blocks|key|798109|text|如果没有冲突，或者如果您愿意接受与输入匹配的第一个键，您可以搜索这些键以找到一个以输入字符串开头的键：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|798110|input+=+"a"
dic+=+{'aaa':111,'bbb':222,'ccc':333}
dic[[k+for+k+in+dic.keys()+if+k.startswith(input)][0]]|code-block|syntax|javascript|798111|输出：|798112|111|798113|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 there can be no conflict, or if you're happy to take the first key that matches the input if there is, you could search the keys to find one which starts with the input string:
<pre class="lang-py prettyprint-override"><code>input = &quot;a&quot;
dic = {'aaa':111,'bbb':222,'ccc':333}
dic[[k for k in dic.keys() if k.startswith(input)][0]]
</code></pre>
Output:
<pre><code>111
</code></pre>

blocks|key|798122|text|尝尝这个。(Python+3.x)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|798123|input+=+"a"
dict+=+{'aaa':111,'bbb':222,'ccc':333}

def+find(inp,+dct):
++++for+k+in+dct.keys():
++++++++if+inp.lower()+in+k.lower():
++++++++++++return+dct[k]
++++return+"NOT+FOUND"

print(find(input,+dict))|code-block|syntax|javascript|798124|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this. (Python 3.x)
<pre><code>input = &quot;a&quot;
dict = {'aaa':111,'bbb':222,'ccc':333}

def find(inp, dct):
 for k in dct.keys():
 if inp.lower() in k.lower():
 return dct[k]
 return &quot;NOT FOUND&quot;

print(find(input, dict))
</code></pre>

blocks|key|1596267|text|一种方法(虽然不是最简单的方法)是对dict进行子类。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1596268|code00.py|1596269|#!/usr/bin/env+python

import+sys


class+LooseKeyDict(dict):
++++def+__matching_keys(self,+key):
++++++++return+[e+for+e+in+self.keys()+if+key+in+e]

++++def+__getitem__(self,+key):
++++++++mc+=+self.__matching_keys(key)
++++++++if+not+mc+or+len(mc)+>+1:
++++++++++++raise+KeyError(key)
++++++++return+super().__getitem__(mc[0])


def+main(*argv):
++++d0+=+LooseKeyDict({"aaa":+111,+"bbb":+222,+"ccc":+333})
++++print(d0["a"])
++++print(d0["aaa"])
++++try:
++++++++print(d0["aaa1"])
++++except+KeyError+as+ke:
++++++++print("KeyError:",+ke)

++++d1+=+LooseKeyDict({"aaa":+111,+"abc":+222})
++++try:
++++++++print(d1["a"])
++++except+KeyError+as+ke:
++++++++print("KeyError:",+ke)


if+__name__+==+"__main__":
++++print("Python+{:s}+{:03d}bit+on+{:s}\n".format("+".join(elem.strip()+for+elem+in+sys.version.split("\n")),
+++++++++++++++++++++++++++++++++++++++++++++++++++64+if+sys.maxsize+>+0x100000000+else+32,+sys.platform))
++++rc+=+main(*sys.argv[1:])
++++print("\nDone.")
++++sys.exit(rc)|code-block|syntax|javascript|1596270|输出|offset|length|style|BOLD|1596271|cfati@CFATI-5510-0:e:\Work\Dev\StackOverflow\q071851970>+"e:\Work\Dev\VEnvs\py_pc064_03.09_test0\Scripts\python.exe“码00.py+Python3.9.9(tag/v3.9.9:ccb0e6a，11月15日2021年，18:08:50)+MSC+v.1929+64位(AMD64)+064位(AMD64)064位在win32+222+222+KeyError上：'aaa1‘KeyError：'a’完成。|blockquote|1596272|Notes|1596273|这只是为了阅读部分。如果您也想为写(或删除)执行此操作，则还必须重写___setitem__+(或__delitem___)。关于[Python.Docs]：数据模型的更多细节|unordered-list-item|1596274|不要命名您的变量输入或dict，因为您正在跟踪内置的名称|1596275|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/reference/datamodel.html#data-model^0|0|0|0|0|2|0|0|0|5|0|1T|I|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|17|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|18|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|19|8|@$K|1A|L|1B|M|N]]|9|@]|A|$]]|$1|O|3|P|5|Q|7|1C|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|1D|8|@$K|1E|L|1F|M|N]]|9|@]|A|$]]|$1|T|3|U|5|V|7|1G|8|@]|9|@$K|1H|L|1I|1|1J]]|A|$]]|$1|W|3|X|5|V|7|1K|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]]|Z|$10|$5|11|12|13|A|$14|15]]]]

