1

我有一个对象列表。每个对象都有两个属性：DispName和MachID。DispName可以开始，theoretical也可以是其他东西。

我需要按以下方式对该列表进行排序：

  • 第一个按字母顺序排列MachID。
    • 在每个MachID子组中，首先是名称开头的对象theoretical
    • 然后其他对象按字母顺序排序。

这是我现在拥有的代码，它可以工作并产生所需的输出，但我想知道我是否可以编写更多的pythonic，也许利用groupby？（我为驼峰找借口）。

from collections import defaultdict, namedtuple
from operator import attrgetter

Mapping = namedtuple('Mapping', ['DispName', 'MachID'])

objectList = [Mapping('map 2 (MT1)', 'MT1'),
          Mapping('theoretical (MT1)', 'MT1'),
          Mapping('map 3 (MT2)', 'MT2'),
          Mapping('theoretical (MT2)', 'MT2'),
          Mapping('map 1 (MT1)', 'MT1'),
          Mapping('map 2 (MT2)', 'MT2')]

def complexSort(objectList):
    objectDict = defaultdict(list)
    sortedMappingList = []
    # group by machine ID 
    for obj in objectList:
        objectDict[obj.MachID].append(obj)
    # loop over the mappings sorted alphabetically by machine ID
    for machID in sorted(objectDict.keys()):
        mappings = objectDict[machID]
        nonTheoreticalMappings = []
        for mapping in mappings:
            if mapping.DispName.startswith('theoretical'):
                # if we encounter the theoretical mapping, add it first
                sortedMappingList.append(mapping)
            else:
                # gather the other mappings in a sublist
                nonTheoreticalMappings.append(mapping)
        # and add that sublist sorted alphabetically
        sortedMappingList.extend(sorted(nonTheoreticalMappings, 
                                     key=attrgetter('DispName')))           
    return sortedMappingList

for mapping in complexSort(objectList):
    print mapping.DispName

产生：

theoretical (MT1)
map 1 (MT1)
map 2 (MT1)
theoretical (MT2)
map 2 (MT2)
map 3 (MT2)
4

3 回答 3

3

只需sorted与key产生您想要的订单的 a 一起使用。由于元组是按字典顺序排列的，key所以产生元组的 a 应该工作得很好。

def sort_key(thing):
    return (thing.MachID, not thing.DispName.startswith('theoretical'))

sorted(objectList, key=sort_key) # returns a list sorted the way you want
于 2013-07-10T17:12:32.600 回答
2 
import collections 
import operator
import itertools as IT

Mapping = collections.namedtuple('Mapping', ['DispName', 'MachID'])

objectList = [Mapping('map 2 (MT1)', 'MT1'),
          Mapping('theoretical (MT1)', 'MT1'),
          Mapping('map 3 (MT2)', 'MT2'),
          Mapping('theoretical (MT2)', 'MT2'),
          Mapping('map 1 (MT1)', 'MT1'),
          Mapping('map 2 (MT2)', 'MT2')]

sortedMappingList = sorted(objectList,
             key=lambda mapping:
                            (mapping.MachID,
                             not mapping.DispName.startswith('theoretical'),
                             mapping.DispName))

for key, group in IT.groupby(sortedMappingList, key=operator.attrgetter('MachID')):
    for g in group:
        print(g.DispName)

产量

theoretical (MT1)
map 1 (MT1)
map 2 (MT1)
theoretical (MT2)
map 2 (MT2)
map 3 (MT2)

这里有一个关于如何使用键功能进行排序的优秀教程。

于 2013-07-10T17:17:11.593 回答
1

您可以创建一个自定义比较器来描述两个映射应如何相互排序。这比你的要干净一些complexSort，因为该函数的唯一职责是比较两个对象，并将实际排序留给 Python。

from collections import namedtuple

Mapping = namedtuple('Mapping', ['DispName', 'MachID'])


def cmp_Mapping(a,b):
    #first, sort alphabetically by MachID
    if a.MachID != b.MachID:
        return cmp(a.MachID, b.MachID)
    else:
        #if MachIDs match, and one starts with "theoretical", it should go first.
        if a.DispName.startswith("theoretical") and not b.DispName.startswith("theoretical"):
            return -1
        elif b.DispName.startswith("theoretical") and not a.DispName.startswith("theoretical"):
            return 1
        #everything else is ordered alphabetically.
        else:
            return cmp(a.DispName, b.DispName)

objectList = [Mapping('map 2 (MT1)', 'MT1'),
          Mapping('theoretical (MT1)', 'MT1'),
          Mapping('map 3 (MT2)', 'MT2'),
          Mapping('theoretical (MT2)', 'MT2'),
          Mapping('map 1 (MT1)', 'MT1'),
          Mapping('map 2 (MT2)', 'MT2')]

for mapping in sorted(objectList, cmp = cmp_Mapping):
    print mapping.DispName

结果：

theoretical (MT1)
map 1 (MT1)
map 2 (MT1)
theoretical (MT2)
map 2 (MT2)
map 3 (MT2)
于 2013-07-10T17:25:56.273 回答