Python 如果多个列表项以字符串形式组合在一起

我有一本关于国家及其字母代码的字典

all_countries = {
    ...
    "Cambodia": "KH",
    "Saint Kitts and Nevis": "KN",
    "Comoros": "KM",
    "Sao Tome and Principe": "ST",
    "Slovakia": "SK",
    "Korea, Republic of": "KR",
    "Slovenia": "SI",
    "Korea, Democratic People's Republic of": "KP",
    "Kuwait": "KW",
    "Senegal": "SN",
    ....
}

我需要在下面的列表中将国家名称替换为字母2代码

....
["Pusan National University Hospital", "49241 Busan", "Korea", "Republic of"]
["Severance Hospital", "03722 Seoul", "Korea", "Republic of"]
["Ospedale degli Infermi", "47923 Rimini", "Italy"]
....

输出应为：

....
["Pusan National University Hospital", "49241 Busan", "KR"]
["Severance Hospital", "03722 Seoul", "KR"]
["Ospedale degli Infermi", "47923 Rimini", "IT"]
....

我有一个脚本来改变它

def get_alpha_code(row):
    for i in row:
        if i in all_countries.keys():
            country_code = all_countries[i]
            row.remove(i)
            return country_code
        elif i in all_countries.values():
            code = i
            row.remove(i)
            return code
    return None

但我如何检查字典键中是否有多个项目？

---

例如韩国

这是工作吗？我没有你们所有的全部数据，但我想我们的想法是把字符串组合成你们的dict格式

all_countries = {
    "Cambodia": "KH",
    "Saint Kitts and Nevis": "KN",
    "Comoros": "KM",
    "Sao Tome and Principe": "ST",
    "Slovakia": "SK",
    "Korea, Republic of": "KR",
    "Slovenia": "SI",
    "Korea, Democratic People's Republic of": "KP",
    "Kuwait": "KW",
    "Senegal": "SN",
    "Italy": "IT",
}


a = [["Pusan National University Hospital", "49241 Busan", "Korea", "Republic of"],
     ["Severance Hospital", "03722 Seoul", "Korea", "Republic of"],
     ["Ospedale degli Infermi", "47923 Rimini", "Italy"]]

def process_data(a):
    ll = []
    for row in a:
        if len(row) == 3:
            c = row[-1]
        elif len(row) == 4:
            c = row[-2]+', '+row[-1]
        rr = row[0:2]+[all_countries[c]]
        ll.append(rr)
    return ll

process_data(a)

---

在您的列表“Korea”中，“Republic of”是两个不同的元素，但在字典中是单键Yes。所以我需要检查一下？这两个不同的元素可能是国家的全称如何识别这两个元素？您不需要更改dict，只需将数据转换为dict之后的格式。