blocks|key|2258896|text|如果您只需要查找以前缀开头的键，则可以使用trie。存在更复杂的数据结构来查找键，其中任何地方都包含一个子字符串，但是它们占用了更多的存储空间，因此这是一种空时交换。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2258897|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Trie^0|L|4|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

If you only need to find keys that start with a prefix then you can use a <a href="http://en.wikipedia.org/wiki/Trie">trie</a>. More complex data structures exist for finding keys that contain a substring anywhere within them, but they take up a lot more space to store so it's a space-time trade-off.

blocks|key|2258914|text|如果您只需要查找以前缀开头的键，则可以使用二进制搜索。像这样的东西就能完成任务：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2258915|import+bisect
words+=+sorted("""
a+b+c+stack+stacey+stackoverflow+stacked+star+stare+x+y+z
""".split())
n+=+len(words)
print+n,+"words"
print+words
print
tests+=+sorted("""
r+s+ss+st+sta+stack+star+stare+stop+su+t
""".split())
for+test+in+tests:
++++i+=+bisect.bisect_left(words,+test)
++++if+words[i]+<+test:+i+%2B=+1
++++print+test,+i
++++while+i+<+n+and+words[i].startswith(test):
++++++++print+i,+words[i]
++++++++i+%2B=+1|code-block|syntax|javascript|2258916|输出：|2258917|12+words
['a',+'b',+'c',+'stacey',+'stack',+'stacked',+'stackoverflow',+'star',+'stare',
'x',+'y',+'z']

r+3
s+3
3+stacey
4+stack
5+stacked
6+stackoverflow
7+star
8+stare
ss+3
st+3
3+stacey
4+stack
5+stacked
6+stackoverflow
7+star
8+stare
sta+3
3+stacey
4+stack
5+stacked
6+stackoverflow
7+star
8+stare
stack+4
4+stack
5+stacked
6+stackoverflow
star+7
7+star
8+stare
stare+8
8+stare
stop+9
su+9
t+9|2258918|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If you only need to find keys that start with a prefix then you can use a binary search. Something like this will do the job:

<pre><code>import bisect
words = sorted("""
a b c stack stacey stackoverflow stacked star stare x y z
""".split())
n = len(words)
print n, "words"
print words
print
tests = sorted("""
r s ss st sta stack star stare stop su t
""".split())
for test in tests:
 i = bisect.bisect_left(words, test)
 if words[i] &lt; test: i += 1
 print test, i
 while i &lt; n and words[i].startswith(test):
 print i, words[i]
 i += 1
</code></pre>

Output:

<pre><code>12 words
['a', 'b', 'c', 'stacey', 'stack', 'stacked', 'stackoverflow', 'star', 'stare',
'x', 'y', 'z']

r 3
s 3
3 stacey
4 stack
5 stacked
6 stackoverflow
7 star
8 stare
ss 3
st 3
3 stacey
4 stack
5 stacked
6 stackoverflow
7 star
8 stare
sta 3
3 stacey
4 stack
5 stacked
6 stackoverflow
7 star
8 stare
stack 4
4 stack
5 stacked
6 stackoverflow
star 7
7 star
8 stare
stare 8
8 stare
stop 9
su 9
t 9
</code></pre>

blocks|key|2617856|text|您可以使用适当的分隔符将所有键连接到一个长字符串中，并使用字符串的find方法。太快了。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2617857|也许这段代码对你有帮助。search方法返回一个字典值列表，其键包含子字符串key。|2617858|class+DictLookupBySubstr(object):
++++def+__init__(self,+dictionary,+separator='\n'):
++++++++self.dic+=+dictionary
++++++++self.sep+=+separator
++++++++self.txt+=+separator.join(dictionary.keys())%2Bseparator

++++def+search(self,+key):
++++++++res+=+[]
++++++++i+=+self.txt.find(key)
++++++++while+i+>=+0:
++++++++++++left+=+self.txt.rfind(self.sep,+0,+i)+%2B+1
++++++++++++right+=+self.txt.find(self.sep,+i)
++++++++++++dic_key+=+self.txt[left:right]
++++++++++++res.append(self.dic[dic_key])
++++++++++++i+=+self.txt.find(key,+right%2B1)
++++++++return+res|code-block|syntax|javascript|2617859|entityMap^0|X|4|0|C|6|12|3|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

You could join all the keys into one long string with a suitable separator character and use the <code>find</code> method of the string. That is pretty fast.

