python3汇总元素列表形式的列表

Question

I have a list of lists

lst = [[0,1],[1,1],[3,1],[1,2],[1,3],[3,5]]

I want to summarize the second items of the sublists for equal first items. The result should look like

lst = [[0,1],[1,6],[3,6]]

I tried something like this:

lst=[[0, 0.25], [1, 0.125], [2, 0.0625], [3, 0.0625], [1, 0.125], [2, 0.0625], [3, 0.03125], [4, 0.03125]]
for listitem in range(len(lst)):
    first=listitem[0]
    second = 0
    for obj in lst:
        if obj[0] == first:
            second += obj[1]
            lst.remove(obj)
    listitem[1] = second

Answer 1

Using defaultdict

    from collections import defaultdict

    lst=[[0, 0.25], [1, 0.125], [2, 0.0625], [3, 0.0625], [1, 0.125], [2, 0.0625], [3, 0.03125], [4, 0.03125]]

    res = defaultdict(int)    
    for el in lst:
        res[el[0]] += el[1] 
    res_list = [[k,v] for (k,v) in res.items()]

    print(res_list)

[[0, 0.25], [1, 0.25], [2, 0.125], [3, 0.09375], [4, 0.03125]]

Answer 2

groupby returns a key and iterator over the items with the same key. It must be passed a list sorted by the same key:

from itertools import groupby
from operator import itemgetter

lst = [[0,1],[1,1],[3,1],[1,2],[1,3],[3,5]]

result = [[k,sum(b for a,b in g)]
           for k,g in groupby(sorted(lst),key=itemgetter(0))]

print(result)

Output:

 [[0, 1], [1, 6], [3, 6]]

Answer 3

Another way to think about this is that the first item in each pair is the key , and the second is the value -- so if you create a dict you can then add the values together as you come across each key -- and it's even easier if you use a defaultdict :

from collections import defaultdict

summary = defaultdict(int)
for sublist in lst:
    key, value = sublist
    summary[key] += value

print(summary.items())

The line summary[key] += value is the workhorse here. What it does:

• looks up the key in summary
• if it doesn't exist add it and use int to create the default value
• return the value (either already there or freshly created)
• add the new value to it
• store it back to summary under key

Answer 4

range(n) produces a list such as [0,1,2...n-1] , so the type of listitem is int rather than list .

lst=[[0, 0.25], [1, 0.125], [2, 0.0625], [3, 0.0625], [1, 0.125], [2, 0.0625], [3, 0.03125], [4, 0.03125]]
for listitem in lst:
    first = listitem[0]
    second = 0
    for obj in lst:
        if obj[0] == first:
            second += obj[1]
            lst.remove(obj)
    listitem[1] = second