Question

我有多本可预测的词典。每个字典包含3个元素。如何从每本词典中提取特定键并找到总值？

player_A = {
   "kills": 15
   "death": 3
   "kda": 5
}

player_B = {
   "kills": 20
   "death": 4
   "kda": 5
}
:
player_ZA = {
   "kills": 35
   "death": 7
   "kda": 5
}

比方说，我想找出所有玩家的总击杀数。我应该如何访问词典中所有名为“ kill”的键并找到这些键的总和？

即。预期的输出类似于：

sum(15 + 20 + ... + 35)

Answer 1

player_a = {"kills": 15, "death": 3, "kda": 5}
player_b = {"kills": 20, "death": 4, "kda": 5}
player_c = {"kills": 35, "death": 7, "kda": 5}

players = [player_a, player_b, player_c]

total_kills = sum(player["kills"] for player in players)

Answer 2

尝试这个：

player_A = {"kills": 15, "death": 3, "kda": 5}

player_B = {"kills": 20, "death": 4, "kda": 5}

player_ZA = {"kills": 35, "death": 7, "kda": 5}


def f(x):
    result = 0
    for key, value in x.items():
        if key.startswith("player"):
            result += value["kills"]
    return result


print(f(globals()))

或者这个：

player_A = {"kills": 15, "death": 3, "kda": 5 }

player_B = { "kills": 20, "death": 4, "kda": 5 }

player_ZA = { "kills": 35, "death": 7, "kda": 5 }


def f(x):
    return sum(value["kills"] for key, value in x.items() if key.startswith("player"))


print(f(globals()))

Answer 3

您也可以执行以下操作：-

player_a = {"kills": 15, "death": 3, "kda": 5}
player_b = {"kills": 20, "death": 4, "kda": 5}
player_c = {"kills": 35, "death": 7, "kda": 5}
player_d = {'death':5}  # If dictionary does not contains 'kills' key.

players = [player_a , player_b , player_c, player_d]

l = []
for player in players:
    try:
        l.append(player['kills'])

    except Exception as e:
        continue
print(f"Total kills are {sum(l)}")

输出

Total kills are 70

希望您能学到新知识。

Answer 4

这是将"kills"值相加的另一种方法（函数式编程）：

player_A = {"kills": 15, "death": 3, "kda": 5}

player_B = {"kills": 20, "death": 4, "kda": 5}

player_ZA = {"kills": 35, "death": 7, "kda": 5}


f = lambda x: sum(map(lambda j: x[j]["kills"], filter(lambda i: i.startswith("player"), x)))


print(f(globals()))

有没有办法从多个字典访问具有相同名称的键？

4 个答案: