blocks|key|2924499|text|a+=+numpy.array([0,+3,+0,+1,+0,+1,+2,+1,+0,+0,+0,+0,+1,+3,+4])
unique,+counts+=+numpy.unique(a,+return_counts=True)
dict(zip(unique,+counts))

#+{0:+7,+1:+4,+2:+1,+3:+2,+4:+1}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2924500|非numpy方式|unstyled|offset|length|style|BOLD|2924501|使用collections.Counter；|CODE|2924502|import+collections,+numpy
a+=+numpy.array([0,+3,+0,+1,+0,+1,+2,+1,+0,+0,+0,+0,+1,+3,+4])
collections.Counter(a)

#+Counter({0:+7,+1:+4,+3:+2,+2:+1,+4:+1})|2924503|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/collections.html#collections.Counter^0|0|0|8|0|2|J|2|J|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|Y|8|@$G|Z|H|10|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|11|8|@$G|12|H|13|I|M]]|9|@$G|14|H|15|1|16]]|A|$]]|$1|N|3|O|5|6|7|17|8|@]|9|@]|A|$B|C]]|$1|P|3|-4|5|F|7|18|8|@]|9|@]|A|$]]]|Q|$R|$5|S|T|U|A|$V|W]]]]

<pre><code>a = numpy.array([0, 3, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 4])
unique, counts = numpy.unique(a, return_counts=True)
dict(zip(unique, counts))

# {0: 7, 1: 4, 2: 1, 3: 2, 4: 1}
</code></pre>
Non-numpy way:
Use <a href="https://docs.python.org/2/library/collections.html#collections.Counter" rel="noreferrer"><code>collections.Counter</code></a>;
<pre><code>import collections, numpy
a = numpy.array([0, 3, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 4])
collections.Counter(a)

# Counter({0: 7, 1: 4, 3: 2, 2: 1, 4: 1})
</code></pre>

blocks|key|2924575|text|那么使用numpy.count_nonzero怎么样呢？|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2924576|>>>+import+numpy+as+np
>>>+y+=+np.array([1,+2,+2,+2,+2,+0,+2,+3,+3,+3,+0,+0,+2,+2,+0])

>>>+np.count_nonzero(y+==+1)
1
>>>+np.count_nonzero(y+==+2)
7
>>>+np.count_nonzero(y+==+3)
3|code-block|syntax|javascript|2924577|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.count_nonzero.html^0|4|J|4|J|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

What about using <a href="https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.count_nonzero.html" rel="noreferrer"><code>numpy.count_nonzero</code></a>, something like 

<pre><code>&gt;&gt;&gt; import numpy as np
&gt;&gt;&gt; y = np.array([1, 2, 2, 2, 2, 0, 2, 3, 3, 3, 0, 0, 2, 2, 0])

&gt;&gt;&gt; np.count_nonzero(y == 1)
1
&gt;&gt;&gt; np.count_nonzero(y == 2)
7
&gt;&gt;&gt; np.count_nonzero(y == 3)
3
</code></pre>

blocks|key|2924595|text|就我个人而言，我倾向于：(y+==+0).sum()和(y+==+1).sum()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2924596|例如。|2924597|import+numpy+as+np
y+=+np.array([0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1])
num_zeros+=+(y+==+0).sum()
num_ones+=+(y+==+1).sum()|code-block|syntax|javascript|2924598|entityMap^0|C|E|R|E|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

Personally, I'd go for:
<code>(y == 0).sum()</code> and <code>(y == 1).sum()</code>

E.g.

<pre><code>import numpy as np
y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
num_zeros = (y == 0).sum()
num_ones = (y == 1).sum()
</code></pre>

blocks|key|2711968|text|对于您的情况，您还可以查看numpy.bincount|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2711969|In+[56]:+a+=+np.array([0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1])

In+[57]:+np.bincount(a)
Out[57]:+array([8,+4])++#count+of+zeros+is+at+index+0+:+8
++++++++++++++++++++++++#count+of+ones+is+at+index+1+:+4|code-block|syntax|javascript|2711970|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.bincount.html^0|D|E|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 your case you could also look into <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.bincount.html">numpy.bincount</a>

<pre><code>In [56]: a = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])

