将十进制数字转换为罗马数字

Question

d_hsp={"1":"I","2":"II","3":"III","4":"IV","5":"V","6":"VI","7":"VII","8":"VIII",
       "9":"IX","10":"X","11":"XI","12":"XII","13":"XIII","14":"XIV","15":"XV",
       "16":"XVI","17":"XVII","18":"XVIII","19":"XIX","20":"XX","21":"XXI",
       "22":"XXII","23":"XXIII","24":"XXIV","25":"XXV"}
HSP_OLD['tryl'] = HSP_OLD['tryl'].replace(d_hsp, regex=True)

HSP_OLD是一个数据帧，tryl是HSP_OLD的一列，tryl中的一些值示例：

SAF/HSP: Secondary diagnosis E code 1

SAF/HSP: Secondary diagnosis E code 11

我用字典代替，它适用于1-10，但是对于11，它将变成“II”，对于12，它将变成“III”。在

Answer 1

您需要保持项目的顺序，并开始搜索最长的子字符串。在

您可以在这里使用OrderDict。要初始化它，请使用元组列表。在初始化时，您可以在这里反转它，但以后也可以这样做。在

import collections
import pandas as pd
# My test data    
HSP_OLD = pd.DataFrame({'tryl':['1. Text', '11. New Text', '25. More here']})

d_hsp_lst=[("1","I"),("2","II"),("3","III"),("4","IV"),("5","V"),("6","VI"),("7","VII"),("8","VIII"), ("9","IX"),("10","X"),("11","XI"),("12","XII"),("13","XIII"),("14","XIV"),("15","XV"), ("16","XVI"),("17","XVII"),("18","XVIII"),("19","XIX"),("20","XX"),("21","XXI"), ("22","XXII"),("23","XXIII"),("24","XXIV"),("25","XXV")]
d_hsp = collections.OrderedDict(d_hsp_lst)  # Creating the OrderedDict
d_hsp = collections.OrderedDict(reversed(d_hsp.items())) # Here, reversing

>>> HSP_OLD['tryl'] = HSP_OLD['tryl'].replace(d_hsp, regex=True)
>>> HSP_OLD
             tryl
0         I. Text
1    XI. New Text
2  XXV. More here

Answer 2

抱歉，我没注意到你不仅仅是在更新字段，还想在结尾处替换一个数字，但是即使是这样，把你的数字正确地转换成罗马数字比映射每一个可能出现的数字要好得多（如果一个数字大于25，你的代码会发生什么？）。所以，有一种方法可以做到：

ROMAN_MAP = [(1000, 'M'), (900, 'CM'), (500, 'D'), (400, 'CD'), (100, 'C'), (90, 'XC'),
             (50, 'L'), (40, 'XL'), (10, 'X'), (9, 'IX'), (5, 'V'), (4, 'IV'), (1, 'I')]

def romanize(data):
    if not data or not isinstance(data, str):  # we know how to work with strings only
        return data
    data = data.rstrip()  # remove potential extra whitespace at the end
    space_pos = data.rfind(" ")  # find the last space before the number
    if space_pos != -1:
        try:
            number = int(data[space_pos + 1:])  # get the number at the end
            roman_number = ""
            for i, r in ROMAN_MAP:  # loop-reduce substitution based on the ROMAN_MAP
                while number >= i:
                    roman_number += r
                    number -= i
            return data[:space_pos + 1] + roman_number  # put everything back together
        except (TypeError, ValueError):
            pass  # couldn't extract a number
    return data

现在，如果我们将数据帧创建为：

^{pr2}$

我们无法轻松地将我们的函数应用于整列：

HSP_OLD['tryl'] = HSP_OLD['tryl'].apply(romanize)

结果是：

                                         tryl
0       SAF/HSP: Secondary diagnosis E code I
1                                        None
2      SAF/HSP: Secondary diagnosis E code XI
3  Something else without a number at the end

当然，您可以根据需要调整romanize()函数来搜索字符串中的任何数字并将其转换为罗马数字—这只是一个如何快速找到字符串末尾的数字的示例。在