我有一個列表列表：['test', 'testlink', 1]['test', 'testlink', 2]['test ', 'testlink', 3]['test ', 'testlink', 4]['test 2', 'test2link', 1]['test 2', 'test2link', 2]['test 3', 'test3link', 1]['test 3', 'test3link', 3]['test 3', 'test3link', 4]我想過濾列表以返回每個唯一第一個元素的第三個元素的最大值。我想要的結果是：['test ', 'testlink', 4]['test 2', 'test2link', 2]['test 3', 'test3link', 4]我正在努力尋找一種方法來做到這一點。下面的代碼塊進入我的排序列表，但在那之后我被卡住了。#create some dummy datarows=[]rows.append(["test","testlink",1])rows.append(["test 2","test2link",1])rows.append(["test 3","test3link",1])rows.append(["test","testlink",2])rows.append(["test","testlink",1])rows.append(["test 2","test2link",1])rows.append(["test 3","test3link",1])rows.append(["test ","testlink",3])rows.append(["test 3","test3link",3])rows.append(["test ","testlink",4])rows.append(["test 3","test3link",4])rows.append(["test 2","test2link",2])#filter out duplicatesnewRows = []for elem in rows: if elem not in newRows: newRows.append(elem)rows = newRows#sort the listsrows = sorted(rows,key=lambda x: (x[0],x[2]))謝謝所有排序給出的答案都可以在我的 Ironpython 環境中工作
2 回答
UYOU
TA貢獻1878條經驗 獲得超4個贊
您可以使用defaultdict：
from collections import defaultdict
l = [['test', 'testlink', 1],
['test', 'testlink', 2],
['test', 'testlink', 3],
['test', 'testlink', 4],
['test 2', 'test2link', 1],
['test 2', 'test2link', 2],
['test 3', 'test3link', 1],
['test 3', 'test3link', 3],
['test 3', 'test3link', 4]]
d = defaultdict(int)
for first, second, third in l:
if d[(first, second)] < third:
d[(first, second)] = third
要形成您想要使用下一行的格式的結果：
res = [[*key, value] for key, value in d.items()]
如果您不想導入defaultdict，可以使用常規的：
d = {}
for first, second, third in l:
item = d.get((first, second), None)
if not item or item < third:
d[(first, second)] = third
慕妹3146593
TA貢獻1820條經驗 獲得超9個贊
使用itertools.groupby（doc）的一種解決方案：
rows=[]
rows.append(["test","testlink",1])
rows.append(["test 2","test2link",1])
rows.append(["test 3","test3link",1])
rows.append(["test","testlink",2])
rows.append(["test","testlink",1])
rows.append(["test 2","test2link",1])
rows.append(["test 3","test3link",1])
rows.append(["test ","testlink",3])
rows.append(["test 3","test3link",3])
rows.append(["test ","testlink",4])
rows.append(["test 3","test3link",4])
rows.append(["test 2","test2link",2])
from itertools import groupby
out = []
for v, g in groupby(sorted(rows, key=lambda k: k[0].strip()), lambda k: k[0].strip()):
item = max(g, key=lambda k: k[2])
out.append(item)
from pprint import pprint
pprint(out)
印刷：
[['test ', 'testlink', 4],
['test 2', 'test2link', 2],
['test 3', 'test3link', 4]]
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