如何使用特定规则对列表进行排序

Question

嗨，我有一些复杂的数据对象，我想对它们进行排序。 下面的简化版本：

class Data(object):
    def __init__(self, s):
        self.s = s

每个数据对象都将放置在特定类别中，以便以后使用。 简化的版本再次在下面

class DataCategory(object):
    def __init__(self, id1, id2, linked_data=None):
        self.id1 = id1
        self.id2 = id2
        self.ld = linked_data

我想按以下编号对数据进行排序，但还有更多规则。 如果从第一个数据收集中使用了一个数据对象，那么我想从第二个收集中使用一个数据对象，如果s的数量相同或更低。 这就是我得到的以及我想要实现的

# order I get
# [['p02g01r05', 5], ['p02g01r01', 4], ['p01g01r05', 4], ['p01g01r01', 3], ['p01g01r02', 2], ['p01g01r03', 2], ['p01g01r06', 2], ['p02g01r02', 2], ['p02g01r03', 2], ['p02g01r04', 2], ['p01g01r04', 1], ['p02g01r06', 1]]
# order I want
# [['p02g01r05', 5], ['p01g01r05', 4], ['p02g01r01', 4], ['p01g01r01', 3], ['p02g01r02', 2], ['p01g01r02', 2], ['p02g01r03', 2], ['p01g01r03', 2], ['p02g01r04', 2], ['p01g01r06', 2], ['p02g01r06', 1]], ['p01g01r04', 1]

到目前为止，这是我创建的，但是我认为我朝这个方向走了。 我认为要替换的索引列表是正确的。

# Some data objects
p01g01r01 = Data(3)
p01g01r02 = Data(2)
p01g01r03 = Data(2)
p01g01r04 = Data(1)
p01g01r05 = Data(4)
p01g01r06 = Data(2)

p02g01r01 = Data(4)
p02g01r02 = Data(2)
p02g01r03 = Data(2)
p02g01r04 = Data(2)
p02g01r05 = Data(5)
p02g01r06 = Data(1)

p01g01 = DataCategory("01", "01", [])
p02g01 = DataCategory("02", "01", [])


# link data to data category
def ldtdc(dc):
    lst = []
    data = "p" + dc.id1 + "g" + dc.id2 + "r"
    for i in range(1, 7):
        if i < 10:
            lst.append(data + "0" + str(i))
        else:
            lst.append(data + str(i))
    return lst

p01g01.ld = ldtdc(p01g01)
p02g01.ld = ldtdc(p02g01)


# /@= This starts to get way too complicated fast ############################
def lstu(ag, dg):
    lst = []
    # data list of first collection
    dlofc = []
    # data list of second collection
    dlosc = []

    # for every data unit that exists in data collection
    for unit in ag.ld:
        # lst.append([unit, globals()[unit].s+10])
        lst.append([unit, globals()[unit].s])
        dlofc.append([unit, globals()[unit].s])

    for unit in dg.ld:
        lst.append([unit, globals()[unit].s])
        dlosc.append([unit, globals()[unit].s])

    # lambda function is used here to sort list by data value ([1] is index of the item)
    lst = sorted(lst, key=lambda x: x[1], reverse=True)
    # current index
    ci = 0

    previous_data = ["last data unit will be stored here", 0]
    # sorted list
    slst = []

    for unit in lst:
        try:
            next_data = lst[ci+1]
        except IndexError:
            next_data = ["endoflist", 0]
        if previous_data[0] == "last data unit will be stored here":
            pass
        elif previous_data[0][:6] == unit[0][:6]:
            if unit[0][:6] not in dlofc[0][0]:
                slst.append([unit[0], unit[1], ci])
            elif unit[0][:6] not in dlosc[0][0]:
                slst.append([unit[0], unit[1], ci])
            else:
                print "Error"

        previous_data = unit
        ci += 1

    print "slist below"
    print slst

    return lst
# \@= END #####################################################################


print p01g01.ld
print p02g01.ld


data_list = lstu(p01g01, p02g01)
print data_list

排序此数据的快速正确的方法是什么？

Answer 1

您是否尝试过先按字符串排序，然后按项目中的数字排序？

>>> items = [['p02g01r05', 5], ['p02g01r01', 4], ['p01g01r05', 4], ['p01g01r01', 3], ['p01g01r02', 2], ['p01g01r03', 2], ['p01g01r06', 2], ['p02g01r02', 2], ['p02g01r03', 2], ['p02g01r04', 2], ['p01g01r04', 1], ['p02g01r06', 1]]
>>> partially_sorted = sorted(items, key=lambda item: item[0], reverse=True)
>>> sorted(partially_sorted, key=lambda item: item[1], reverse=True)
[['p02g01r05', 5], ['p02g01r01', 4], ['p01g01r05', 4], ['p01g01r01', 3], ['p02g01r04', 2], ['p02g01r03', 2], ['p02g01r02', 2], ['p01g01r06', 2], ['p01g01r03', 2], ['p01g01r02', 2], ['p02g01r06', 1], ['p01g01r04', 1]]

Answer 2

找到解决方案。 新的lstu函数：

# replaced lambda with normal function
def get_key(item):
    return item[1]


def lstu(ag, dg):
    # ag list
    agslst = []
    # dg list
    dgslst = []

    # for every unit in first data collection
    for unit in ag.u:
        agslst.append([unit, globals()[unit].s])
    # sorted first data collection list
    agslst = sorted(agslst, key=get_key, reverse=True)
    print agslst

    for unit in dg.u:
        dgslst.append([unit, globals()[unit].s])
    # 2nd collection sorted list
    dgslst = sorted(dgslst, key=get_key, reverse=True)
    print dgslst

    lst = []
    # last item
    li = ["Empty", 0]

    for item in range(0, len(agslst)+len(dgslst)+1):
        if agslst and dgslst:
            if agslst[0][1] == dgslst[0][1]:
                if li[0][:6] == agslst[0][0][:6]:
                    li = dgslst.pop(0)
                    lst.append(li)
                else:
                    li = agslst.pop(0)
                    lst.append(li)

            elif agslst[0][1] > dgslst[0][1]:
                li = agslst.pop(0)
                lst.append(li)
            else:
                li = dgslst.pop(0)
                lst.append(li)

    return lst

这样，我就可以满足前面提到的新（和最终）清单的要求

输出：

[['p02g01r05', 5], ['p01g01r05', 4], ['p02g01r01', 4], ['p01g01r01', 3], ['p02g01r02', 2], ['p01g01r02', 2], ['p02g01r03', 2], ['p01g01r03', 2], ['p02g01r04', 2], ['p01g01r06', 2], ['p02g01r06', 1]], ['p01g01r04', 1]]

我愿意接受任何优化建议。