如何優化兩個元組列表的組合並刪除其重復項？

Question

從這里開始， 如果每個元組中的第二項都是重復的 ， 如何從元組列表中刪除元素？ ，我能夠從1個元組列表中刪除元組中第二個元素的重復項。

假設我有2個元組列表：

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')]

如果第二個元素相同（score_from_alist * score_from_blist），我需要合並分數，以實現所需的輸出：

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，我是通過這樣做來實現clist的，但是當alist和blist包含大約5500個以上的元組時，這需要5秒鍾以上的時間，其中第二個元素每個都包含20至40個單詞。 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

Answer 1

我將使用字典/集合來消除重復並提供快速查找：

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)

在這里， blist用於消除每個句子的除首出現外的所有內容，並按句子提供快速查找。

如果您不關心clist的順序，則可以在某種程度上簡化結構：

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()))

Answer 2

如果元組中的第一項按降序排序，則在單個列表中存在重復項的情況下，將元組中具有最高第一項的元組保留下來；如果元組中的對應第二項是相同：

# 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)

輸出：

[(0.510496368389, 'this is a foo bar sentence'),
 (0.10523121352499999, 'this is not really a foo bar')]