One way (although not the simplest one) would be subclassing dict.
code00.py:
<pre class="lang-py prettyprint-override"><code>#!/usr/bin/env python

import sys


class LooseKeyDict(dict):
 def __matching_keys(self, key):
 return [e for e in self.keys() if key in e]

 def __getitem__(self, key):
 mc = self.__matching_keys(key)
 if not mc or len(mc) &gt; 1:
 raise KeyError(key)
 return super().__getitem__(mc[0])


def main(*argv):
 d0 = LooseKeyDict({&quot;aaa&quot;: 111, &quot;bbb&quot;: 222, &quot;ccc&quot;: 333})
 print(d0[&quot;a&quot;])
 print(d0[&quot;aaa&quot;])
 try:
 print(d0[&quot;aaa1&quot;])
 except KeyError as ke:
 print(&quot;KeyError:&quot;, ke)

 d1 = LooseKeyDict({&quot;aaa&quot;: 111, &quot;abc&quot;: 222})
 try:
 print(d1[&quot;a&quot;])
 except KeyError as ke:
 print(&quot;KeyError:&quot;, ke)


if __name__ == &quot;__main__&quot;:
 print(&quot;Python {:s} {:03d}bit on {:s}\n&quot;.format(&quot; &quot;.join(elem.strip() for elem in sys.version.split(&quot;\n&quot;)),
 64 if sys.maxsize &gt; 0x100000000 else 32, sys.platform))
 rc = main(*sys.argv[1:])
 print(&quot;\nDone.&quot;)
 sys.exit(rc)
</code></pre>
Output:
<blockquote>
<pre class="lang-py prettyprint-override"><code>[cfati@CFATI-5510-0:e:\Work\Dev\StackOverflow\q071851970]&gt; &quot;e:\Work\Dev\VEnvs\py_pc064_03.09_test0\Scripts\python.exe&quot; code00.py
Python 3.9.9 (tags/v3.9.9:ccb0e6a, Nov 15 2021, 18:08:50) [MSC v.1929 64 bit (AMD64)] 064bit on win32

222
222
KeyError: 'aaa1'
KeyError: 'a'

Done.
</code></pre>
</blockquote>
Notes:
<ul>
<li>This is only for the read part. If you want to also do it for write (or delete), you'd have to also override __setitem__ (or __delitem__). More details on <a href="https://docs.python.org/reference/datamodel.html#data-model" rel="nofollow noreferrer">[Python.Docs]: Data model</a></li>
<li>Don't name your variables input or dict, as you're shadowing built-in names</li>
</ul>

I have a dict and input like that, and now I want to it can search the input even though it is only part of key words, so how to solve it?
<pre><code>input = &quot;a&quot;
dict = {'aaa':111,'bbb':222,'ccc':333}
print(dict[input])
</code></pre>
P.S Assuming there is no conflict between input and dict keywords, for example,
<pre><code>input ='a'
dict = {'aaa':111,'abc':222}
</code></pre>
is impossible

How to check the dict results if the input is only part of the keyword?

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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我有这样的一个切分和输入，现在我想它可以搜索输入，即使它只是关键字的一部分，那么如何解决它呢？input = "a"dict = {'aaa':111,'bbb':222,'ccc':333}print(dict[input])假设输入和关键词之间没有冲突，例如，input ='a'dict = {'aaa':111,...

问如果输入只是关键字的一部分，如何检查dict结果？
EN

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问如果输入只是关键字的一部分，如何检查dict结果？EN