blocks|key|562122|text|您可以使用apply并传递一个lambda函数，该函数给出的索引绝对值最小，然后从该索引的列表中获取项。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|562123|df['Nearest']+=+df['Data'].apply(lambda+x:+mylist[(min([(idx,abs(x+-+v))+for+idx,+v+in+enumerate(mylist)],+key=+lambda+x:+x[1]))[0]])


+++Data++Nearest
0++++11+++++++11
1++++22+++++++23
2++++33+++++++23
3++++43+++++++44
4++++52+++++++44
5++++63+++++++66|code-block|syntax|javascript|562124|entityMap^0|5|5|F|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

You can use <code>apply</code> and pass a <code>lambda</code> function which takes gives the index with minimum absolute difference, then just take the item from the list for that index.
<pre class="lang-py prettyprint-override"><code>df['Nearest'] = df['Data'].apply(lambda x: mylist[(min([(idx,abs(x - v)) for idx, v in enumerate(mylist)], key= lambda x: x[1]))[0]])


 Data Nearest
0 11 11
1 22 23
2 33 23
3 43 44
4 52 44
5 63 66
</code></pre>

blocks|key|564692|text|这可能管用，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|564693|def+nearest(a,+x):
++++n+=+[abs(i-x)+for+i+in+a]
++++idx+=+n.index(min(n))
++++return+a[idx]

def+nearest_alternative(a,+v):
++++idx+=+(np.abs(a-v)).argmin()
++++return+array[idx]

df['nearest']+=+df['Data'].apply(lambda+x:+nearest(mylist,+x))|code-block|syntax|javascript|564694|备选方案：|564695|O/P：|564696|+++Data++nearest
0++++11+++++++11
1++++22+++++++23
2++++33+++++++23
3++++43+++++++44
4++++52+++++++44
5++++63+++++++66|564697|解释：|564698|564699|nearest函数将接收mylist和每个数据元素。|unordered-list-item|offset|length|style|CODE|564700|计算x和mylist|564701|返回元素之间的差异，该元素具有最小值|564702|564703|entityMap^0|0|0|0|0|0|0|0|0|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|15|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|16|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|18|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|19|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|1A|8|@$S|1B|T|1C|U|V]]|9|@]|A|$]]|$1|W|3|X|5|R|7|1D|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|R|7|1E|8|@]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|11|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|12|$]]

This could work,
<pre><code>def nearest(a, x):
 n = [abs(i-x) for i in a]
 idx = n.index(min(n))
 return a[idx]

def nearest_alternative(a, v):
 idx = (np.abs(a-v)).argmin()
 return array[idx]

df['nearest'] = df['Data'].apply(lambda x: nearest(mylist, x))
</code></pre>
alternative:
O/P:
<pre><code> Data nearest
0 11 11
1 22 23
2 33 23
3 43 44
4 52 44
5 63 66
</code></pre>
Explanation:
<ul>
<li><code>nearest</code> function will receive mylist and each element of Data.</li>
<li>Compute difference between x and mylist</li>
<li>Return element which has minimum value</li>
</ul>

blocks|key|562146|text|df1=pd.DataFrame(mylist,columns=['Nearest']).sort_values(by='Nearest')#Create+DataFrame+from+list

pd.merge_asof(df,+df1,left_on="Data",+right_on="Nearest",direction="forward")#Merge+asof

++++Data++Nearest
0++++11+++++++11
1++++22+++++++23
2++++33+++++++44
3++++43+++++++44
4++++52+++++++66
5++++63+++++++66|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|562147|或|unstyled|562148|print(pd.merge_asof(df,+df1,left_on="Data",+right_on="Nearest",direction="nearest"))



++++Data++Nearest
0++++11+++++++11
1++++22+++++++23
2++++33+++++++23
3++++43+++++++44
4++++52+++++++44
5++++63+++++++66|562149|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>df1=pd.DataFrame(mylist,columns=['Nearest']).sort_values(by='Nearest')#Create DataFrame from list

pd.merge_asof(df, df1,left_on=&quot;Data&quot;, right_on=&quot;Nearest&quot;,direction=&quot;forward&quot;)#Merge asof

 Data Nearest
0 11 11
1 22 23
2 33 44
3 43 44
4 52 66
5 63 66
</code></pre>
Or
<pre><code>print(pd.merge_asof(df, df1,left_on=&quot;Data&quot;, right_on=&quot;Nearest&quot;,direction=&quot;nearest&quot;))



 Data Nearest
0 11 11
1 22 23
2 33 23
3 43 44
4 52 44
5 63 66
</code></pre>

I have a <code>list</code> of some numbers. I want to find closest value from list for entire column:
<pre><code>mylist=[11,44,23,66,100]
df = pd.DataFrame(data={'Data':[11,22,33,43,52,63]})
df&gt;&gt;
 Data
0 11
1 22
2 33
3 43
4 52
5 63
</code></pre>
My desired output:
<pre><code> Data Nearest
0 11 11
1 22 23
2 33 23
3 43 44
4 52 66
5 63 66
</code></pre>
I have tried by using <code>min</code> function with <code>iteraration</code> but its slow

Find nearest from given list for entire column

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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我有一些数字的list。我希望从list中为整个列找到最接近的值：mylist=[11,44,23,66,100]df = pd.DataFrame(data={'Data':[11,22,33,43,52,63]})df>>    Data0   111   222   333   434   525   63我想要...

问为整个列从给定列表中查找最近的
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问为整个列从给定列表中查找最近的EN