如何从另一个列表列表中的条件项生成列表列表

Question

I have a list of lists and am trying to make another list of lists from specific items within the first list of lists

listOne = [[1,1,9,9],[1,4,9,6],[2,1,12,12]]
listTwo = []

for every inner list with the same numbers in positions 0 and 2 , append to listTwo only the inner list with the largest value in position 3

for example, inner list 0 and inner list 1 both have a 1 in position 0 and a 9 in position 2 , but inner list 0 has a 9 in position 3 and inner list 1 has a 6 in position 3 so I want to append inner list 1 and not inner list 9 to listTwo . Since inner list 2 is the only list with a 2 in position 0 and a 12 in position 1 , it need not be compared to anything else, and can be appended to listTwo.

I'm thinking something like:

for items in listOne : 
    #for all items where items[0] and items[2] are equal :
        #tempList = []
        #tempList.append(items)
        #tempList.sort(key = lambda x: (x[3]))
        #for value in tempList[0] :
            #listTwo.append(all lists with value in tempList[0])

but I'm not sure how to implement this without a lot of really bad looking code, any suggestions for a "pythonic" way of sorting these lists?

Answer 1

Perhaps throwing everything into a dictionary? Something like this:

def strangeFilter(listOne):
    listTwo = []
    d = {}

    for innerList in listOne:
        positions = (innerList[0],innerList[2])
        if positions not in d:
            d[positions] = []
        d[positions].append(innerList)

    for positions in d:
        listTwo.append(max(d[positions], key= lambda x: x[3]))

    return listTwo

Not sure how much of a 'pythonic' solution this is, but it uses python-defined structures alright and has a nice time order of O(n)

Answer 2

If you're looking to write concise python you will want to use list comprehensions wherever possible. Your description was a little confusing, but something like

list_two = [inner_list for inner_list in list_one if inner_list[0] == inner_list[2]]

will get you all of the inner lists in which the 0 and 2 indices match. Then you can search all these to find the one with the largest 3 index, assuming there aren't any ties

list_three = [0,0,0,0]
for i in list_two:
    if i[3] > list_three[3]:
        list_three = i

Answer 3

Sort the list on items zero and two of the inner-lists . Using itertools.groupby extract the item in each group that has a maximum value at position 3.

import operator, itertools

# a couple of useful callables for the key functions
zero_two = operator.itemgetter(0,2)
three = operator.itemgetter(3)

a = [[2,1,12,22],[1,1,9,9],[2,1,12,10],
     [1,4,9,6],[8,8,8,1],[2,1,12,12],
     [1,3,9,8],[2,1,12,15],[8,8,8,0]
     ]

a.sort(key = zero_two)
for key, group in itertools.groupby(a, zero_two):
    print(key, max(group, key = three))

'''
>>> 
(1, 9) [1, 1, 9, 9]
(2, 12) [2, 1, 12, 22]
(8, 8) [8, 8, 8, 1]
>>>
'''
result = [max(group, key = three) for key, group in itertools.groupby(a, zero_two)]

You could also sort on items zero, two, three. Then group by items zero and two and extract the last item of the group.

zero_two_three = operator.itemgetter(0,2,3)
zero_two = operator.itemgetter(0,2)
last_item = operator.itemgetter(-1)
a.sort(key = zero_two_three)
for key, group in itertools.groupby(a, zero_two):
    print(key, last_item(list(group)))