_depth) + result else: return result 金额大写换了几个字符 , 几乎和上面一样
def fun(number, recursive_depth=0): str_number = str(number) if len(str_number) > 4: str_number = str_number[-4:] bits = "零 壹 贰 叁 肆 伍 陆 柒 捌 玖".split(" ") units = " 拾 佰 仟".split(" ") large_unit = ' 万 亿 万'.split(" ") # 可扩展,以万为单位 number_len = len(str_number) result = "" for i in range(number_len): result += bits[int(str_number[i])] if str_number[i] != "0": result += units[number_len - i - 1] # 去除连续的零 while "零零" in result: result = result.replace("零零", "零") # 去除尾部的零 if result[-1] == "零": result = result[:-1] # 调整10~20之间的数 if result[:2] == "一十": result = result[1:] # 字符串连接上大单位 result += large_unit[recursive_depth] # 判断是否递归 if len(str(number)) > 4: recursive_depth += 1 return fun(str(number)[:-4], recursive_depth) + result else: return result 调用 print(fun(240350056))# 贰亿肆仟零叁拾伍万零伍拾陆