如何在dict中计算相同的索引？

Question

这是我的代码示例：

budi = {"Name" : "Budi", "Gender" : "Male", "Age" : 18}
ahmad = {"Name" : "Ahmad", "Gender" : "Male", "Age" : 7}
ika = {"Name" : "Ika", "Gender" : "Female", "Age" : 18}

marged = [budi, ahmad, ika]

我希望结果显示相同的值或索引，例如：

male : 2

Answer 1

始终可以使用^{}和条件理解：

num_males = sum(1 for subdict in marged if subdict['Gender'] == 'Male')

它迭代marged中的所有词典（这不应该称为merged吗？）并检查'Gender'是否为'Male'。您还可以通过以下方式检查5岁到10岁之间的人数：

num_age_5_to_10 = sum(1 for subdict in marged if 5 <= subdict["Age"] <= 10)

当然，也可以用布尔条件替换1（因为True等价于1，而False等价于0）：

num_males = sum(subdict['Gender'] == 'Male' for subdict in marged)
num_age_5_to_10 = sum(5 <= subdict["Age"] <= 10 for subdict in marged)

Answer 2

我认为你有一份记录清单。 你要做的是统计“性别”是“男性”的记录。你知道吗

您可以通过一个小函数来实现这一点：

budi = {"Name" : "Budi", "Gender" : "Male", "Age" : 18}
ahmad = {"Name" : "Ahmad", "Gender" : "Male", "Age" : 7}
ika = {"Name" : "Ika", "Gender" : "Female", "Age" : 18}

merged = [budi, ahmad, ika]

def count_items(records, key, value):
    values = [r[key] for r in records if key in r]
    return values.count(value)

示例：

print(count_items(merged, "Gender", "Male"))
# -> 2

print(count_items(merged, "Age", 7))
# -> 1

当然，您可以简化：

males = [r["Gender"] for r in merged].count("Male")