In [57]: np.bincount(a)
Out[57]: array([8, 4]) #count of zeros is at index 0 : 8
 #count of ones is at index 1 : 4
</code></pre>

blocks|key|138502|text|y+=+np.array([0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|138503|如果您知道它们只是0和1|unstyled|offset|length|style|CODE|138504|np.sum(y)|138505|给出了1的数量。np.sum(1-y)给出了零。|138506|对于一般情况，如果您想计算0而不是0(但可能是2或3)：|138507|np.count_nonzero(y)|138508|给出非零的数字。|138509|但是如果你需要更复杂的东西，我不认为numpy会提供一个很好的count选项。在这种情况下，请转到集合：|138510|import+collections
collections.Counter(y)
>+Counter({0:+8,+1:+4})|138511|它的行为就像一个字典|138512|collections.Counter(y)[0]
>+8|138513|entityMap^0|0|9|1|B|1|0|0|8|B|0|D|1|0|0|0|V|5|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|15|8|@$G|16|H|17|I|J]|$G|18|H|19|I|J]]|9|@]|A|$]]|$1|K|3|L|5|6|7|1A|8|@]|9|@]|A|$B|C]]|$1|M|3|N|5|F|7|1B|8|@$G|1C|H|1D|I|J]]|9|@]|A|$]]|$1|O|3|P|5|F|7|1E|8|@$G|1F|H|1G|I|J]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1H|8|@]|9|@]|A|$B|C]]|$1|S|3|T|5|F|7|1I|8|@]|9|@]|A|$]]|$1|U|3|V|5|F|7|1J|8|@$G|1K|H|1L|I|J]]|9|@]|A|$]]|$1|W|3|X|5|6|7|1M|8|@]|9|@]|A|$B|C]]|$1|Y|3|Z|5|F|7|1N|8|@]|9|@]|A|$]]|$1|10|3|11|5|6|7|1O|8|@]|9|@]|A|$B|C]]|$1|12|3|-4|5|F|7|1P|8|@]|9|@]|A|$]]]|13|$]]

<pre><code>y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
</code></pre>

If you know that they are just <code>0</code> and <code>1</code>:

<pre><code>np.sum(y)
</code></pre>

gives you the number of ones. <code>np.sum(1-y)</code> gives the zeroes.

For slight generality, if you want to count <code>0</code> and not zero (but possibly 2 or 3):

<pre><code>np.count_nonzero(y)
</code></pre>

gives the number of nonzero.

But if you need something more complicated, I don't think numpy will provide a nice <code>count</code> option. In that case, go to collections:

<pre><code>import collections
collections.Counter(y)
&gt; Counter({0: 8, 1: 4})
</code></pre>

This behaves like a dict

<pre><code>collections.Counter(y)[0]
&gt; 8
</code></pre>

blocks|key|139772|text|将数组y转换为list+l，然后执行l.count(1)和l.count(0)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|139773|>>>+y+=+numpy.array([0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1])
>>>+l+=+list(y)
>>>+l.count(1)
4
>>>+l.count(0)
8+|code-block|syntax|javascript|139774|entityMap^0|3|1|C|1|I|A|T|A|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

Convert your array <code>y</code> to list <code>l</code> and then do <code>l.count(1)</code> and <code>l.count(0)</code>

<pre><code>&gt;&gt;&gt; y = numpy.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
&gt;&gt;&gt; l = list(y)
&gt;&gt;&gt; l.count(1)
4
&gt;&gt;&gt; l.count(0)
8 
</code></pre>

blocks|key|139922|text|如果你确切地知道你要找的是哪个数字，你可以使用下面的代码；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|139923|lst+=+np.array([1,1,2,3,3,6,6,6,3,2,1])
(lst+==+2).sum()|code-block|syntax|javascript|139924|返回数组中出现2的次数。|139925|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|$]]

If you know exactly which number you're looking for, you can use the following;

<pre><code>lst = np.array([1,1,2,3,3,6,6,6,3,2,1])
(lst == 2).sum()
</code></pre>

returns how many times 2 is occurred in your array.

blocks|key|140932|text|过滤并使用len|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|140933|使用len可能是另一种选择。|140934|A+=+np.array([1,0,1,0,1,0,1])|code-block|syntax|javascript|140935|假设我们想要0的出现次数。|140936|A[A==0]++#+Return+the+array+where+item+is+0,+array([0,+0,+0])|140937|现在，使用len将其包装起来。|140938|len(A[A==0])++#+3
len(A[A==1])++#+4
len(A[A==7])++#+0,+because+there+isn't+such+item.|140939|entityMap^0|5|3|0|2|3|0|0|6|1|0|0|5|3|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$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|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|S|3|T|5|J|7|1A|8|@]|D|@]|E|$K|L]]|$1|U|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|V|$]]

<h1>Filter and use <code>len</code></h1>
Using <code>len</code> could be another option.
<pre><code>A = np.array([1,0,1,0,1,0,1])
</code></pre>
Say we want the number of occurrences of <code>0</code>.
<pre><code>A[A==0] # Return the array where item is 0, array([0, 0, 0])
</code></pre>
Now, wrap it around with <code>len</code>.
<pre><code>len(A[A==0]) # 3
len(A[A==1]) # 4
len(A[A==7]) # 0, because there isn't such item.
</code></pre>

blocks|key|139851|text|老实说，我发现转换成pandas+Series或DataFrame是最容易的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|139852|import+pandas+as+pd
import+numpy+as+np

df+=+pd.DataFrame({'data':np.array([0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1])})
print+df['data'].value_counts()|code-block|syntax|javascript|139853|或者是Robert+Muil建议的这句很好的一行：|139854|pd.Series([0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1]).value_counts()|139855|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|$]]

