python - 形成嵌套列表并计算字段总和

Question

记录以这种形式存储在数据库中:


克鲁
姓名
一种
乙


q
一
7
6

q
二
6
9

q
五
6
10

电子
三
8
7

电子
四
10
4


我正在编写一个函数来计算不同 clu 变量(例如 q 和 e)的给定值的总和。也就是说，首先确定哪些 name 值对应于 e 的值(它们是 3 和 4)。然后是对字典“e_dict”的调用，它存储了与不同名称对应的某些值。该函数应确定对于组 e，必须将“名称”键 3 和 4 的值相加并给出结果。
字典 e_dict 示例:

{'one': {'u_mean': 4.25, 'c_mean': 4.25}, 'three': {'u_mean': 4.5, 'c_mean': 4.5}, 'two': {'u_mean': 4.583333333333334, 'c_mean': 4.583333333333334}, 'four': {'u_mean': 4.5625, 'c_mean': 4.5625}, 'five': {'u_mean': 4.65, 'c_mean': 4.65}}

结果应该是这样的:

 {'e': {'u_mean': 4.531, 'c_mean': 4.531}, 'q': {'u_mean': 4.49443, 'c_mean': 4.49443}}

也就是说，字段都是 u_mean 并找到它们的平均值，并且它们也被添加。
我的函数的完整代码:

def group_names():
    st, c_clus, n_names = [], [], []
    for h in Utilizations.objects.values('clu', 'name', 'a', 'b'):
        st.append((h.get('clu'), h.get('name'), h.get('a'), h.get('b')))
        c_clus.append(h.get('clu'))
        n_names.append(h.get('name'))
    """получение названий"""
    names, clus = [], []
    for nam in n_names:
        if nam not in names:
            names.append(nam)
    for cl in c_clus:
        if cl not in clus:
            clus.append(cl)
    clu, e = {}, {}
    u_load, u_max = {}, {}
    mean_all, u_load_mean, u_max_mean = 0, 0, 0
    for nam in names:
        hs = Utilizations.objects.filter(name=nam)
        o, p = 0, 0
        for h in hs:
            o += h.a
            p += h.b
            u_load[nam] = o / 2 + 1
            u_max[nam] = p / 2 + 1
        u_max_mean = mean(u_max.values())
        u_load_mean = mean(u_load.values())
        mean_all = (u_max_mean + u_load_mean) / 2
        e[nam] = {'u_mean': mean_all, 'c_mean': mean_all}
    for cl in clus:
        for nam in names:
            s = Utilizations.objects.filter(name=nam, clu=cl)
            for h in hs:
                clu[nam] = cl
    return clu

结果以这种形式分组:{'one': 'q', 'two': 'q', 'five': 'q', 'three': 'e', 'four': 'e'}
而且我不知道接下来要做什么(

Answer 1

我不知道您的原始数据是如何存储的(我不认识“Utilizations.objects.values”)，但这里的代码将根据简单的列表列表计算这些平均值:

data = [
    ['q','one',7,6],
    ['q','two',7,6],
    ['q','five',7,6],
    ['e','three',7,6],
    ['e','four',7,6]
]

e_dict = {
        'one': {'u_mean': 4.25, 'c_mean': 4.25},
        'three': {'u_mean': 4.5, 'c_mean': 4.5},
        'two': {'u_mean': 4.583333333333334, 'c_mean': 4.583333333333334},
        'four': {'u_mean': 4.5625, 'c_mean': 4.5625},
        'five': {'u_mean': 4.65, 'c_mean': 4.65}
}

def group_names():
    sums = {}
    counts = {}
    for h in data:
        if h[0] not in sums:
            sums[h[0]] = { "u_mean": 0, "c_mean": 0 }
            counts[h[0]] = 0
        for k,v in e_dict[h[1]].items():
            sums[h[0]][k] += v
        counts[h[0]] += 1
    for k,v in sums.items():
        sums[k]['u_mean'] /= counts[k]
        sums[k]['c_mean'] /= counts[k]
    return sums

print(group_names())

输出:

{'q': {'u_mean': 4.4944444444444445, 'c_mean': 4.4944444444444445}, 'e': {'u_mean': 4.53125, 'c_mean': 4.53125}}