将十进制转换为罗马数字 [英] Convert decimal to Roman numerals

问题描述

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，这是tryl中值的一些示例:

HSP_OLD is a dataframe, tryl is one column of HSP_OLD, and here's some example of values in tryl:

SAF/HSP: Secondary diagnosis E code 1

SAF/HSP: Secondary diagnosis E code 11

我使用字典进行替换，它适用于1-10，但对于11，它将变为"II"，对于12，它将变为"III".

I use a dictionary to replace, it works for 1-10, but for 11, it will become "II" , for 12, it will become "III".

推荐答案

您需要保持项目的顺序，并从最长的子字符串开始搜索.

You need to keep the order of the items, and start searching with the longest substring.

您可以在此处使用OrderDict.要初始化它，请使用元组列表.初始化时，您可以在此处将其反转，但是以后也可以进行.

You may use an OrderDict here. To initialize it, use a list of tuples. You may reverse it already here, when initializing, but you can do it later, too.

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

这篇关于将十进制转换为罗马数字的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！