Python：如果列表中的第一个元素重复并且第二个元素在列表系列中最低，则删除列表中的列表

Question

I have a list of lists. Each list has the same number of elements. I'd like to delete an entire list if a new list supersedes the old one based on a numeric key in the nth element that all the lists have. This numeric key is an increment of 1 starting from 1. Highest key is desired.

all = [[123, 1],[456, 1],[789, 1],[123,2],[456, 2],[789,1]]

The last element in each list is the key: 2 supersedes 1 etc... the output desired is:

[[123,2],[456,2],[789,1]]

Answer 1

for x in list(all):
    for y in list(all):
        if y[0] == x[0] and y[1] <= x[1] and y is not x:
            all.remove(y)

Answer 2

Would something like a dictionary work better here?

all = [[123, 1],[456, 1],[789, 1],[123,2],[456, 2],[789,1]]

as_dict = {}
for item in all:
    if not (item[0] in as_dict and as_dict[item[0]] > item[1]):
        as_dict[item[0]] = item[1]

print(as_dict)
# Returns {123: 2, 456: 2, 789: 1}

In fact, if you know that the second numbers in each pair will never decrease (eg, you won't see something like [123,0] appear later in the list after [123,2] ) then just converting the list to a dictionary with dict() should accomplish the same thing. Then you could convert it back to a list if you want.

d = dict(all)  # This is {123: 2, 456: 2, 789: 1}
newlist = [ [k,d[k]] for k in d] # This is [[123, 2], [456, 2], [789, 1]]