blocks|key|634673|text|(A==B).all()|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|634674|测试数组(A==B)的所有值是否为真。|unstyled|634675|注意:也许您还想测试A和B形状，比如A.shape+==+B.shape|offset|length|style|CODE|634676|特殊情况和替代方案(来自dbaupp的回答和yoavram的评论)|BOLD|634677|需要注意的是：|634678|634679|此解决方案在特定情况下可能具有奇怪的行为:如果A或B中的一个为空，而另一个包含单个元素，则它将返回True。由于某种原因，比较操作符会返回一个空数组，对于该数组，all运算符返回的风险是，如果A和B形状不同且不可广播，则此方法将引发错误。|unordered-list-item|634680|634681|总而言之，如果你对A和B+shape有疑问，或者只是想安全起见:使用一个专门的函数：|634682|np.array_equal(A,B)++#+test+if+same+shape,+same+elements+values
np.array_equiv(A,B)++#+test+if+broadcastable+shape,+same+elements+values
np.allclose(A,B,...)+#+test+if+same+shape,+elements+have+close+enough+values|634683|entityMap^0|0|0|I|I|0|0|9|0|0|0|N|1|P|1|1D|4|29|3|2O|1|2Q|1|0|0|9|1|B|1|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|13|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|14|8|@$I|15|J|16|K|L]]|9|@]|A|$]]|$1|M|3|N|5|F|7|17|8|@$I|18|J|19|K|O]]|9|@]|A|$]]|$1|P|3|Q|5|F|7|1A|8|@]|9|@]|A|$]]|$1|R|3|-4|5|F|7|1B|8|@]|9|@]|A|$]]|$1|S|3|T|5|U|7|1C|8|@$I|1D|J|1E|K|L]|$I|1F|J|1G|K|L]|$I|1H|J|1I|K|L]|$I|1J|J|1K|K|L]|$I|1L|J|1M|K|L]|$I|1N|J|1O|K|L]]|9|@]|A|$]]|$1|V|3|-4|5|F|7|1P|8|@]|9|@]|A|$]]|$1|W|3|X|5|F|7|1Q|8|@$I|1R|J|1S|K|L]|$I|1T|J|1U|K|L]]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1V|8|@]|9|@]|A|$B|C]]|$1|10|3|-4|5|F|7|1W|8|@]|9|@]|A|$]]]|11|$]]

<pre><code>(A==B).all()
</code></pre>

test if all values of array (A==B) are True.

Note: maybe you also want to test A and B shape, such as <code>A.shape == B.shape</code>

Special cases and alternatives (from dbaupp's answer and yoavram's comment)

It should be noted that:

<ul>
<li>this solution can have a strange behavior in a particular case: if either <code>A</code> or <code>B</code> is empty and the other one contains a single element, then it return <code>True</code>. For some reason, the comparison <code>A==B</code> returns an empty array, for which the <code>all</code> operator returns <code>True</code>.</li>
<li>Another risk is if <code>A</code> and <code>B</code> don't have the same shape and aren't broadcastable, then this approach will raise an error.</li>
</ul>

In conclusion, if you have a doubt about <code>A</code> and <code>B</code> shape or simply want to be safe: use one of the specialized functions:

<pre><code>np.array_equal(A,B) # test if same shape, same elements values
np.array_equiv(A,B) # test if broadcastable shape, same elements values
np.allclose(A,B,...) # test if same shape, elements have close enough values
</code></pre>

blocks|key|1369035|text|(A==B).all()解决方案非常简洁，但是有一些用于此任务的内置函数。即array_equal、allclose和array_equiv。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1369036|(不过，使用timeit进行的一些快速测试似乎表明(A==B).all()方法是最快的，这有点奇怪，因为它必须分配一个全新的数组。)|1369037|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.array_equal.html|1|http://docs.scipy.org/doc/numpy/reference/generated/numpy.allclose.html#numpy.allclose|2|http://docs.scipy.org/doc/numpy/reference/generated/numpy.array_equiv.html#numpy.array_equiv^0|0|C|12|B|1E|8|1N|B|12|B|0|1E|8|1|1N|B|2|0|6|6|P|C|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]|$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@$9|12|A|13|1|14]|$9|15|A|16|1|17]|$9|18|A|19|1|1A]]|E|$]]|$1|F|3|G|5|6|7|1B|8|@$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]|P|$5|K|L|M|E|$N|Q]]|R|$5|K|L|M|E|$N|S]]]]

The <code>(A==B).all()</code> solution is very neat, but there are some built-in functions for this task. Namely <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.array_equal.html" rel="noreferrer"><code>array_equal</code></a>, <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.allclose.html#numpy.allclose" rel="noreferrer"><code>allclose</code></a> and <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.array_equiv.html#numpy.array_equiv" rel="noreferrer"><code>array_equiv</code></a>.

