从这里开始,如果每个元组中的第二个项目是重复的,我如何从元组列表中删除元素?,我可以从 1 个元组列表中删除元组中第二个元素的重复项。
假设我有 2 个元组列表:
alist = [(0.7897897,'this is a foo bar sentence'),
(0.653234, 'this is a foo bar sentence'),
(0.353234, 'this is a foo bar sentence'),
(0.325345, 'this is not really a foo bar'),
(0.323234, 'this is a foo bar sentence'),]
blist = [(0.64637,'this is a foo bar sentence'),
(0.534234, 'i am going to foo bar this sentence'),
(0.453234, 'this is a foo bar sentence'),
(0.323445, 'this is not really a foo bar')]
如果第二个元素相同(score_from_alist * score_from_blist),我需要合并分数并获得所需的输出:
clist = [(0.51,'this is a foo bar sentence'), # 0.51 = 0.789 * 0.646
(0.201, 'this is not really a foo bar')] # 0.201 = 0.325 * 0.323
目前,我正在通过这样做来实现 clist,但是当我的 alist 和 blist 有大约 5500 多个元组时,它需要 5 多秒,其中第二个元素每个大约有 20-40 个单词。有什么方法可以使以下功能更快?
def overlapMatches(alist, blist):
start_time = time.time()
clist = []
overlap = set()
for d in alist:
for dn in blist:
if d[1] == dn[1]:
score = d[0]*dn[0]
overlap.add((score,d[1]))
for s in sorted(overlap, reverse=True)[:20]:
clist.append((s[0],s[1]))
print "overlapping matches takes", time.time() - start_time
return clist