将具有相同键字符串的多个字典键值相加

Question

假设我的字典是这样的：

{
    "432701228292636694" : {
        "64AEdcb" : {
            "inviter" : "282764296075870208",
            "uses" : 40
        },
        "9WVmh3Q" : {
            "inviter" : "173219378492473345",
            "uses" : 26
        },
        "AE8ugEN" : {
            "inviter" : "285548207189000204",
            "uses" : 4
        },
        "EmCddbN" : {
            "inviter" : "193926566881918978",
            "uses" : 2
        },
        "GKnPWyW" : {
            "inviter" : "147756076299845632",
            "uses" : 1
        },
        "JnmyFKE" : {
            "inviter" : "173219378492473345",
            "uses" : 10
        },
        "Km4wZTz" : {
            "inviter" : "161132853256650752",
            "uses" : 9
        },
        "M8uYMNb" : {
            "inviter" : "193896125969858561",
            "uses" : 8
        },
        "e9vTXcz" : {
            "inviter" : "273852837988007936",
            "uses" : 1
        },
        "gTB7w4R" : {
            "inviter" : "327660257423917066",
            "uses" : 1
        },
        "hG2KyZa" : {
            "inviter" : "315630485114191872",
            "uses" : 2
        },
        "pdZNPFa" : {
            "inviter" : "141431182792458241",
            "uses" : 537
        },
        "rZwH4QF" : {
            "inviter" : "384664458628169728",
            "uses" : 4
        },
        "s9FXVAU" : {
            "inviter" : "125092433657593856",
            "uses" : 2
        },
        "wrPbkGz" : {
            "inviter" : "277605262486011904",
            "uses" : 3
        }
    }
}

我想把具有相同inviter的每个键的总数相加，得到uses的总和。在上面的数据中，inviter“173219378492473345”键使用26和10。我想把它合并成36。你知道吗

我怎样才能做到这一点？我能够写一个列表理解语句，可以让我得到每一个的价值，但把它们加在一起，得到一个总数是我迷路的地方。你知道吗

下面的代码是我已经编写的：

codes = sorted(self.invites_db[server.id], key=lambda x: (self.invites_db[server.id][x]['uses']), reverse=True)
inviters = [server.get_member(self.invites_db[server.id][x]['inviter']).name + "#" + server.get_member(self.invites_db[server.id][x]['inviter']).discriminator for x in codes if self.invites_db[server.id][x]['inviter'] != "141431182792458241"]
uses = [self.invites_db[server.id][x]['uses'] for x in codes if self.invites_db[server.id][x]['inviter'] != "141431182792458241"]

这将给我一个所有inviters和uses的列表，但我不希望有两个相同的inviter，而是希望它们作为一个添加在一起。我怎样才能让他们使用zip以文本格式并排显示这些数据？你知道吗

Answer 1

collections.Counter对于那些“计数我或零”的东西/积累问题（你不能用列表理解来解决）：

result = collections.Counter()
for subdict in d["432701228292636694"].values():
    result[subdict["inviter"]] += subdict["uses"]

结果：

>>> result
{'125092433657593856': 2,
 '141431182792458241': 537,
 '147756076299845632': 1,
 '161132853256650752': 9,
 '173219378492473345': 36,
 '193896125969858561': 8,
 '193926566881918978': 2,
 '273852837988007936': 1,
 '277605262486011904': 3,
 '282764296075870208': 40,
 '285548207189000204': 4,
 '315630485114191872': 2,
 '327660257423917066': 1,
 '384664458628169728': 4}

Answer 2

假设dict存储为变量d：

o = {}
for s in d.values():
    for i in s.values():
        o[i['inviter']] = o.get(i['inviter'], 0) + i['uses']

o将变成：

{'125092433657593856': 2,
 '141431182792458241': 537,
 '147756076299845632': 1,
 '161132853256650752': 9,
 '173219378492473345': 36,
 '193896125969858561': 8,
 '193926566881918978': 2,
 '273852837988007936': 1,
 '277605262486011904': 3,
 '282764296075870208': 40,
 '285548207189000204': 4,
 '315630485114191872': 2,
 '327660257423917066': 1,
 '384664458628169728': 4}

Answer 3

现在不是使用简单词典的好时机吗？你知道吗

uses = {}
for _,v in test["432701228292636694"].items():
    if v["inviter"] in uses:
        uses[v["inviter"]] += v["uses"]
    else:
        uses[v["inviter"]] = v["uses"]

print (uses)

---

{'282764296075870208': 40, '173219378492473345': 36, '285548207189000204': 4, '193926566881918978': 2, '147756076299845632': 1, '161132853256650752': 9, '193896125969858561': 8, '273852837988007936': 1, '327660257423917066': 1, '315630485114191872': 2, '141431182792458241': 537, '384664458628169728': 4, '125092433657593856': 2, '277605262486011904': 3}