[英]How to optimize the combination of 2 lists of tuples and remove their duplicates?
从这里开始, 如果每个元组中的第二项都是重复的 , 如何从元组列表中删除元素? ,我能够从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,以实现所需的输出:
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,我是通过这样做来实现clist的,但是当alist和blist包含大约5500个以上的元组时,这需要5秒钟以上的时间,其中第二个元素每个都包含20至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
我将使用字典/集合来消除重复并提供快速查找:
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')]
bdict = {k:v for v,k in reversed(blist)}
clist = []
cset = set()
for v,k in alist:
if k not in cset:
b = bdict.get(k, None)
if b is not None:
clist.append((v * b, k))
cset.add(k)
print(clist)
在这里, blist
用于消除每个句子的除首出现外的所有内容,并按句子提供快速查找。
如果您不关心clist
的顺序,则可以在某种程度上简化结构:
bdict = {k:v for v,k in reversed(blist)}
cdict = {}
for v,k in alist:
if k not in cdict:
b = bdict.get(k, None)
if b is not None:
cdict[k] = v * b
print(list((k,v) for v,k in cdict.items()))
如果元组中的第一项按降序排序,则在单个列表中存在重复项的情况下,将元组中具有最高第一项的元组保留下来;如果元组中的对应第二项是相同:
# remove duplicates (take the 1st item among duplicates)
a, b = [{sentence: score for score, sentence in reversed(lst)}
for lst in [alist, blist]]
# merge (leave only tuples that have common 2nd items (sentences))
clist = [(a[s]*b[s], s) for s in a.viewkeys() & b.viewkeys()]
clist.sort(reverse=True) # sort by (score, sentence) in descending order
print(clist)
输出:
[(0.510496368389, 'this is a foo bar sentence'),
(0.10523121352499999, 'this is not really a foo bar')]
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