blocks|key|1583047|text|使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1583048|norm+=+[float(i)/sum(raw)+for+i+in+raw]|code-block|syntax|javascript|1583049|根据总和进行标准化，以确保总和始终为1.0+(或尽可能接近)。|1583050|使用|1583051|norm+=+[float(i)/max(raw)+for+i+in+raw]|1583052|根据最大值进行归一化|1583053|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Use : 

<pre><code>norm = [float(i)/sum(raw) for i in raw]
</code></pre>

to normalize against the sum to ensure that the sum is always 1.0 (or as close to as possible).

use 

<pre><code>norm = [float(i)/max(raw) for i in raw]
</code></pre>

to normalize against the maximum

blocks|key|1798485|text|如果你的列表有负数，这就是你规范化它的方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1798486|a+=+range(-30,31,5)
norm+=+[(float(i)-min(a))/(max(a)-min(a))+for+i+in+a]|code-block|syntax|javascript|1798487|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|$]]

if your list has negative numbers, this is how you would normalize it

<pre><code>a = range(-30,31,5)
norm = [(float(i)-min(a))/(max(a)-min(a)) for i in a]
</code></pre>

blocks|key|1583057|text|尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1583058|normed+=+[i/sum(raw)+for+i+in+raw]

normed
[0.25,+0.5,+0.25]|code-block|syntax|javascript|1583059|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:

<pre><code>normed = [i/sum(raw) for i in raw]

normed
[0.25, 0.5, 0.25]
</code></pre>

blocks|key|1583429|text|你要规范化的列表有多长？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1583430|def+psum(it):
++++"This+function+makes+explicit+how+many+calls+to+sum()+are+done."
++++print+"Another+call!"
++++return+sum(it)

raw+=+[0.07,0.14,0.07]
print+"How+many+calls+to+sum()?"
print+[+r/psum(raw)+for+r+in+raw]

print+"\nAnd+now?"
s+=+psum(raw)
print+[+r/s+for+r+in+raw]

#+if+one+doesn't+want+auxiliary+variables,+it+can+be+done+inside
#+a+list+comprehension,+but+in+my+opinion+it's+quite+Baroque++++
print+"\nAnd+now?"
print+[+r/s++for+s+in+[psum(raw)]+for+r+in+raw]|code-block|syntax|javascript|1583431|输出|1583432|#+How+many+calls+to+sum()?
#+Another+call!
#+Another+call!
#+Another+call!
#+[0.25,+0.5,+0.25]
#+
#+And+now?
#+Another+call!
#+[0.25,+0.5,+0.25]
#+
#+And+now?
#+Another+call!
#+[0.25,+0.5,+0.25]|1583433|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|$]]

How long is the list you're going to normalize?

<pre><code>def psum(it):
 "This function makes explicit how many calls to sum() are done."
 print "Another call!"
 return sum(it)

raw = [0.07,0.14,0.07]
print "How many calls to sum()?"
print [ r/psum(raw) for r in raw]

print "\nAnd now?"
s = psum(raw)
print [ r/s for r in raw]

# if one doesn't want auxiliary variables, it can be done inside
# a list comprehension, but in my opinion it's quite Baroque 
print "\nAnd now?"
print [ r/s for s in [psum(raw)] for r in raw]
</code></pre>

Output

<pre><code># How many calls to sum()?
# Another call!
# Another call!
# Another call!
# [0.25, 0.5, 0.25]
# 
# And now?
# Another call!
# [0.25, 0.5, 0.25]
# 
# And now?
# Another call!
# [0.25, 0.5, 0.25]
</code></pre>

blocks|key|1583337|text|据我所知，标准库中没有任何函数可以做到这一点，但绝对有一些模块有这样的函数。但是，它非常简单，您可以只编写自己的函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1583338|def+normalize(lst):
++++s+=+sum(lst)
++++return+map(lambda+x:+float(x)/s,+lst)|code-block|syntax|javascript|1583339|示例输出：|1583340|>>>+normed+=+normalize(raw)
>>>+normed
[0.25,+0.5,+0.25]|1583341|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|$]]

There isn't any function in the standard library (to my knowledge) that will do it, but there are absolutely modules out there which have such functions. However, its easy enough that you can just write your own function:

<pre><code>def normalize(lst):
 s = sum(lst)
 return map(lambda x: float(x)/s, lst)
</code></pre>

Sample output:

<pre><code>&gt;&gt;&gt; normed = normalize(raw)
&gt;&gt;&gt; normed
[0.25, 0.5, 0.25]
</code></pre>

blocks|key|1798729|text|如果你考虑使用numpy，你可以得到一个更快的解决方案。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1798730|import+random,+time
import+numpy+as+np

a+=+random.sample(range(1,+20000),+10000)
since+=+time.time();+b+=+[i/sum(a)+for+i+in+a];+print(time.time()-since)
#+0.7956490516662598

since+=+time.time();+c=np.array(a);d=c/sum(a);+print(time.time()-since)
#+0.001413106918334961|code-block|syntax|javascript|1798731|entityMap^0|7|5|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|$]]

If you consider using <code>numpy</code>, you can get a faster solution.