(Although, some quick testing with <code>timeit</code> seems to indicate that the <code>(A==B).all()</code> method is the fastest, which is a little peculiar, given it has to allocate a whole new array.)

blocks|key|1370470|text|让我们使用下面这段代码来测量性能。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1370471|import+numpy+as+np
import+time

exec_time0+=+[]
exec_time1+=+[]
exec_time2+=+[]

sizeOfArray+=+5000
numOfIterations+=+200

for+i+in+xrange(numOfIterations):

++++A+=+np.random.randint(0,255,(sizeOfArray,sizeOfArray))
++++B+=+np.random.randint(0,255,(sizeOfArray,sizeOfArray))

++++a+=+time.clock()+
++++res+=+(A==B).all()
++++b+=+time.clock()
++++exec_time0.append(+b+-+a+)

++++a+=+time.clock()+
++++res+=+np.array_equal(A,B)
++++b+=+time.clock()
++++exec_time1.append(+b+-+a+)

++++a+=+time.clock()+
++++res+=+np.array_equiv(A,B)
++++b+=+time.clock()
++++exec_time2.append(+b+-+a+)

print+'Method:+(A==B).all(),+++++++',+np.mean(exec_time0)
print+'Method:+np.array_equal(A,B),',+np.mean(exec_time1)
print+'Method:+np.array_equiv(A,B),',+np.mean(exec_time2)|code-block|syntax|javascript|1370472|输出|offset|length|style|BOLD|1370473|Method:+(A==B).all(),++++++++0.03031857
Method:+np.array_equal(A,B),+0.030025185
Method:+np.array_equiv(A,B),+0.030141515|1370474|根据上面的结果，numpy.array_equal方法似乎比==运算符和()方法的组合更快，并且通过比较numpy方法和()方法，最快的方法似乎是numpy方法。|1370475|entityMap^0|0|0|0|2|0|0|8|H|U|2|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|R|$]]

Let's measure the performance by using the following piece of code.

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

exec_time0 = []
exec_time1 = []
exec_time2 = []

sizeOfArray = 5000
numOfIterations = 200

for i in xrange(numOfIterations):

 A = np.random.randint(0,255,(sizeOfArray,sizeOfArray))
 B = np.random.randint(0,255,(sizeOfArray,sizeOfArray))

 a = time.clock() 
 res = (A==B).all()
 b = time.clock()
 exec_time0.append( b - a )

 a = time.clock() 
 res = np.array_equal(A,B)
 b = time.clock()
 exec_time1.append( b - a )

 a = time.clock() 
 res = np.array_equiv(A,B)
 b = time.clock()
 exec_time2.append( b - a )

print 'Method: (A==B).all(), ', np.mean(exec_time0)
print 'Method: np.array_equal(A,B),', np.mean(exec_time1)
print 'Method: np.array_equiv(A,B),', np.mean(exec_time2)
</code></pre>

Output

<pre><code>Method: (A==B).all(), 0.03031857
Method: np.array_equal(A,B), 0.030025185
Method: np.array_equiv(A,B), 0.030141515
</code></pre>

According to the results above, the numpy methods seem to be faster than the combination of the == operator and the all() method and by comparing the numpy methods the fastest one seems to be the numpy.array_equal method.

blocks|key|2384413|text|如果您想检查两个数组是否具有相同的shape和elements，您应该使用np.array_equal，因为它是文档中推荐的方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2384414|2384415|性能方面，不要期望任何一个相等检查都会击败另一个，因为没有太多的空间来优化comparing+two+elements。只是为了，我还是做了一些测试。|blockquote|2384416|2384417|2384418|import+numpy+as+np
import+timeit

A+=+np.zeros((300,+300,+3))
B+=+np.zeros((300,+300,+3))
C+=+np.ones((300,+300,+3))

timeit.timeit(stmt='(A==B).all()',+setup='from+__main__+import+A,+B',+number=10**5)
timeit.timeit(stmt='np.array_equal(A,+B)',+setup='from+__main__+import+A,+B,+np',+number=10**5)
timeit.timeit(stmt='np.array_equiv(A,+B)',+setup='from+__main__+import+A,+B,+np',+number=10**5)
>+51.5094
>+52.555
>+52.761|code-block|syntax|javascript|2384419|所以几乎是相等的，不需要谈论速度。|2384420|(A==B).all()的行为与下面的代码片段非常相似：|2384421|x+=+[1,2,3]
y+=+[1,2,3]
print+all([x[i]==y[i]+for+i+in+range(len(x))])
>+True|2384422|entityMap^0|H|5|N|8|11|E|0|0|11|M|0|0|0|0|0|0|C|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|15|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|16|8|@$9|17|A|18|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|19|8|@]|D|@]|E|$]]|$1|K|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|1B|8|@]|D|@]|E|$O|P]]|$1|Q|3|R|5|6|7|1C|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|1D|8|@$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|U|3|V|5|N|7|1G|8|@]|D|@]|E|$O|P]]|$1|W|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|X|$]]

If you want to check if two arrays have the same <code>shape</code> AND <code>elements</code> you should use <code>np.array_equal</code> as it is the method recommended in the documentation.

