python - 在python中创建字典的字典

Question

问题

我在 for 循环的每次迭代中得到单独的字典，但是当第二次迭代运行时，它没有将第二个字典添加到第一个字典，而是用第二个字典替换了第一个字典。

class Nation:
    def __init__(self, short_name, long_name, iso_code, iso_short, iso_long, capital):
        self.short_name = short_name
        self.long_name = long_name
        self.iso_code = iso_code
        self.iso_short = iso_short
        self.iso_long = iso_long
        self.capital = capital

def to_dictionaries(z):
    a = {}
    sD = {}
    for i in range(len(z)):
        sD["long_name"] = z[i].long_name
        sD["iso_code"] = z[i].iso_code
        sD["iso_short"] = z[i].iso_short
        sD["iso_long"] = z[i].iso_long
        sD["capital"] = z[i].capital
        a.update(sD)
    return a

new_nation_1 = Nation("Albania", "Republic of Albania", 8, "AL", "ALB", "Tirana")
new_nation_2 = Nation("Angola", "Republic of Angola", 24, "AO", "AGO", "Luanda")
nation_list = [new_nation_1, new_nation_2]
print(to_dictionaries(nation_list))

期望的输出:

 {"Albania": {"long_name": "Republic of Albania", "iso_code": 8, "iso_short": "AL", "iso_long": "ALB", "capital": "Tirana"},
    "Angola": {"long_name": "Republic of Angola", "iso_code": 24, "iso_short": "AO", "iso_long": "AGO", "capital": "Luanda"}}

实际输出:

{'long_name': 'Republic of Angola', 'iso_code': 24, 'iso_short': 'AO', 'iso_long': 'AGO', 'capital': 'Luanda'}

我该如何解决这个问题？

Answer 1

改变

a.update(sD)

到

a[z[i].short_name] = sD

因为您需要使用short_name 作为键。

另一个建议是将 sD = {} 放在 for block 中。所以想:

def to_dictionaries(z):
    a = {}
    for i in range(len(z)):
        sD = {}
        sD["long_name"] = z[i].long_name
        sD["iso_code"] = z[i].iso_code
        sD["iso_short"] = z[i].iso_short
        sD["iso_long"] = z[i].iso_long
        sD["capital"] = z[i].capital
        a[z[i].short_name] = sD
    return a