python - 在 Python 中优雅地将排序推广到字典中？

Question

列表推导式是一个很好的结构，可以以一种可以优雅地管理列表创建的方式概括列表的工作。 Python 中有类似的管理字典的工具吗？

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

这似乎很笨拙。在 Python 中几乎所有事情都有优雅的方法可以完成，我倾向于认为我错误地使用了 Python。

更重要的是，我希望能够更好地概括这个函数，以便我可以指定字典中应该有多少“层”(不使用多个 lambda 并以 Lisp 风格解决问题)。我想要一个功能:

# 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)

这可以用 Python 实现吗？

谢谢!即使没有办法在 Python 中优雅地做到这一点，任何使上述代码更优雅的建议都将非常感激。

::编辑::

就上下文而言，属性对象、z_matrix 和original_zmatrix 都是Numpy 数组的列表。

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 矩阵可能如下所示:

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

样本是一个 dict{num => dict {num => dict {num => tuple(attributes, z_matrix)}}}，具有一行 z 矩阵。

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?

您尝试过使用字典理解吗？

看到这个great question about dictionary comperhansions