blocks|key|1994148|text|numpy.where()是我的最爱。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1994149|>>>+x+=+numpy.array([1,0,2,0,3,0,4,5,6,7,8])
>>>+numpy.where(x+==+0)[0]
array([1,+3,+5])|code-block|syntax|javascript|1994150|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.where.html^0|0|D|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]]]]

<a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.where.html">numpy.where()</a> is my favorite.

<pre><code>&gt;&gt;&gt; x = numpy.array([1,0,2,0,3,0,4,5,6,7,8])
&gt;&gt;&gt; numpy.where(x == 0)[0]
array([1, 3, 5])
</code></pre>

blocks|key|1994177|text|您可以使用以下命令搜索任何标量条件：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1994178|>>>+a+=+np.asarray([0,1,2,3,4])
>>>+a+==+0+#+or+whatver
array([+True,+False,+False,+False,+False],+dtype=bool)|code-block|syntax|javascript|1994179|它会将数组作为条件的布尔掩码返回。|1994180|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|$]]

You can search for any scalar condition with:

<pre><code>&gt;&gt;&gt; a = np.asarray([0,1,2,3,4])
&gt;&gt;&gt; a == 0 # or whatver
array([ True, False, False, False, False], dtype=bool)
</code></pre>

Which will give back the array as an boolean mask of the condition.

blocks|key|1994718|text|如果你使用的是一维数组，就会有语法上的甜头：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1994719|>>>+x+=+numpy.array([1,0,2,0,3,0,4,5,6,7,8])
>>>+numpy.flatnonzero(x+==+0)
array([1,+3,+5])|code-block|syntax|javascript|1994720|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 you are working with a one-dimensional array there is a syntactic sugar:

<pre class="lang-python prettyprint-override"><code>&gt;&gt;&gt; x = numpy.array([1,0,2,0,3,0,4,5,6,7,8])
&gt;&gt;&gt; numpy.flatnonzero(x == 0)
array([1, 3, 5])
</code></pre>

blocks|key|1781042|text|还可以通过在条件的布尔掩码上使用nonzero()来使用它，因为False也是一种零。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1781043|>>>+x+=+numpy.array([1,0,2,0,3,0,4,5,6,7,8])

>>>+x==0
array([False,+True,+False,+True,+False,+True,+False,+False,+False,+False,+False],+dtype=bool)

>>>+numpy.nonzero(x==0)[0]
array([1,+3,+5])|code-block|syntax|javascript|1781044|它与mtrw的方法完全相同，但它与问题更相关;)|1781045|entityMap^0|G|9|W|5|0|0|2|4|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

You can also use <code>nonzero()</code> by using it on a boolean mask of the condition, because <code>False</code> is also a kind of zero.

<pre><code>&gt;&gt;&gt; x = numpy.array([1,0,2,0,3,0,4,5,6,7,8])

&gt;&gt;&gt; x==0
array([False, True, False, True, False, True, False, False, False, False, False], dtype=bool)

&gt;&gt;&gt; numpy.nonzero(x==0)[0]
array([1, 3, 5])
</code></pre>

It's doing exactly the same as <code>mtrw</code>'s way, but it is more related to the question ;)

blocks|key|1781128|text|import+numpy+as+np

x+=+np.array([1,0,2,3,6])
non_zero_arr+=+np.extract(x>0,x)

min_index+=+np.amin(non_zero_arr)
min_value+=+np.argmin(non_zero_arr)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1781129|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>import numpy as np

x = np.array([1,0,2,3,6])
non_zero_arr = np.extract(x&gt;0,x)

min_index = np.amin(non_zero_arr)
min_value = np.argmin(non_zero_arr)
</code></pre>

blocks|key|1994949|text|那就是np.argwhere|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1994950|import+numpy+as+np
arr+=+np.array([[1,2,3],+[0,+1,+0],+[7,+0,+2]])
np.argwhere(arr+==+0)|code-block|syntax|javascript|1994951|它以行的形式返回所有找到的索引：|1994952|array([[1,+0],++++#+Indices+of+the+first+zero
+++++++[1,+2],++++#+Indices+of+the+second+zero
+++++++[2,+1]],+++#+Indices+of+the+third+zero
++++++dtype=int64)|1994953|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/numpy/reference/generated/numpy.argwhere.html#numpy-argwhere^0|3|B|3|B|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

There is <a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.argwhere.html#numpy-argwhere" rel="noreferrer"><code>np.argwhere</code></a>,

<pre><code>import numpy as np
arr = np.array([[1,2,3], [0, 1, 0], [7, 0, 2]])
np.argwhere(arr == 0)
</code></pre>

which returns all found indices as rows:

