元组部分匹配

发布于 2024-12-01 04:49:55 字数 585 浏览 2 评论 0原文

我有一个元组的元组和一个元组。我有兴趣知道第一个元组的哪些元素与第二个元组(如果有)匹配,也考虑部分匹配。

这是一个过滤函数来演示我的意思。

def f(repo):
    pattern = (None, None, '1.3')
    for idx, item in enumerate(pattern):
        if item != None and item != repo[idx]:
            return False
    return True

>>> repo = (('framework', 'django', '1.3'), ('cms', 'fein', '1.3'), ('cms', 'django-cms', '2.2'))
>>> filter(f, repo)
(('framework', 'django', '1.3'), ('cms', 'fein', '1.3'))

过滤器在这种形式下是无用的,因为模式不能作为参数从外部提供(我想使用相同的函数来检查不同的输入)。有办法解决这个问题吗?

并且,为了更好地解决原始问题,可以采用另一种算法吗?

I have a tuple of tuples and a tuple. I'm interested to know which elements of the first tuple match the second tuple (if any), considering partial matches too.

This is a filter function to demonstrate what I mean.

def f(repo):
    pattern = (None, None, '1.3')
    for idx, item in enumerate(pattern):
        if item != None and item != repo[idx]:
            return False
    return True

>>> repo = (('framework', 'django', '1.3'), ('cms', 'fein', '1.3'), ('cms', 'django-cms', '2.2'))
>>> filter(f, repo)
(('framework', 'django', '1.3'), ('cms', 'fein', '1.3'))

The filter is useless in this form because the pattern can't be provided externally as an argument (I want to use the same function to check different inputs). Is there a way to fix this?

And, what could be another algorithm to embrace for a better approach to the original problem?

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。

评论(6

再见回来 2024-12-08 04:49:55

为什么不使用内置的 过滤器

>>> filter(lambda x: x[2] == '1.3', repo)
<<< (('framework', 'django', '1.3'), ('cms', 'fein', '1.3'))

...或 列表理解

>>> [x for x in repo if x[2] == '1.3']
<<< [('framework', 'django', '1.3'), ('cms', 'fein', '1.3')]

如果你想换行它变成一个函数:

types = {'desc': 0, 'name': 1, 'version': 2}
def repo_filter(type, critera, repo=repo, types=types):
    return [x for x in repo if x[types[type]] == critera]

>>> repo_filter('version', '1.3')
<<< [('framework', 'django', '1.3'), ('cms', 'fein', '1.3')]

Why don't you use the built-in filter:

>>> filter(lambda x: x[2] == '1.3', repo)
<<< (('framework', 'django', '1.3'), ('cms', 'fein', '1.3'))

...or a list comprehension:

>>> [x for x in repo if x[2] == '1.3']
<<< [('framework', 'django', '1.3'), ('cms', 'fein', '1.3')]

If you wanted to wrap it up into a function:

types = {'desc': 0, 'name': 1, 'version': 2}
def repo_filter(type, critera, repo=repo, types=types):
    return [x for x in repo if x[types[type]] == critera]

>>> repo_filter('version', '1.3')
<<< [('framework', 'django', '1.3'), ('cms', 'fein', '1.3')]
羞稚 2024-12-08 04:49:55

您可以使用闭包将模式绑定到函数中:

def matcher(pattern):
    def f(repo):
        return all(p is None or r == p for r, p in zip(repo, pattern))
    return f

>>> repo = (('framework', 'django', '1.3'), ('cms', 'fein', '1.3'), ('cms', 'django-cms', '2.2'))
>>> pattern = (None, None, '1.3')
>>> filter(matcher(pattern), repo)
(('framework', 'django', '1.3'), ('cms', 'fein', '1.3'))

我还提供了一个不同的表达式来比较元组。

You can use a closure to bind the pattern into the function:

def matcher(pattern):
    def f(repo):
        return all(p is None or r == p for r, p in zip(repo, pattern))
    return f

>>> repo = (('framework', 'django', '1.3'), ('cms', 'fein', '1.3'), ('cms', 'django-cms', '2.2'))
>>> pattern = (None, None, '1.3')
>>> filter(matcher(pattern), repo)
(('framework', 'django', '1.3'), ('cms', 'fein', '1.3'))

I've also provided a different expression for comparing the tuples.