Perhaps this code is helpful to you. The <code>search</code> method returns a list of dictionary values whose keys contain the substring <code>key</code>.

<pre><code>class DictLookupBySubstr(object):
 def __init__(self, dictionary, separator='\n'):
 self.dic = dictionary
 self.sep = separator
 self.txt = separator.join(dictionary.keys())+separator

 def search(self, key):
 res = []
 i = self.txt.find(key)
 while i &gt;= 0:
 left = self.txt.rfind(self.sep, 0, i) + 1
 right = self.txt.find(self.sep, i)
 dic_key = self.txt[left:right]
 res.append(self.dic[dic_key])
 i = self.txt.find(key, right+1)
 return res
</code></pre>

blocks|key|2258941|text|也许使用has_key也可以解决这个问题。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2258942|http://docs.python.org/release/2.5.2/lib/typesmapping.html|2258943|entityMap|0|LINK|mutability|MUTABLE|url^0|4|7|0|0|1M|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@]|D|@$9|S|A|T|1|U]]|E|$]]|$1|H|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|G]]]]

Perhaps using has_key solve this too.

<a href="http://docs.python.org/release/2.5.2/lib/typesmapping.html" rel="nofollow">http://docs.python.org/release/2.5.2/lib/typesmapping.html</a>

blocks|key|2258963|text|dpath可以很容易地为您解决这个问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2258964|http://github.com/akesterson/dpath-python|offset|length|2258965|$+easy_install+dpath
>>>+for+(path,+value)+in+dpath.util.search(MY_DICT,+"glob/to/start/{}".format(userinput),+yielded=True):
>>>+...++++#+(do+something+with+the+path+and+value)|code-block|syntax|javascript|2258966|您可以传递一个eglob+(‘path//to//某事/0-9a-z’)进行高级搜索。|2258967|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|15|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@$D|V|E|W|1|X]]|A|$]]|$1|F|3|G|5|H|7|Y|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|Z|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|C]]]]

dpath can solve this for you easily.

<a href="http://github.com/akesterson/dpath-python" rel="nofollow">http://github.com/akesterson/dpath-python</a>

<pre><code>$ easy_install dpath
&gt;&gt;&gt; for (path, value) in dpath.util.search(MY_DICT, "glob/to/start/{}".format(userinput), yielded=True):
&gt;&gt;&gt; ... # (do something with the path and value)
</code></pre>

You can pass an eglob ('path//to//something/[0-9a-z]') for advanced searching.

I want to know how I could perform some kind of index on keys from a python dictionary. The dictionary holds approx. 400,000 items, so I am trying to avoid a linear search.

Basically, I am trying to find if the <code>userinput</code> is inside any of the dict keys.

<pre><code>for keys in dict:
 if userinput in keys:
 DoSomething()
 break
</code></pre>

That would be an example of what I am trying to do. Is there a way to search in a more direct way, without a loop ? or what would be a more efficient way.

Clarification: The <code>userinput</code> is not exactly what the key will be, eg <code>userinput</code> could be <code>log</code>, whereas the key is <code>logfile</code>

Edit: any list/cache creation, pre-processing or organisation that can be done prior to searching is acceptable. The only thing that needs to be quick is the search for the key.

Search of Dictionary Keys python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想知道如何对python字典中的键执行某种索引。这本字典大约有几分。400,000项，所以我试图避免线性搜索。基本上，我正在试图找出userinput是否在任何一个丁字键中。for keys in dict:    if userinput in keys:        DoSomething()        b...

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