<pre><code>array([[1, 0], # Indices of the first zero
 [1, 2], # Indices of the second zero
 [2, 1]], # Indices of the third zero
 dtype=int64)
</code></pre>

blocks|key|1781198|text|我会这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1781199|>>>+x+=+np.array([[1,0,0],+[0,2,0],+[1,1,0]])
>>>+x
array([[1,+0,+0],
+++++++[0,+2,+0],
+++++++[1,+1,+0]])
>>>+np.nonzero(x)
(array([0,+1,+2,+2]),+array([0,+1,+0,+1]))

#+if+you+want+it+in+coordinates
>>>+x[np.nonzero(x)]
array([1,+2,+1,+1])
>>>+np.transpose(np.nonzero(x))
array([[0,+0],
+++++++[1,+1],
+++++++[2,+0],
+++++++[2,+1])|code-block|syntax|javascript|1781200|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|$]]

I would do it the following way:

<pre><code>&gt;&gt;&gt; x = np.array([[1,0,0], [0,2,0], [1,1,0]])
&gt;&gt;&gt; x
array([[1, 0, 0],
 [0, 2, 0],
 [1, 1, 0]])
&gt;&gt;&gt; np.nonzero(x)
(array([0, 1, 2, 2]), array([0, 1, 0, 1]))

# if you want it in coordinates
&gt;&gt;&gt; x[np.nonzero(x)]
array([1, 2, 1, 1])
&gt;&gt;&gt; np.transpose(np.nonzero(x))
array([[0, 0],
 [1, 1],
 [2, 0],
 [2, 1])
</code></pre>

blocks|key|1781266|text|您可以使用numpy.nonzero来查找零。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1781267|>>>+import+numpy+as+np
>>>+x+=+np.array([1,0,2,0,3,0,0,4,0,5,0,6]).reshape(4,+3)
>>>+np.nonzero(x==0)++#+this+is+what+you+want
(array([0,+1,+1,+2,+2,+3]),+array([1,+0,+2,+0,+2,+1]))
>>>+np.nonzero(x)
(array([0,+0,+1,+2,+3,+3]),+array([0,+2,+1,+1,+0,+2]))|code-block|syntax|javascript|1781268|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|$]]

You can use numpy.nonzero to find zero.

<pre><code>&gt;&gt;&gt; import numpy as np
&gt;&gt;&gt; x = np.array([1,0,2,0,3,0,0,4,0,5,0,6]).reshape(4, 3)
&gt;&gt;&gt; np.nonzero(x==0) # this is what you want
(array([0, 1, 1, 2, 2, 3]), array([1, 0, 2, 0, 2, 1]))
&gt;&gt;&gt; np.nonzero(x)
(array([0, 0, 1, 2, 3, 3]), array([0, 2, 1, 1, 0, 2]))
</code></pre>

blocks|key|1995093|text|import+numpy+as+np
arr+=+np.arange(10000)
arr[8000:8900]+=+0

%25timeit+np.where(arr+==+0)[0]
%25timeit+np.argwhere(arr+==+0)
%25timeit+np.nonzero(arr==0)[0]
%25timeit+np.flatnonzero(arr==0)
%25timeit+np.amin(np.extract(arr+!=+0,+arr))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1995094|23.4+µs+±+1.5+µs+per+loop+(mean+±+std.+dev.+of+7+runs,+10000+loops+each)
34.5+µs+±+680+ns+per+loop+(mean+±+std.+dev.+of+7+runs,+10000+loops+each)
23.2+µs+±+447+ns+per+loop+(mean+±+std.+dev.+of+7+runs,+10000+loops+each)
27+µs+±+506+ns+per+loop+(mean+±+std.+dev.+of+7+runs,+10000+loops+each)
109+µs+±+669+ns+per+loop+(mean+±+std.+dev.+of+7+runs,+10000+loops+each)|1995095|unstyled|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$B|C]]|$1|F|3|-4|5|G|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre class="lang-py prettyprint-override"><code>import numpy as np
arr = np.arange(10000)
arr[8000:8900] = 0

%timeit np.where(arr == 0)[0]
%timeit np.argwhere(arr == 0)
%timeit np.nonzero(arr==0)[0]
%timeit np.flatnonzero(arr==0)
%timeit np.amin(np.extract(arr != 0, arr))
</code></pre>
<pre><code>23.4 µs ± 1.5 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
34.5 µs ± 680 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
23.2 µs ± 447 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
27 µs ± 506 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
109 µs ± 669 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
</code></pre>

NumPy has the efficient function/method <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.nonzero.html" rel="noreferrer"><code>nonzero()</code></a> to identify the indices of non-zero elements in an <code>ndarray</code> object. What is the most efficient way to obtain the indices of the elements that do have a value of zero?

Find indices of elements equal to zero in a NumPy array

NumPy具有有效的函数/方法来标识ndarray对象中非零元素的索引。获取值为零的元素的索引的最有效方法是什么？

问在NumPy数组中查找等于零的元素的索引
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