在Python中将排序优雅地归纳为字典？

Question

列表理解是一个很好的结构，它可以概括使用列表，从而可以优雅地管理列表的创建。在Python中有没有类似的管理词典的工具？你知道吗

我有以下功能：

    # takes in 3 lists of lists and a column specification by which to group
def custom_groupby(atts, zmat, zmat2, col):
    result = dict()
    for i in range(0, len(atts)):
        val = atts[i][col]
        row = (atts[i], zmat[i], zmat2[i])
        try:
            result[val].append(row)
        except KeyError:
            result[val] = list()
            result[val].append(row)
    return result

    # organises samples into dictionaries using the groupby
def organise_samples(attributes, z_matrix, original_z_matrix):
    strucdict = custom_groupby(attributes, z_matrix, original_z_matrix, 'SecStruc')

    strucfrontdict = dict()
    for k, v in strucdict.iteritems():
        strucfrontdict[k] = custom_groupby([x[0] for x in strucdict[k]],                                            
                                [x[1] for x in strucdict[k]], [x[2] for x in strucdict[k]], 'Front')

    samples = dict()
    for k in strucfrontdict:
        samples[k] = dict()
        for k2 in strucfrontdict[k]:
            samples[k][k2] = dict()
            samples[k][k2] = custom_groupby([x[0] for x in strucfrontdict[k][k2]],
                    [x[1] for x in strucfrontdict[k][k2]], [x[2] for x in strucfrontdict[k][k2]], 'Back')
    return samples

这似乎很难处理。在Python中，有很多优雅的方法可以完成几乎所有的事情，我倾向于认为我使用Python是错误的。你知道吗

更重要的是，我希望能够更好地概括这个函数，这样我就可以指定字典中应该有多少“层”（而不需要使用几个lambda和用Lisp风格处理问题）。我想要一个函数：

# organises samples into a dictionary by specified columns
# number of layers could also be assumed by number of criterion
def organise_samples(number_layers, list_of_strings_for_column_ids)

在Python中可以这样做吗？你知道吗

谢谢你！即使在Python中没有一种优雅的方法，任何使上述代码更优雅的建议都将非常感激。你知道吗

*编辑：：

对于context，attributes object、z_matrix和original_zmatrix都是Numpy数组的列表。你知道吗

属性可能如下所示：

Type,Num,Phi,Psi,SecStruc,Front,Back
11,181,-123.815,65.4652,2,3,19
11,203,148.581,-89.9584,1,4,1
11,181,-123.815,65.4652,2,3,19
11,203,148.581,-89.9584,1,4,1
11,137,-20.2349,-129.396,2,0,1
11,163,-34.75,-59.1221,0,1,9

Z矩阵可能都是这样的：

CA-1, CA-2, CA-CB-1, CA-CB-2, N-CA-CB-SG-1, N-CA-CB-SG-2
-16.801, 28.993, -1.189, -0.515, 118.093, 74.4629
-24.918, 27.398, -0.706, 0.989, 112.854, -175.458
-1.01, 37.855, 0.462, 1.442, 108.323, -72.2786
61.369, 113.576, 0.355, -1.127, 111.217, -69.8672

Samples是一个dict{num=>；dict{num=>；dict{num=>；tuple（attributes，z\u matrix）}}，具有一行z矩阵。你知道吗

Answer 1

The list comprehension is a great structure for generalising working with lists in such a way that the creation of lists can be managed elegantly. Is there a similar tool for managing Dictionaries in Python?

你试过用词典理解吗？你知道吗

看这个great question about dictionary comperhansions