Question

我正在尝试确定参数值的长度方差，并在相应参数/值组合的集合之后打印方差值。

例如，date和date=2007-04-14中date=2007-08-19的方差值为0. id_eve中id_eve=479989的值{{1} }}和id_eve=47将是2.88。

从Group values with common domain and page values开始，我们有一组URL被解析，以提供一组网址的参数/值。

示例数据集：

id_eve=479

由以下Python代码解析：

www.domain.com/page?id_eve=479989&adm=no
www.domain.com/page?id_eve=47&adm=yes
www.domain.com/page?id_eve=479
domain.com/cal?view=month
domain.com/cal?view=day
ww2.domain.com/cal?date=2007-04-14
ww2.domain.com/cal?date=2007-08-19
www.domain.edu/some/folder/image.php?l=adm&y=5&id=2&page=http%3A//support.domain.com/downloads/index.asp&unique=12345
blog.news.org/news/calendar.php?view=day&date=2011-12-10
www.domain.edu/some/folder/image.php?l=adm&y=5&id=2&page=http%3A//.domain.com/downloads/index.asp&unique=12345
blog.news.org/news/calendar.php?view=month&date=2011-12-10

提供：

from collections import defaultdict
from urllib import quote
from urlparse import parse_qsl, urlparse

urls = defaultdict(list)
with open('links.txt') as f:
    for url in f:
        parsed_url = urlparse(url.strip())
        params = parse_qsl(parsed_url.query, keep_blank_values=True)
        for key, value in params:
            urls[parsed_url.path].append("%s=%s" % (key, quote(value)))

# printing results
for url, params in urls.iteritems():
    print url
    for param in params:
        print param

所需的额外部分是每个参数/值组合打印参数值的长度变化，以便匹配参数与上面输出中定义的类似URL（希望清楚地阅读）。

用于分组网址的组参数
计算参数值的长度
确定长度的变化

所以期望的输出是：

ww2.domain.com/cal
date=2007-04-14
date=2007-08-19
www.domain.edu/some/folder/image.php
l=adm
y=5
id=2
page=http%3A//support.domain.com/downloads/index.asp
unique=12345
l=adm
y=5
id=2
page=http%3A//.domain.com/downloads/index.asp
unique=12345
domain.com/cal
view=month
view=day
www.domain.com/page
id_eve=479989
adm=no
id_eve=47
adm=yes
id_eve=479
blog.news.org/news/calendar.php
view=day
date=2011-12-10
view=month
date=2011-12-10

Answer 1

from collections import defaultdict
from urllib import quote
from urlparse import parse_qsl, urlparse

我们需要能够计算方差：

def variance(values):
    mean = sum(values) / float(len(values))
    return sum((elem - mean)**2 for elem in values) / float(len(values))

我们希望按“密钥”分组，因此我们不会将"%s=%s"添加到defaultdict。

urls = defaultdict(lambda: defaultdict(list))
with open('links.txt') as f:
    for url in f:
        parsed_url = urlparse(url.strip())
        params = parse_qsl(parsed_url.query, keep_blank_values=True)
        for key, value in params:
            urls[parsed_url.path][key].append(quote(value))

然后我们可以通过打印东西

for domain, keys in urls.items():
    print domain
    for key, values in keys.items():
        for value in values:
            print "%s=%s" % (key, value)

        if len(values) > 1:
            print variance(map(len, values))

确定长度的变化

1 个答案: