我们可以通过多种方式加快匹配速度。我假设在你的代码中str1
是数据集中的名称,并且str2
是一个地理名称字符串。为了测试代码,我根据您问题中的数据制作了两个小数据集。我写了两个匹配函数best_match
and first_match
使用您当前的string_similarity
函数,这样我们就可以看到我的策略给出了相同的结果。best_match
检查所有地理名称字符串,如果超过给定的阈值分数,则返回分数最高的字符串,否则返回None
. first_match
(可能)更快:它只返回超过阈值的第一个地理名称字符串,或者None
如果它找不到,那么如果它没有找到匹配项,那么它仍然必须搜索整个地理名称列表。
在我的改进版本中,我们为每个生成二元组str1
一次,而不是重新生成二元组str1
对于每个str2
我们将其与它进行比较。我们提前计算所有的地理名称二元组,将它们存储在由字符串索引的字典中,这样我们就不必为每个地理名称二元组重新生成它们str
。此外,我们将地理名称二元组存储为集合。这使得计算hit_count
更快,因为集合成员资格测试比对字符串列表进行线性扫描要快得多。这geodict
还需要存储每个二元组的长度:集合不包含重复项,因此二元组集合的长度可能小于二元组列表,但我们需要列表长度才能正确计算分数。
# Some fake data
geonames = [
'Slettmarkmountains Jotunheimen Norway',
'Fairy Glen Skye Scotland UK',
'Emigrant Wilderness California',
'Yosemite National Park',
'Half Dome Yosemite National Park',
]
mynames = [
'Jotunheimen Norway',
'Fairy Glen',
'Slettmarkmountains Jotunheimen Norway',
'Bryce Canyon',
'Half Dome',
]
def get_bigrams(string):
"""
Take a string and return a list of bigrams.
"""
s = string.lower()
return [s[i:i+2] for i in range(len(s) - 1)]
def string_similarity(str1, str2):
"""
Perform bigram comparison between two strings
and return a percentage match in decimal form.
"""
pairs1 = get_bigrams(str1)
pairs2 = get_bigrams(str2)
union = len(pairs1) + len(pairs2)
hit_count = 0
for x in pairs1:
for y in pairs2:
if x == y:
hit_count += 1
break
return (2.0 * hit_count) / union
# Find the string in geonames which is the best match to str1
def best_match(str1, thresh=0.2):
score, str2 = max((string_similarity(str1, str2), str2) for str2 in geonames)
if score < thresh:
str2 = None
return score, str2
# Find the 1st string in geonames that matches str1 with a score >= thresh
def first_match(str1, thresh=0.2):
for str2 in geonames:
score = string_similarity(str1, str2)
if score >= thresh:
return score, str2
return None
print('Best')
for mystr in mynames:
print(mystr, ':', best_match(mystr))
print()
print('First')
for mystr in mynames:
print(mystr, ':', best_match(mystr))
print()
# Put all the geoname bigrams into a dict
geodict = {}
for s in geonames:
bigrams = get_bigrams(s)
geodict[s] = (set(bigrams), len(bigrams))
def new_best_match(str1, thresh=0.2):
pairs1 = get_bigrams(str1)
pairs1_len = len(pairs1)
score, str2 = max((2.0 * sum(x in pairs2 for x in pairs1) / (pairs1_len + pairs2_len), str2)
for str2, (pairs2, pairs2_len) in geodict.items())
if score < thresh:
str2 = None
return score, str2
def new_first_match(str1, thresh=0.2):
pairs1 = get_bigrams(str1)
pairs1_len = len(pairs1)
for str2, (pairs2, pairs2_len) in geodict.items():
score = 2.0 * sum(x in pairs2 for x in pairs1) / (pairs1_len + pairs2_len)
if score >= thresh:
return score, str2
return None
print('New Best')
for mystr in mynames:
print(mystr, ':', new_best_match(mystr))
print()
print('New First')
for mystr in mynames:
print(mystr, ':', new_first_match(mystr))
print()
output
Best
Jotunheimen Norway : (0.6415094339622641, 'Slettmarkmountains Jotunheimen Norway')
Fairy Glen : (0.5142857142857142, 'Fairy Glen Skye Scotland UK')
Slettmarkmountains Jotunheimen Norway : (1.0, 'Slettmarkmountains Jotunheimen Norway')
Bryce Canyon : (0.1875, None)
Half Dome : (0.41025641025641024, 'Half Dome Yosemite National Park')
First
Jotunheimen Norway : (0.6415094339622641, 'Slettmarkmountains Jotunheimen Norway')
Fairy Glen : (0.5142857142857142, 'Fairy Glen Skye Scotland UK')
Slettmarkmountains Jotunheimen Norway : (1.0, 'Slettmarkmountains Jotunheimen Norway')
Bryce Canyon : (0.1875, None)
Half Dome : (0.41025641025641024, 'Half Dome Yosemite National Park')
New Best
Jotunheimen Norway : (0.6415094339622641, 'Slettmarkmountains Jotunheimen Norway')
Fairy Glen : (0.5142857142857142, 'Fairy Glen Skye Scotland UK')
Slettmarkmountains Jotunheimen Norway : (1.0, 'Slettmarkmountains Jotunheimen Norway')
Bryce Canyon : (0.1875, None)
Half Dome : (0.41025641025641024, 'Half Dome Yosemite National Park')
New First
Jotunheimen Norway : (0.6415094339622641, 'Slettmarkmountains Jotunheimen Norway')
Fairy Glen : (0.5142857142857142, 'Fairy Glen Skye Scotland UK')
Slettmarkmountains Jotunheimen Norway : (1.0, 'Slettmarkmountains Jotunheimen Norway')
Bryce Canyon : None
Half Dome : (0.41025641025641024, 'Half Dome Yosemite National Park')
new_first_match
相当简单。线路
for str2, (pairs2, pairs2_len) in geodict.items():
循环遍历中的每个项目geodict
提取每个字符串、二元组和真实二元长度。
sum(x in pairs2 for x in pairs1)
计算有多少个二元组pairs1
是成员pairs2
set.
因此,对于每个地理名称字符串,我们计算相似度分数,如果它 >= 阈值(默认值为 0.2),则返回它。你可以给它一个不同的默认值thresh
,或通过thresh
当你调用它时。
new_best_match
有点复杂。 ;)
((2.0 * sum(x in pairs2 for x in pairs1) / (pairs1_len + pairs2_len), str2)
for str2, (pairs2, pairs2_len) in geodict.items())
是一个生成器表达式。它循环geodict
项目并创建一个(score, str2)
每个地理名称字符串的元组。然后我们将该生成器表达式提供给max
函数,返回得分最高的元组。
这是一个版本new_first_match
这实现了 juvian 在评论中提出的建议。它可能会节省一点时间。此版本还避免测试任一二元组是否为空。
def new_first_match(str1, thresh=0.2):
pairs1 = get_bigrams(str1)
pairs1_len = len(pairs1)
if not pairs1_len:
return None
hiscore = 0
for str2, (pairs2, pairs2_len) in geodict.items():
if not pairs2_len:
continue
total_len = pairs1_len + pairs2_len
bound = 2.0 * pairs1_len / total_len
if bound >= hiscore:
score = 2.0 * sum(x in pairs2 for x in pairs1) / total_len
if score >= thresh:
return score, str2
hiscore = max(hiscore, score)
return None
一个更简单的变化是不打扰计算hiscore
并比较bound
to thresh
.