<pre><code>import random, time
import numpy as np

a = random.sample(range(1, 20000), 10000)
since = time.time(); b = [i/sum(a) for i in a]; print(time.time()-since)
# 0.7956490516662598

since = time.time(); c=np.array(a);d=c/sum(a); print(time.time()-since)
# 0.001413106918334961
</code></pre>

blocks|key|1584127|text|对于那些想要使用scikit-learn的用户，您可以使用|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1584128|from+sklearn.preprocessing+import+normalize

x+=+[1,2,3,4]
normalize([x])+#+array([[0.18257419,+0.36514837,+0.54772256,+0.73029674]])
normalize([x],+norm="l1")+#+array([[0.1,+0.2,+0.3,+0.4]])
normalize([x],+norm="max")+#+array([[0.25,+0.5+,+0.75,+1.]])|code-block|syntax|javascript|1584129|entityMap|0|LINK|mutability|MUTABLE|url|https://scikit-learn.org/stable/modules/preprocessing.html#normalization^0|8|C|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

For ones who wanna use <a href="https://scikit-learn.org/stable/modules/preprocessing.html#normalization" rel="nofollow noreferrer">scikit-learn</a>, you can use
<pre class="lang-py prettyprint-override"><code>from sklearn.preprocessing import normalize

x = [1,2,3,4]
normalize([x]) # array([[0.18257419, 0.36514837, 0.54772256, 0.73029674]])
normalize([x], norm=&quot;l1&quot;) # array([[0.1, 0.2, 0.3, 0.4]])
normalize([x], norm=&quot;max&quot;) # array([[0.25, 0.5 , 0.75, 1.]])
</code></pre>

blocks|key|1584008|text|如果要处理数据，很多时候pandas是简单的键|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1584009|这段特定的代码将把raw放到一列中，然后逐行对其进行规范化。(但是我们可以把它放到一行中，也可以逐行逐列地完成！只需更改axis值，其中0表示行，1表示列。)|1584010|import+pandas+as+pd


raw+=+[0.07,+0.14,+0.07]++

raw_df+=+pd.DataFrame(raw)
normed_df+=+raw_df.div(raw_df.sum(axis=0),+axis=1)
normed_df|code-block|syntax|javascript|1584011|其中normed_df将显示为：|1584012|++++0
0+++0.25
1+++0.50
2+++0.25|1584013|然后还可以继续处理数据！|1584014|entityMap^0|C|6|0|9|3|1O|4|0|0|2|9|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|16|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|17|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|T|$]]

If working with data, many times <code>pandas</code> is the simple key

This particular code will put the <code>raw</code> into one column, then normalize by column per row. (But we can put it into a row and do it by row per column, too! Just have to change the <code>axis</code> values where 0 is for row and 1 is for column.)

<pre><code>import pandas as pd


raw = [0.07, 0.14, 0.07] 

raw_df = pd.DataFrame(raw)
normed_df = raw_df.div(raw_df.sum(axis=0), axis=1)
normed_df
</code></pre>

where <code>normed_df</code> will display like:

<pre><code> 0
0 0.25
1 0.50
2 0.25
</code></pre>

and then can keep playing with the data, too!

blocks|key|1799318|text|这是一个效率不是很低的线性函数，类似于顶部的答案(只执行一次求和)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1799319|norm+=+(lambda+the_sum:[float(i)/the_sum+for+i+in+raw])(sum(raw))|code-block|syntax|javascript|1799320|对于包含负数的列表，也可以使用类似的方法|1799321|norm+=+(lambda+the_max,+the_min:+[(float(i)-the_min)/(the_max-the_min)+for+i+in+raw])(max(raw),min(raw))|1799322|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|$]]

Here is a not-terribly-inefficient one liner similar to the top answer (only performs summation once)
<pre class="lang-py prettyprint-override"><code>norm = (lambda the_sum:[float(i)/the_sum for i in raw])(sum(raw))
</code></pre>
A similar method can be done for a list with negative numbers
<pre class="lang-py prettyprint-override"><code>norm = (lambda the_max, the_min: [(float(i)-the_min)/(the_max-the_min) for i in raw])(max(raw),min(raw))
</code></pre>

blocks|key|1799069|text|使用scikit-learn|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1799070|from+sklearn.preprocessing+import+MinMaxScaler
data+=+np.array([1,2,3]).reshape(-1,+1)
scaler+=+MinMaxScaler()
scaler.fit(data)
print(scaler.transform(data))|code-block|syntax|javascript|1799071|entityMap|0|LINK|mutability|MUTABLE|url|https://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.MinMaxScaler.html^0|2|C|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Use <a href="https://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.MinMaxScaler.html" rel="nofollow noreferrer">scikit-learn</a>:
<pre><code>from sklearn.preprocessing import MinMaxScaler
data = np.array([1,2,3]).reshape(-1, 1)
scaler = MinMaxScaler()
scaler.fit(data)
print(scaler.transform(data))
</code></pre>

I need to normalize a list of values to fit in a probability distribution, i.e. between 0.0 and 1.0.

I understand how to normalize, but was curious if Python had a function to automate this.

I'd like to go from:

<pre><code>raw = [0.07, 0.14, 0.07] 
</code></pre>

to 

<pre><code>normed = [0.25, 0.50, 0.25]
</code></pre>

Normalizing a list of numbers in Python

我需要规格化值列表以适应概率分布，即介于0.0和1.0之间。我知道如何规范化，但我很好奇Python是否有自动化的函数。我想从下面说起：raw = [0.07, 0.14, 0.07]  至normed = [0.25, 0.50, 0.25]

问在Python中规范化数字列表
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回答 10

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