<blockquote>
 Performance-wise don't expect that any equality check will beat another, as there is not much room to optimize <code>comparing two elements</code>. Just for the sake, i still did some tests. 
</blockquote>

<pre><code>import numpy as np
import timeit

A = np.zeros((300, 300, 3))
B = np.zeros((300, 300, 3))
C = np.ones((300, 300, 3))

timeit.timeit(stmt='(A==B).all()', setup='from __main__ import A, B', number=10**5)
timeit.timeit(stmt='np.array_equal(A, B)', setup='from __main__ import A, B, np', number=10**5)
timeit.timeit(stmt='np.array_equiv(A, B)', setup='from __main__ import A, B, np', number=10**5)
&gt; 51.5094
&gt; 52.555
&gt; 52.761
</code></pre>

So pretty much equal, no need to talk about the speed. 

The <code>(A==B).all()</code> behaves pretty much as the following code snippet:

<pre><code>x = [1,2,3]
y = [1,2,3]
print all([x[i]==y[i] for i in range(len(x))])
&gt; True
</code></pre>

blocks|key|1370501|text|通常两个数组会有一些小的数值误差，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1370502|您可以使用numpy.allclose(A,B)，而不是(A==B).all()。这将返回布尔值True/False|offset|length|style|CODE|1370503|entityMap^0|0|5|J|S|C|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|I|$]]

Usually two arrays will have some small numeric errors,

You can use <code>numpy.allclose(A,B)</code>, instead of <code>(A==B).all()</code>. This returns a bool True/False

blocks|key|2273250|text|现在使用np.array_equal。来自文档：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2273251|np.array_equal([1,+2],+[1,+2])
True
np.array_equal(np.array([1,+2]),+np.array([1,+2]))
True
np.array_equal([1,+2],+[1,+2,+3])
False
np.array_equal([1,+2],+[1,+4])
False|code-block|syntax|javascript|2273252|entityMap^0|4|E|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|$]]

Now use <code>np.array_equal</code>. From documentation:

<pre><code>np.array_equal([1, 2], [1, 2])
True
np.array_equal(np.array([1, 2]), np.array([1, 2]))
True
np.array_equal([1, 2], [1, 2, 3])
False
np.array_equal([1, 2], [1, 4])
False
</code></pre>

blocks|key|2273269|text|在其他答案之上，您现在可以使用断言：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2273270|numpy.testing.assert_array_equal(x,+y)|code-block|syntax|javascript|2273271|您还可以使用类似的函数，如numpy.testing.assert_almost_equal()|offset|length|style|CODE|2273272|https://numpy.org/doc/stable/reference/generated/numpy.testing.assert_array_equal.html|2273273|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|D|Z|0|0|2E|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|X|8|@$I|Y|J|Z|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|10|8|@]|9|@$I|11|J|12|1|13]]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|N]]]]

On top of the other answers, you can now use an assertion:
<pre><code>numpy.testing.assert_array_equal(x, y)
</code></pre>
You also have similar function such as <code>numpy.testing.assert_almost_equal()</code>
<a href="https://numpy.org/doc/stable/reference/generated/numpy.testing.assert_array_equal.html" rel="nofollow noreferrer">https://numpy.org/doc/stable/reference/generated/numpy.testing.assert_array_equal.html</a>

What is the simplest way to compare two NumPy arrays for equality (where equality is defined as: A = B iff for all indices i: <code>A[i] == B[i]</code>)?

Simply using <code>==</code> gives me a boolean array:

<pre><code> &gt;&gt;&gt; numpy.array([1,1,1]) == numpy.array([1,1,1])

array([ True, True, True], dtype=bool)
</code></pre>

Do I have to <code>and</code> the elements of this array to determine if the arrays are equal, or is there a simpler way to compare?

Comparing two NumPy arrays for equality, element-wise

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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比较两个NumPy数组相等性的最简单方法是什么(相等性定义为:A=B当且仅当对于所有索引i：A[i] == B[i])？只需使用==就可以得到一个布尔数组： >>> numpy.array([1,1,1]) == numpy.array([1,1,1])array([ True,  True,  True], dtyp...

问逐个元素比较两个NumPy数组是否相等
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问逐个元素比较两个NumPy数组是否相等EN