Question

这是我能想到的两个版本。当两个单词都是常见的时（例如“是”和“the”，版本1的n1 * n2缩放将是一个问题），并且对恶意输入（例如只有两个单词的文件）更健壮，V2是优选的。但是对于更有趣的查询（比如“大”和“动物”），v1同样快，我可以想到更真实的语义问题，而v2根本不会起作用，而v1会起作用。有没有办法加快速度？

导入时间 T1 = timeit.default_timer（）

def distance（version，filename，wordOne，wordTwo）：

f = open(filename, 'rU')
text = f.read()
f.close()
index = 0
distance = index
version = int(version)
print 'inputs', filename, wordOne, wordTwo
countOne = 0
countTwo = 0

print 'version', version

if version == 1:
    word_pos = {}
    for word in text.split():
        if word in [wordOne, wordTwo]:
            if word in word_pos.keys():
                word_pos[word].append(index)
            else:
                word_pos[word] = [index]

        index += 1

    countOne = len(word_pos[wordOne])
    countTwo = len(word_pos[wordTwo])

    distances = []
    low = 0
    high = index
    for posOne in word_pos[wordOne]:
        for posTwo in word_pos[wordTwo]:
            #shrink innner loop by distance?:
            #for posTwo in range(int(posOne-distance), (posOne+distance)):
            #if abs(posOne-posTwo) < distance:
            #distance = abs(posOne-posTwo)
            distances.append(abs(posOne-posTwo))
    distance = min(distances)

elif version == 2:
    switch = 0
    indexOne = 0
    indexTwo = 0
    distance = len(text)
    for word in text.split():

        if word == wordOne:
            indexOne = index
            countOne += 1
        if word == wordTwo:
            indexTwo = index
            countTwo += 1

        if indexOne != 0 and indexTwo != 0:
            if distance > abs(indexOne-indexTwo):
                distance = abs(indexOne - indexTwo)

        index += 1

t2 = timeit.default_timer()
print 'Delta t:', t2 - t1

print 'number of words in text:', index
print 'number of occurrences of',wordOne+':', countOne
print 'number of occurrences of',wordTwo+':', countTwo
if countOne < 1 or countTwo < 1:
    print 'not all words are present'
    return 1

print 'Shortest distance between \''+wordOne+'\' and \''+wordTwo+'\' is', distance, 'words'
return distance

Answer 1

v2中昂贵的部分是if indexOne != 0 ...块。一旦找到wordOne和wordTwo，文本中的剩余单词就会被调用多次。使用switch变量（我发现你有意使用它:)它可以移动它，如果阻塞到if word == wordOne和if word == wordTwo。在这种情况下，块的调用小于n1 + n2次。

这是代码。请注意，我们不再需要检查索引。

elif version == 3:
    last_word_is_one = None
    indexOne = 0
    indexTwo = 0
    countOne = 0
    countTwo = 0
    distance = len(text)
    for word in text.split():

        if word == wordOne:
            indexOne = index
            countOne += 1

            if last_word_is_one == False:
                if distance > abs(indexOne-indexTwo):
                    distance = abs(indexOne - indexTwo)

            last_word_is_one = True

        if word == wordTwo:
            indexTwo = index
            countTwo += 1

            if last_word_is_one == True:
                if distance > abs(indexOne-indexTwo):
                    distance = abs(indexOne - indexTwo)

            last_word_is_one = False

        index += 1

两个单词之间的最小距离（python）

1 个答案: