blocks|key|1191585|text|您可以使用"i“(如idiv，imul..)版本，而且看起来还不错：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1191586|image+/=+(image.max()/255.0)|code-block|syntax|javascript|1191587|对于另一种情况，您可以编写一个函数来按列规范化n维数组：|1191588|def+normalize_columns(arr):
++++rows,+cols+=+arr.shape
++++for+col+in+xrange(cols):
++++++++arr[:,col]+/=+abs(arr[:,col]).max()|1191589|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|$]]

You can use the "i" (as in idiv, imul..) version, and it doesn't look half bad:

<pre><code>image /= (image.max()/255.0)
</code></pre>

For the other case you can write a function to normalize an n-dimensional array by colums:

<pre><code>def normalize_columns(arr):
 rows, cols = arr.shape
 for col in xrange(cols):
 arr[:,col] /= abs(arr[:,col]).max()
</code></pre>

blocks|key|1191618|text|也可以使用sklearn重新缩放。优点是，除了以数据为中心进行均值调整外，您还可以调整归一化标准差，并且可以在任一轴上、按要素或按记录执行此操作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1191619|from+sklearn.preprocessing+import+scale
X+=+scale(+X,+axis=0,+with_mean=True,+with_std=True,+copy=True+)|code-block|syntax|javascript|1191620|关键字参数axis、with_mean和with_std是自解释的，并以其默认状态显示。如果参数copy设置为False，则它将就地执行操作。文档here。|1191621|entityMap|0|LINK|mutability|MUTABLE|url|http://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.scale.html^0|5|7|0|0|5|4|A|9|K|8|1C|4|1J|5|21|4|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|M|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

You can also rescale using <code>sklearn</code>. The advantages are that you can adjust normalize the standard deviation, in addition to mean-centering the data, and that you can do this on either axis, by features, or by records.

<pre><code>from sklearn.preprocessing import scale
X = scale( X, axis=0, with_mean=True, with_std=True, copy=True )
</code></pre>

The keyword arguments <code>axis</code>, <code>with_mean</code>, <code>with_std</code> are self explanatory, and are shown in their default state. The argument <code>copy</code> performs the operation in-place if it is set to <code>False</code>. Documentation <a href="http://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.scale.html">here</a>.

blocks|key|1407035|text|如果数组同时包含正数据和负数据，我会使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1407036|import+numpy+as+np

a+=+np.random.rand(3,2)

#+Normalised+[0,1]
b+=+(a+-+np.min(a))/np.ptp(a)

#+Normalised+[0,255]+as+integer:+don't+forget+the+parenthesis+before+astype(int)
c+=+(255*(a+-+np.min(a))/np.ptp(a)).astype(int)++++++++

#+Normalised+[-1,1]
d+=+2.*(a+-+np.min(a))/np.ptp(a)-1|code-block|syntax|javascript|1407037|如果数组包含nan，一种解决方案可能是将它们删除，如下所示：|offset|length|style|CODE|1407038|def+nan_ptp(a):
++++return+np.ptp(a[np.isfinite(a)])

b+=+(a+-+np.nanmin(a))/nan_ptp(a)|1407039|但是，根据上下文的不同，您可能希望以不同的方式对待nan。例如，对该值进行插值，例如用0替换，或者引发错误。|1407040|最后，值得一提的是，即使这不是OP的问题，standardization|1407041|e+=+(a+-+np.mean(a))+/+np.std(a)|1407042|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Standard_score^0|0|0|6|3|0|0|P|3|0|L|F|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@$I|15|J|16|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|18|8|@$I|19|J|1A|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@$I|1C|J|1D|1|1E]]|A|$]]|$1|S|3|T|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

If the array contains both positive and negative data, I'd go with:
<pre><code>import numpy as np

a = np.random.rand(3,2)

# Normalised [0,1]
b = (a - np.min(a))/np.ptp(a)

# Normalised [0,255] as integer: don't forget the parenthesis before astype(int)
c = (255*(a - np.min(a))/np.ptp(a)).astype(int) 

# Normalised [-1,1]
d = 2.*(a - np.min(a))/np.ptp(a)-1
</code></pre>
If the array contains <code>nan</code>, one solution could be to just remove them as:
<pre><code>def nan_ptp(a):
 return np.ptp(a[np.isfinite(a)])

b = (a - np.nanmin(a))/nan_ptp(a)
</code></pre>
However, depending on the context you might want to treat <code>nan</code> differently. E.g. interpolate the value, replacing in with e.g. 0, or raise an error.
Finally, worth mentioning even if it's not OP's question, <a href="https://en.wikipedia.org/wiki/Standard_score" rel="noreferrer">standardization</a>:
<pre><code>e = (a - np.mean(a)) / np.std(a)
</code></pre>

blocks|key|1407064|text|一个简单的解决方案是使用sklearn.preprocessing库提供的定标器。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1407065|scaler+=+sk.MinMaxScaler(feature_range=(0,+250))
scaler+=+scaler.fit(X)
X_scaled+=+scaler.transform(X)
#+Checking+reconstruction
X_rec+=+scaler.inverse_transform(X_scaled)|code-block|syntax|javascript|1407066|错误X_rec-X将为零。您可以根据需要调整feature_range，甚至可以使用标准缩放器sk.StandardScaler()|1407067|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

A simple solution is using the scalers offered by the sklearn.preprocessing library. 

<pre><code>scaler = sk.MinMaxScaler(feature_range=(0, 250))
scaler = scaler.fit(X)
X_scaled = scaler.transform(X)
# Checking reconstruction
X_rec = scaler.inverse_transform(X_scaled)
</code></pre>

The error X_rec-X will be zero. You can adjust the feature_range for your needs, or even use a standart scaler sk.StandardScaler()

blocks|key|1407103|text|我尝试跟踪this，但得到了错误|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1407104|TypeError:+ufunc+'true_divide'+output+(typecode+'d')+could+not+be+coerced+to+provided+output+parameter+(typecode+'l')+according+to+the+casting+rule+''same_kind''|code-block|syntax|javascript|1407105|我试图规范化的numpy数组是一个integer数组。他们似乎不赞成在versions+>+1.10中进行类型转换，你必须使用numpy.true_divide()来解决这个问题。|style|CODE|1407106|arr+=+np.array(img)
arr+=+np.true_divide(arr,[255.0],out=None)|1407107|img是一个PIL.Image对象。|1407108|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/1735122/8507120^0|5|4|0|0|0|7|5|H|7|1A|4|1R|J|0|0|0|3|6|9|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|13|8|@$A|14|B|15|K|L]|$A|16|B|17|K|L]|$A|18|B|19|K|L]|$A|1A|B|1B|K|L]]|9|@]|C|$]]|$1|M|3|N|5|F|7|1C|8|@]|9|@]|C|$G|H]]|$1|O|3|P|5|6|7|1D|8|@$A|1E|B|1F|K|L]|$A|1G|B|1H|K|L]]|9|@]|C|$]]|$1|Q|3|-4|5|6|7|1I|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]]]

I tried following <a href="https://stackoverflow.com/a/1735122/8507120">this</a>, and got the error 

<pre><code>TypeError: ufunc 'true_divide' output (typecode 'd') could not be coerced to provided output parameter (typecode 'l') according to the casting rule ''same_kind''
</code></pre>

The <code>numpy</code> array I was trying to normalize was an <code>integer</code> array. It seems they deprecated type casting in versions > <code>1.10</code>, and you have to use <code>numpy.true_divide()</code> to resolve that.

<pre><code>arr = np.array(img)
arr = np.true_divide(arr,[255.0],out=None)
</code></pre>

<code>img</code> was an <code>PIL.Image</code> object.

blocks|key|1407134|text|This+answer+to+a+similar+question为我解决了这个问题|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1407135|np.interp(a,+(a.min(),+a.max()),+(-1,+%2B1))|code-block|syntax|javascript|1407136|entityMap|0|LINK|mutability|MUTABLE|url|https://codereview.stackexchange.com/a/185794|1|https://codereview.stackexchange.com/questions/185785/scale-numpy-array-to-certain-range^0|0|B|0|H|G|1|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]|$A|W|B|X|1|Y]]|C|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|10|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]|Q|$5|L|M|N|C|$O|R]]]]

<a href="https://codereview.stackexchange.com/a/185794">This answer</a> to a <a href="https://codereview.stackexchange.com/questions/185785/scale-numpy-array-to-certain-range">similar question</a> solved the problem for me with
<pre><code>np.interp(a, (a.min(), a.max()), (-1, +1))
</code></pre>

After doing some processing on an audio or image array, it needs to be normalized within a range before it can be written back to a file. This can be done like so:

<pre><code># Normalize audio channels to between -1.0 and +1.0
audio[:,0] = audio[:,0]/abs(audio[:,0]).max()
audio[:,1] = audio[:,1]/abs(audio[:,1]).max()

# Normalize image to between 0 and 255
image = image/(image.max()/255.0)
</code></pre>

Is there a less verbose, convenience function way to do this? <code>matplotlib.colors.Normalize()</code> doesn't seem to be related.

How to normalize a NumPy array to within a certain range?

在对音频或图像数组进行一些处理后，需要在一定范围内进行归一化，然后才能将其写回文件。这可以像这样做：# Normalize audio channels to between -1.0 and +1.0audio[:,0] = audio[:,0]/abs(audio[:,0]).max()audio[:,1] = a...

问如何将NumPy数组归一化到一定范围内？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何将NumPy数组归一化到一定范围内？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何将NumPy数组归一化到一定范围内？
EN