形成嵌套列表并计算字段的总和

Question

记录以以下形式存储在数据库中：

^{tb1}$

我正在编写一个函数，用于计算不同clu变量（例如，q和e）的给定值之和。也就是说，首先确定name的哪些值对应于e的值（它们是3和4）。然后是字典“e_dict”，它存储与不同名称对应的某些值。函数应确定，对于e组，有必要添加“名称”键3和4的值，并给出结果

词典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，这些字段都是u_平均值，找到它们的平均值，并将它们相加

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

结果是以这种形式分组：{'1'：'q'，'2'：'q'，'5'：'q'，'3'：'e'，'4'：'e'}

我不知道下一步该怎么办

Answer 1

您可以使用pandas进行以下操作：

输入：

data = {'clu': {0: 'q', 1: 'q', 2: 'q', 3: 'e', 4: 'e'}, 'name': {0: 'one', 1: 'two', 2: 'five', 3: 'three', 4: 'four'}, 'a': {0: 7, 1: 6, 2: 6, 3: 8, 4: 10}, 'b': {0: 6, 1: 9, 2: 10, 3: 7, 4: 4}}
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}}

代码：

import pandas as pd
df = pd.DataFrame(data)
df_e = pd.DataFrame(e_dict).T.rename_axis('name').reset_index()

df = df.merge(df_e, on=['name'])
df = df.groupby(['clu']).agg({'u_mean':'mean', 'c_mean':'mean'})
df.to_dict(orient='index')

输出：

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

解释：

Pandas允许以表格形式处理数据，类似于示例中显示的数据。 第一个表（DataFrame）是df，第二个表是包含查找值（e_dict）但经过位预处理（转置和重命名列）的表

然后我们根据两个表的名称值合并这两个表，以便在第一个表中有相应的u_mean和c_mean值

现在，我们用clue值对值进行分组，并用mean值对值进行聚合

最后，我们将该表作为字典返回

Answer 2

我不知道原始数据是如何存储的（我不知道“Utilizations.objects.values”），但下面的代码将根据简单的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}}