迷离° 2024-12-08 04:49:55
In [43]: [r for r in repo if all((p is None or q==p) for q,p in zip(r,pattern))]
Out[43]: [('framework', 'django', '1.3'), ('cms', 'fein', '1.3')]
In [43]: [r for r in repo if all((p is None or q==p) for q,p in zip(r,pattern))]
Out[43]: [('framework', 'django', '1.3'), ('cms', 'fein', '1.3')]
毁梦 2024-12-08 04:49:55
def my_filter(pattern, repo):
    def f
        pattern = (None, None, '1.3')
        for idx, item in enumerate(pattern):
            if item != None and item != repo[idx]:
                return False
        return True
    return filter(f, repo)


 my_filter((None, None, '1.3'), repo)
def my_filter(pattern, repo):
    def f
        pattern = (None, None, '1.3')
        for idx, item in enumerate(pattern):
            if item != None and item != repo[idx]:
                return False
        return True
    return filter(f, repo)


 my_filter((None, None, '1.3'), repo)
温柔嚣张 2024-12-08 04:49:55

怎么样:

def f(repo, pattern=None):
    if not pattern:
        pattern = (None, None, '1.3')
    for idx, item in enumerate(pattern):
        if item and item != repo[idx]:
            return False
    return True

repo = (('framework', 'django', '1.3'), ('cms', 'fein', '1.3'), ('cms', 'django-cms', '2.2'))

[x for x in repo if f(x)]
>>>[('framework', 'django', '1.3'), ('cms', 'fein', '1.3')]    

[x for x in repo if f(x, ('cms',None, None))]
>>> [('cms', 'fein', '1.3'), ('cms', 'django-cms', '2.2')]

What about:

def f(repo, pattern=None):
    if not pattern:
        pattern = (None, None, '1.3')
    for idx, item in enumerate(pattern):
        if item and item != repo[idx]:
            return False
    return True

repo = (('framework', 'django', '1.3'), ('cms', 'fein', '1.3'), ('cms', 'django-cms', '2.2'))

[x for x in repo if f(x)]
>>>[('framework', 'django', '1.3'), ('cms', 'fein', '1.3')]    

[x for x in repo if f(x, ('cms',None, None))]
>>> [('cms', 'fein', '1.3'), ('cms', 'django-cms', '2.2')]
◇流星雨 2024-12-08 04:49:55

你可以使用下面的表达式:

repo = (('framework', 'django', '1.3'), ('cms', 'fein', '1.3'), ('cms', 'django-cms', '2.2'))
p = (None, None, '1.3')
matches = [i for i in repo if i[0]==p[0] or i[1]==p[1] or i[2]==p[2]]

或者使用闭包,比如这样:

def matcher(pattern):
    def pattern_matcher(repo):
        for idx, item in enumerate(pattern):
            if item is not None and item != repo[idx]:
                return False
        return True
    return pattern_matcher

然后可以像这样调用:

filter(matcher(pattern), repo)

You can use the following expression:

repo = (('framework', 'django', '1.3'), ('cms', 'fein', '1.3'), ('cms', 'django-cms', '2.2'))
p = (None, None, '1.3')
matches = [i for i in repo if i[0]==p[0] or i[1]==p[1] or i[2]==p[2]]

or use closure, such as this:

def matcher(pattern):
    def pattern_matcher(repo):
        for idx, item in enumerate(pattern):
            if item is not None and item != repo[idx]:
                return False
        return True
    return pattern_matcher

and then can be invoked like this:

filter(matcher(pattern), repo)
~没有更多了~
我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
原文