blocks|key|491222|text|>>>+takeClosest+=+lambda+num,collection:min(collection,key=lambda+x:abs(x-num))
>>>+takeClosest(5,[4,1,88,44,3])
4|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|491223|lambda是编写“匿名”函数(没有名称的函数)的一种特殊方式。您可以为它指定任何想要的名称，因为lambda是一个表达式。|unstyled|offset|length|491224|以上内容的“长”写法是：|491225|def+takeClosest(num,collection):
+++return+min(collection,key=lambda+x:abs(x-num))|491226|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/tutorial/controlflow.html#lambda-forms^0|0|0|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|V|8|@]|9|@$G|W|H|X|1|Y]]|A|$]]|$1|I|3|J|5|F|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$B|C]]|$1|M|3|-4|5|F|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

<pre><code>&gt;&gt;&gt; takeClosest = lambda num,collection:min(collection,key=lambda x:abs(x-num))
&gt;&gt;&gt; takeClosest(5,[4,1,88,44,3])
4
</code></pre>

A <a href="http://docs.python.org/tutorial/controlflow.html#lambda-forms" rel="noreferrer">lambda</a> is a special way of writing an "anonymous" function (a function that doesn't have a name). You can assign it any name you want because a lambda is an expression.

The "long" way of writing the the above would be:

<pre><code>def takeClosest(num,collection):
 return min(collection,key=lambda x:abs(x-num))
</code></pre>

blocks|key|491175|text|遍历列表并将当前最接近的数字与abs(currentNumber+-+myNumber)进行比较|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|491176|def+takeClosest(myList,+myNumber):
++++closest+=+myList[0]
++++for+i+in+range(1,+len(myList)):
++++++++if+abs(i+-+myNumber)+<+closest:
++++++++++++closest+=+i
++++return+closest|code-block|syntax|javascript|491177|entityMap^0|F|T|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Iterate over the list and compare the current closest number with <code>abs(currentNumber - myNumber)</code>:

<pre><code>def takeClosest(myList, myNumber):
 closest = myList[0]
 for i in range(1, len(myList)):
 if abs(i - myNumber) &lt; closest:
 closest = i
 return closest
</code></pre>

blocks|key|274400|text|def+closest(list,+Number):
++++aux+=+[]
++++for+valor+in+list:
++++++++aux.append(abs(Number-valor))

++++return+aux.index(min(aux))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|274401|此代码将为您提供列表中最接近的数字的索引。|unstyled|274402|总体来说，KennyTM提供的解决方案是最好的，但在您不能使用它的情况下(如brython)，此函数可以完成工作|274403|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>def closest(list, Number):
 aux = []
 for valor in list:
 aux.append(abs(Number-valor))

 return aux.index(min(aux))
</code></pre>

This code will give you the index of the closest number of Number in the list.

The solution given by KennyTM is the best overall, but in the cases you cannot use it (like brython), this function will do the work

blocks|key|491511|text|def+find_nearest(array,+value):
++++array+=+np.asarray(array)
++++idx+=+(np.abs(array+-+value)).argmin()
++++return+array[idx]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|491512|使用以下命令运行它|unstyled|491513|price_near_to=find_nearest(df['Close'],+df['Close'][-2])|491514|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>def find_nearest(array, value):
 array = np.asarray(array)
 idx = (np.abs(array - value)).argmin()
 return array[idx]
</code></pre>
<h1>run it by using</h1>
<pre><code>price_near_to=find_nearest(df['Close'], df['Close'][-2])
</code></pre>

blocks|key|274492|text|如果我可以添加到@Lauritz's+answer|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|274493|为了不出现运行错误，不要忘记在bisect_left行之前添加一个条件：|style|CODE|274494|if+(myNumber+>+myList[-1]+or+myNumber+<+myList[0]):
++++return+False|code-block|syntax|javascript|274495|因此，完整的代码如下所示：|274496|from+bisect+import+bisect_left

def+takeClosest(myList,+myNumber):
++++"""
++++Assumes+myList+is+sorted.+Returns+closest+value+to+myNumber.
++++If+two+numbers+are+equally+close,+return+the+smallest+number.
++++If+number+is+outside+of+min+or+max+return+False
++++"""
++++if+(myNumber+>+myList[-1]+or+myNumber+<+myList[0]):
++++++++return+False
++++pos+=+bisect_left(myList,+myNumber)
++++if+pos+==+0:
++++++++++++return+myList[0]
++++if+pos+==+len(myList):
++++++++++++return+myList[-1]
++++before+=+myList[pos+-+1]
++++after+=+myList[pos]
++++if+after+-+myNumber+<+myNumber+-+before:
+++++++return+after
++++else:
+++++++return+before|274497|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/12141511/9857220^0|8|H|0|0|F|B|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|6|7|12|8|@$A|13|B|14|F|G]]|9|@]|C|$]]|$1|H|3|I|5|J|7|15|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|16|8|@]|9|@]|C|$]]|$1|O|3|P|5|J|7|17|8|@]|9|@]|C|$K|L]]|$1|Q|3|-4|5|6|7|18|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]]]

If I may add to <a href="https://stackoverflow.com/a/12141511/9857220">@Lauritz's answer</a>

In order not to have a run error
don't forget to add a condition before the <code>bisect_left</code> line:

<pre><code>if (myNumber &gt; myList[-1] or myNumber &lt; myList[0]):
 return False
</code></pre>

so the full code will look like:

<pre><code>from bisect import bisect_left

def takeClosest(myList, myNumber):
 """
 Assumes myList is sorted. Returns closest value to myNumber.
 If two numbers are equally close, return the smallest number.
 If number is outside of min or max return False
 """
 if (myNumber &gt; myList[-1] or myNumber &lt; myList[0]):
 return False
 pos = bisect_left(myList, myNumber)
 if pos == 0:
 return myList[0]
 if pos == len(myList):
 return myList[-1]
 before = myList[pos - 1]
 after = myList[pos]
 if after - myNumber &lt; myNumber - before:
 return after
 else:
 return before
</code></pre>

blocks|key|274525|text|def+takeClosest(myList,+myNumber):
++++newlst+=+[]
++++for+i+in+myList:
++++++++newlst.append(i+-+myNumber)
++++lstt+=+[abs(ele)+for+ele+in+newlst]
++++print(myList[lstt.index(min(lstt))])

myList+=+[4,+1,+88,+44,+3]
myNumber+=+5
takeClosest(myList,myNumber)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|274526|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>def takeClosest(myList, myNumber):
 newlst = []
 for i in myList:
 newlst.append(i - myNumber)
 lstt = [abs(ele) for ele in newlst]
 print(myList[lstt.index(min(lstt))])

myList = [4, 1, 88, 44, 3]
myNumber = 5
takeClosest(myList,myNumber)
</code></pre>

Given a list of integers, I want to find which number is the closest to a number I give in input:

<pre><code>&gt;&gt;&gt; myList = [4, 1, 88, 44, 3]
&gt;&gt;&gt; myNumber = 5
&gt;&gt;&gt; takeClosest(myList, myNumber)
...
4
</code></pre>

Is there any quick way to do this?

From list of integers, get number closest to a given value

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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给定一个整数列表，我想找出哪个数字最接近我在输入中给出的数字：>>> myList = [4, 1, 88, 44, 3]>>> myNumber = 5>>> takeClosest(myList, myNumber)...4有什么快速的方法可以做到这一点吗？

问从整数列表中，获取最接近给定值的数字
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