数字转人民币大写-python实现
# 算法说明:要求字符串输入,现将字符串差费为整数部分和小数部分生成list[整数部分,小数部分]
# 将整数部分拆分为:[亿,万,仟]三组字符串组成的List:['0000','0000','0000'](根据实际输入生成阶梯List)
# 例如:600190000010.70整数部分拆分为:['600','1900','0010']
# 然后对list中每个字符串分组进行大写化再合并
# 最后处理小数部分的大写化
class Cnumber: cdict = {} gdict = {} xdict = {} def __init__(self): self.cdict = {1: u'', 2: u'拾', 3: u'佰', 4: u'仟'} self.xdict = {1: u'元', 2: u'万', 3: u'亿', 4: u'兆'} # 数字标识符 self.gdict = {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'} @staticmethod def csplit(cdata): # 拆分函数,将整数字符串拆分成[亿,万,仟]的list g = len(cdata) % 4 csdata = [] lx = len(cdata) - 1 if g > 0: csdata.append(cdata[0:g]) k = g while k <= lx: csdata.append(cdata[k:k + 4]) k += 4 return csdata def cschange(self, cki): # 对[亿,万,仟]的list中每个字符串分组进行大写化再合并 lenki = len(cki) lk = lenki chk = u'' for i in range(lenki): if int(cki[i]) == 0: if i < lenki - 1: if int(cki[i + 1]) != 0: chk = chk + self.gdict[int(cki[i])] else: chk = chk + self.gdict[int(cki[i])] + self.cdict[lk] lk -= 1 return chk def cwchange(self, data): cdata = str(data).split('.') cki = cdata[0] ckj = cdata[1] chk = u'' cski = self.csplit(cki) # 分解字符数组[亿,万,仟]三组List:['0000','0000','0000'] ikl = len(cski) # 获取拆分后的List长度 # 大写合并 for i in range(ikl): if self.cschange(cski[i]) == '': # 有可能一个字符串全是0的情况 chk = chk + self.cschange(cski[i]) # 此时不需要将数字标识符引入 else: chk = chk + self.cschange(cski[i]) + self.xdict[ikl - i] # 合并:前字符串大写+当前字符串大写+标识符 # 处理小数部分 lenkj = len(ckj) if lenkj == 1: # 若小数只有1位 if int(ckj[0]) == 0: chk = chk + u'整' else: chk = chk + self.gdict[int(ckj[0])] + u'角整' else: # 若小数有两位的四种情况 if int(ckj[0]) == 0 and int(ckj[1]) != 0: chk = chk + u'零' + self.gdict[int(ckj[1])] + u'分' elif int(ckj[0]) == 0 and int(ckj[1]) == 0: chk = chk + u'整' elif int(ckj[0]) != 0 and int(ckj[1]) != 0: chk = chk + self.gdict[int(ckj[0])] + u'角' + self.gdict[int(ckj[1])] + u'分' else: chk = chk + self.gdict[int(ckj[0])] + u'角整' return chk if __name__ == '__main__': pt = Cnumber() print(pt.cwchange('600190101000.08'))