Question

I have the ZZ tuple for 3 lists that each includes 7 lists. I would like to sort ZZ for each of three lists use descending order for the first element. If two first elements have the same values, I would like to first do descending order for them based on their differences. Second, if one of these lists has more cost give it priority. In the following you can see the lists:

ZZ= [[[500, 475], [500, 1307], [1500, 1476], [500, 537], [0, 527], [500, 452], [1000, 446]], [[350, 475], [700, 1307], [1050, 1476], [0, 537], [350, 527], [350, 452], [700, 446]], [[400, 475], [1200, 1307], [1200, 1476], [200, 537], [400, 527], [400, 452], [800, 446]]]
Cost=[600, 110, 35, 110, 200, 120, 80]

my code is:

YY=sorted(enumerate(ZZ[s-1]),key=lambda x: ( x[1], -x[0]), reverse=True)
index_list= [b[0] for b in sorted(enumerate(ZZ[s-1]),key=lambda x: ( x[1]),   reverse=True)]

the results of this code is:

YY = [(2, [1500, 1476]), (6, [1000, 446]), (1, [500, 1307]), (3, [500, 537]), (0, [500, 475]), (5, [500, 452]), (4, [0, 527])]

However I would have the following list:

YY = [(2, [1500, 1476]), (6, [1000, 446]), (0, [500, 475]), (5, [500, 452]), (3, [500, 537]),  (1, [500, 1307]), (4, [0, 527])]

Answer 1

如果要根据多个条件对列表进行排序，可以使用返回元组的lamda函数来执行此操作

一个小例子：

ZZ= [[500, 475], [500, 1307], [1500, 1476], [500, 537], [0, 527], [500, 452], [1000, 446]]
YY = sorted(ZZ,reverse=True,key=lambda y: (y[0],y[0]-y[1]))

y[0]是第一个值，y[0]-y[1]是差异。如果您想引入第二个列表（成本），您可以将两个列表和（如果需要）合并到位置。

ZZ= [[500, 475], [500, 1307], [1500, 1476], [500, 537], [0, 527], [500, 452], [1000, 446]]
Cost=[600, 110, 35, 110, 200, 120, 80]
ZZc=[ZZ[z]+[Cost[z]]+[z] for z in range(len(ZZ))]
print(ZZc)

打印

[[500, 475, 600, 0], [500, 1307, 110, 1], [1500, 1476, 35, 2], [500, 537, 110, 3], [0, 527, 200, 4], [500, 452, 120, 5], [1000, 446, 80, 6]]

现在您可以使用

对其进行排序

YY = sorted(ZZc,reverse=True,key=lambda y: (y[0],y[0]-y[1],y[2]))

或者如果你想排序1.值1,2。成本和3.差异你可以用它来排序：

YY = sorted(ZZc,reverse=True,key=lambda y: (y[0],y[2],y[0]-y[1]))

获取问题中的元组列表，脚本将是：

ZZ= [[[500, 475], [500, 1307], [1500, 1476], [500, 537], [0, 527], [500, 452], [1000, 446]], [[350, 475], [700, 1307], [1050, 1476], [0, 537], [350, 527], [350, 452], [700, 446]], [[400, 475], [1200, 1307], [1200, 1476], [200, 537], [400, 527], [400, 452], [800, 446]]] 
Cost=[600, 110, 35, 110, 200, 120, 80]

YY=[]
for Zi in ZZ:
    ZZi=[Zi[z]+[Cost[z]]+[z] for z in range(len(Zi))]
    YY.append(sorted(ZZi,reverse=True,key=lambda y: (y[0],y[2],y[0]-y[1])))

for Yi in YY:
    print([(y[3],[y[0],y[1]]) for y in Yi])

我原来的答案是因为我知道你可以用元组排序（它有效，但不能使用ist）：

您可以使用

对其进行排序

ZZ= [[[500, 475], [500, 1307], [1500, 1476], [500, 537], [0, 527], [500, 452], [1000, 446]], [[350, 475], [700, 1307], [1050, 1476], [0, 537], [350, 527], [350, 452], [700, 446]], [[400, 475], [1200, 1307], [1200, 1476], [200, 537], [400, 527], [400, 452], [800, 446]]] 
Cost=[600, 110, 35, 110, 200, 120, 80]

for Zi in ZZ:
    ZZi=[Zi[z]+[Cost[z]]+[z] for z in range(len(Zi))]
    YY = sorted(ZZi,reverse=True,key=lambda zlist: (int(zlist[0])*1000000)+(int(zlist[0]-zlist[1])*1000)+zlist[2])
    print [(y[3],[y[0],y[1]]) for y in YY]

Sort elements of a list with two or more criteria

1 个答案: