1 # 时间转中文 2 3 a = "20180509" 4 def time_text(stime): 5 data_list = [] 6 time_info = {'零': 0, '壹': 1, '贰': 2, '叁': 3, '肆': 4, '伍': 5, '陆': 6, '柒': 7, '捌': 8, '玖': 9, } 7 new_dict = {v: k for k, v in time_info.items()} 8 for i in stime: 9 data_list.append(new_dict[int(i)]) 10 str_time = ''.join(data_list) 11 print(str_time) 12 time_text(a) 13 14 15 # 时间转中文 16 17 # 列表推倒式 20180509 18 def time_text(stime): 19 time_info = {'零': 0, '壹': 1, '贰': 2, '叁': 3, '肆': 4, '伍': 5, '陆': 6, '柒': 7, '捌': 8, '玖': 9, } 20 new_dict = {v: k for k, v in time_info.items()} 21 ret = [new_dict[int(i)] for i in stime ] 22 result = ''.join(ret) 23 print(result) 24 time_text(a) 25 26 27 28 29 30 31 32 33 34 # 钱转中文 35 a = '2018' 36 37 38 class cnumber: 39 cdict = {} 40 gdict = {} 41 xdict = {} 42 43 def __init__(self): 44 self.cdict = {1: u'', 2: u'拾', 3: u'佰', 4: u'仟'} 45 self.xdict = {1: u'元', 2: u'万', 3: u'亿', 4: u'兆'} # 数字标识符 46 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'} 47 48 def csplit(self, cdata): # 拆分函数，将整数字符串拆分成[亿，万，仟]的list 49 g = len(cdata) % 4 50 csdata = [] 51 lx = len(cdata) - 1 52 if g > 0: 53 csdata.append(cdata[0:g]) 54 k = g 55 while k <= lx: 56 csdata.append(cdata[k:k + 4]) 57 k += 4 58 return csdata 59 60 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 61 lenki = len(cki) 62 i = 0 63 lk = lenki 64 chk = u'' 65 for i in range(lenki): 66 if int(cki[i]) == 0: 67 if i < lenki - 1: 68 if int(cki[i + 1]) != 0: 69 chk = chk + self.gdict[int(cki[i])] 70 else: 71 chk = chk + self.gdict[int(cki[i])] + self.cdict[lk] 72 lk -= 1 73 return chk 74 75 def cwchange(self, data): 76 cdata = str(data).split('.') 77 78 cki = cdata[0] 79 ckj = cdata[1] 80 i = 0 81 chk = u'' 82 cski = self.csplit(cki) # 分解字符数组[亿，万，仟]三组List:['0000','0000','0000'] 83 ikl = len(cski) # 获取拆分后的List长度 84 # 大写合并 85 for i in range(ikl): 86 if self.cschange(cski[i]) == '': # 有可能一个字符串全是0的情况 87 chk = chk + self.cschange(cski[i]) # 此时不需要将数字标识符引入 88 else: 89 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 90 # 处理小数部分 91 lenkj = len(ckj) 92 if lenkj == 1: # 若小数只有1位 93 if int(ckj[0]) == 0: 94 chk = chk + u'整' 95 else: 96 chk = chk + self.gdict[int(ckj[0])] + u'角整' 97 else: # 若小数有两位的四种情况 98 if int(ckj[0]) == 0 and int(ckj[1]) != 0: 99 chk = chk + u'零' + self.gdict[int(ckj[1])] + u'分' 100 elif int(ckj[0]) == 0 and int(ckj[1]) == 0: 101 chk = chk + u'整' 102 elif int(ckj[0]) != 0 and int(ckj[1]) != 0: 103 chk = chk + self.gdict[int(ckj[0])] + u'角' + self.gdict[int(ckj[1])] + u'分' 104 else: 105 chk = chk + self.gdict[int(ckj[0])] + u'角整' 106 return chk 107 108 109 if __name__ == '__main__': 110 pt = cnumber() 111 print(pt.cwchange('2018.00'))
# 一行搞定，时间转换中文 a = "3435465765" ret = a.replace("0","零").replace("1","壹").replace("2","贰").replace("3","叁").replace("4","肆").replace("5","伍").replace("6","陆").replace("7","柒").replace("8","捌").replace("9","玖") print(ret) # 转换钱 a = {"0": "零", "1": "一", "2": "二", "3": "三", "4": "四", "5": "五", "6": "六", "7": "七", "8": "八", "9": "九"} b = {"0": "", "1": "十", "2": "百", "3": "千", "4": "万", "5": "十", "6": "百", "7": "千", "8": "亿"} c= '120180509' print("".join([a[i] for i in c])) d = [] length = len(c) for i, v in enumerate(c): d.append(a[v]) if v != "0": d.append(b[str(length - (i + 1))]) print("".join(d))