我有一个程序,我创造了转换成罗马数字,它成功地工作。现在,我尝试添加使用argparse
转换为希腊数字的想法,它不会经过列表中的第一个元组:("α", 1)
。我使用的算法和我以前转换成罗马的算法一样,但是由于某种原因它对我的希腊语不起作用?为什么这不能像我预期的那样起作用呢?你知道吗
# <~~ coding=utf-8 ~~>
import argparse
opts = argparse.ArgumentParser()
opts.add_argument("-g", "--greek", type=int,
help="Convert to Medieval Numerals")
opts.add_argument("-r", "--roman", type=int,
help="Convert to Roman Numerals")
args = opts.parse_args()
ROMAN_NUMERAL_TABLE = [
("~M", 1000000), ("~D", 500000), ("~C", 100000),
("~L", 50000), ("~X", 10000), ("~V", 5000), # "~" indicates a Macron
("M", 1000), ("CM", 900), ("D", 500),
("CD", 400), ("C", 100), ("XC", 90),
("L", 50), ("XL", 40), ("X", 10),
("IX", 9), ("V", 5), ("IV", 4),
("I", 1)
]
GREEK_NUMERAL_TABLE = [
("α", 1), ("β", 2), ("γ", 3),
("δ", 4), ("ε", 5), ("Ϝ", 6),
("ζ", 7), ("η", 8), ("θ", 9),
("ι", 10), ("κ", 20), ("λ", 30),
("μ", 40), ("ν", 50), ("ξ", 60),
("ο", 70), ("π", 80), ("ϙ", 90),
("ρ", 100), ("σ", 200), ("τ", 300),
("υ", 400), ("φ", 500), ("χ", 600),
("ψ", 700), ("ω", 800), ("ϡ͵", 900),
("α", 1000), ("β", 2000), ("γ", 3000),
("δ", 4000), ("ε", 5000), ("ϛ", 6000),
("ζ", 7000), ("η", 8000), ("θ", 9000) # The Greeks weren't very creative
]
def convert_init(number, convert_to=None):
display_numerals = []
for numeral, value in convert_to:
count = number // value
number -= count * value
display_numerals.append(numeral * count)
return ''.join(display_numerals)
if __name__ == '__main__':
if args.greek:
data = convert_init(int(args.greek), convert_to=GREEK_NUMERAL_TABLE)
with open("greek_numerals.txt", "a+") as file_data:
file_data.write(data)
elif args.roman:
data = convert_init(int(args.roman), convert_to=ROMAN_NUMERAL_TABLE)
with open("roman_numerals.txt", "a+") as file_data:
file_data.write(data)
else:
raise NotImplementedError("{} is not implemented yet".format(args))
罗马数字输出:
124542
~C~X~XMMMMDXLII # Works
希腊数字输出:
124542
ααααααααααααααααααααααααααααααααααααααααααααααααααααα # Only outputs the 1
...
希腊文列表按升序排列,而罗马文列表按降序排列。你知道吗
这导致希腊语处理的第一个数字是
1
,因此它打印n
乘以字母α
,而不是先减少较大的数字。你知道吗所以,在迭代列表时,只需按降序排序即可:
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