# Copyright 2024 Hung-Shin Lee (hungshinlee@gmail.com) # Apache 2.0 import itertools import re c_basic = "零一二三四五六七八九" d2c = {str(d): c for d, c in enumerate(c_basic)} d2c["."] = "點" def num4year(matched): def _num4year(num): return "{}".format("".join([c_basic[int(i)] for i in num])) matched_str = matched.group(0) for m in matched.groups(): matched_str = matched_str.replace(m, _num4year(m)) return matched_str def num2chines_simple(matched): return "{}".format("".join([d2c[i] for i in matched])) def num4percent(matched): matched = matched.group(1) return "百分之{}".format(num2chinese(matched[:-1])) def num4cellphone(matched): matched = matched.group(1) matched = matched.replace(" ", "").replace("-", "") return "".join([c_basic[int(i)] for i in matched]) def num4er(matched): # 2 to 二 matched = matched.group(1) return matched.replace("2", "二") def num4liang(matched): # 2 to 兩 matched = matched.group(1) return matched.replace("2", "兩") def num4general(matched): num = matched.group(1) if re.match("[A-Za-z-─]", num[0]): if len(num[1:]) < 3: # MP3 or F-16 return "{}{}".format(num[0], num2chinese(num[1:])) else: # AM104 return "{}{}".format(num[0], num2chines_simple(num[1:])) else: if re.match("[0-9]", num[0]): return "{}".format(num2chinese(num)) else: return "{}{}".format(num[0], num2chinese(num[1:])) def parse_num(text: str) -> str: # year text = re.sub("([0-9]{4})[到至]([0-9]{4})年", num4year, text) text = re.sub("([0-9]{4})年", num4year, text) # percentage text = re.sub(r"([0-9]+\.?[0-9]?%)", num4percent, text) # cellphone text = re.sub(r"([0-9]{4}\s?-\s?[0-9]{6})", num4cellphone, text) # single 2 to 二 text = re.sub(r"([^\d]2[診樓月號])", num4er, text) text = re.sub(r"([初]2[^\d])", num4er, text) # single 2 to 兩 text = re.sub(r"([^\d]2[^\d])", num4liang, text) # general number text = re.sub(r"([^0-9]?[0-9]+\.?[0-9]?)", num4general, text) return text def num2chinese(num, big=False, simp=False, o=False, twoalt=True) -> str: """ Converts numbers to Chinese representations. https://gist.github.com/gumblex/0d65cad2ba607fd14de7 `big` : use financial characters. `simp` : use simplified characters instead of traditional characters. `o` : use 〇 for zero. `twoalt`: use 两/兩 for two when appropriate. Note that `o` and `twoalt` is ignored when `big` is used, and `twoalt` is ignored when `o` is used for formal representations. """ # check num first nd = str(num) if abs(float(nd)) >= 1e48: raise ValueError("number out of range") elif "e" in nd: raise ValueError("scientific notation is not supported") c_symbol = "正负点" if simp else "正負點" if o: # formal twoalt = False if big: c_basic = "零壹贰叁肆伍陆柒捌玖" if simp else "零壹貳參肆伍陸柒捌玖" c_unit1 = "拾佰仟" c_twoalt = "贰" if simp else "貳" else: c_basic = "〇一二三四五六七八九" if o else "零一二三四五六七八九" c_unit1 = "十百千" if twoalt: c_twoalt = "两" if simp else "兩" else: c_twoalt = "二" c_unit2 = "万亿兆京垓秭穰沟涧正载" if simp else "萬億兆京垓秭穰溝澗正載" def revuniq(l): return "".join(k for k, g in itertools.groupby(reversed(l))) nd = str(num) result = [] if nd[0] == "+": result.append(c_symbol[0]) elif nd[0] == "-": result.append(c_symbol[1]) if "." in nd: integer, remainder = nd.lstrip("+-").split(".") else: integer, remainder = nd.lstrip("+-"), None if int(integer): splitted = [integer[max(i - 4, 0) : i] for i in range(len(integer), 0, -4)] intresult = [] for nu, unit in enumerate(splitted): # special cases if int(unit) == 0: # 0000 intresult.append(c_basic[0]) continue elif nu > 0 and int(unit) == 2: # 0002 intresult.append(c_twoalt + c_unit2[nu - 1]) continue ulist = [] unit = unit.zfill(4) for nc, ch in enumerate(reversed(unit)): if ch == "0": if ulist: # ???0 ulist.append(c_basic[0]) elif nc == 0: ulist.append(c_basic[int(ch)]) elif nc == 1 and ch == "1" and all([i == "0" for i in unit[: nc + 1]]): # special case for tens # edit the 'elif' if you don't like # 十四, 三千零十四, 三千三百一十四 ulist.append(c_unit1[0]) elif nc > 1 and ch == "2": ulist.append(c_twoalt + c_unit1[nc - 1]) else: ulist.append(c_basic[int(ch)] + c_unit1[nc - 1]) # print(ulist) ustr = revuniq(ulist) if nu == 0: intresult.append(ustr) else: intresult.append(ustr + c_unit2[nu - 1]) result.append(revuniq(intresult).strip(c_basic[0])) else: result.append(c_basic[0]) if remainder: result.append(c_symbol[2]) result.append("".join(c_basic[int(ch)] for ch in remainder)) return "".join(result) if __name__ == "__main__": text = "若手機仔幾多號?吾手機仔係0964-498042。" print(f"{text} -> {parse_num(text)}")