Honestly I find it easiest to convert to a pandas Series or DataFrame:

<pre><code>import pandas as pd
import numpy as np

df = pd.DataFrame({'data':np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])})
print df['data'].value_counts()
</code></pre>

Or this nice one-liner suggested by Robert Muil:

<pre><code>pd.Series([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1]).value_counts()
</code></pre>

blocks|key|141058|text|没有人建议在minlength+=+np.size(input)中使用numpy.bincount(input,+minlength)，但它似乎是一个很好的解决方案，而且绝对是最快的|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|141059|In+[1]:+choices+=+np.random.randint(0,+100,+10000)

In+[2]:+%25timeit+[+np.sum(choices+==+k)+for+k+in+range(min(choices),+max(choices)%2B1)+]
100+loops,+best+of+3:+2.67+ms+per+loop

In+[3]:+%25timeit+np.unique(choices,+return_counts=True)
1000+loops,+best+of+3:+388+µs+per+loop

In+[4]:+%25timeit+np.bincount(choices,+minlength=np.size(choices))
100000+loops,+best+of+3:+16.3+µs+per+loop|code-block|syntax|javascript|141060|这是numpy.unique(x,+return_counts=True)和numpy.bincount(x,+minlength=np.max(x))之间的疯狂加速！|141061|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/numpy/reference/generated/numpy.bincount.html^0|6|Q|Z|W|Z|W|0|0|0|2|Z|12|12|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$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|@$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

No one suggested to use <a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.bincount.html" rel="noreferrer"><code>numpy.bincount(input, minlength)</code></a> with <code>minlength = np.size(input)</code>, but it seems to be a good solution, and definitely the fastest:

<pre><code>In [1]: choices = np.random.randint(0, 100, 10000)

In [2]: %timeit [ np.sum(choices == k) for k in range(min(choices), max(choices)+1) ]
100 loops, best of 3: 2.67 ms per loop

In [3]: %timeit np.unique(choices, return_counts=True)
1000 loops, best of 3: 388 µs per loop

In [4]: %timeit np.bincount(choices, minlength=np.size(choices))
100000 loops, best of 3: 16.3 µs per loop
</code></pre>

That's a crazy speedup between <code>numpy.unique(x, return_counts=True)</code> and <code>numpy.bincount(x, minlength=np.max(x))</code> !

blocks|key|2711972|text|我会使用np.where：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2711973|how_many_0+=+len(np.where(a==0.)[0])
how_many_1+=+len(np.where(a==1.)[0])|code-block|syntax|javascript|2711974|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'd use np.where:

<pre><code>how_many_0 = len(np.where(a==0.)[0])
how_many_1 = len(np.where(a==1.)[0])
</code></pre>

blocks|key|2924600|text|y.tolist().count(val)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2924601|使用val+0或1|2924602|因为python列表有一个本机函数count，所以在使用该函数之前转换为list是一个简单的解决方案。|2924603|entityMap^0|0|L|0|0|H|5|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|K|$]]

<code>y.tolist().count(val)</code>

with val 0 or 1

Since a python list has a native function <code>count</code>, converting to list before using that function is a simple solution.

blocks|key|2712501|text|如果您对最快的执行感兴趣，您预先知道要查找哪个(S)值，并且您的数组是1D，或者您对展平数组的结果感兴趣(在这种情况下，函数的输入应该是np.ravel(arr)而不只是arr)，那么Numba就是您的朋友：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2712502|import+numba+as+nb


@nb.jit
def+count_nb(arr,+value):
++++result+=+0
++++for+x+in+arr:
++++++++if+x+==+value:
++++++++++++result+%2B=+1
++++return+result|code-block|syntax|javascript|2712503|或者，对于非常大的数组，其中并行化可能是有益的：|2712504|@nb.jit(parallel=True)
def+count_nbp(arr,+value):
++++result+=+0
++++for+i+in+nb.prange(arr.size):
++++++++if+arr[i]+==+value:
++++++++++++result+%2B=+1
++++return+result|2712505|将它们与np.count_nonzero()和np.unique()-based解决方案进行基准测试(这也存在创建临时阵列的问题，这一问题可以避免|2712506|import+numpy+as+np


def+count_np(arr,+value):
++++return+np.count_nonzero(arr+==+value)|2712507|import+numpy+as+np


def+count_np2(arr,+value):
++++uniques,+counts+=+np.unique(a,+return_counts=True)
++++counter+=+dict(zip(uniques,+counts))
++++return+counter[value]+if+value+in+counter+else+0+|2712508|对于通过以下方式生成的输入：|2712509|def+gen_input(n,+a=0,+b=100):
++++return+np.random.randint(a,+b,+n)|2712510|将获得以下绘图(第二行绘图是对较快方法的缩放)：|2712511|​|2712512|📷|atomic|2712513|2712514|2712515|结果表明，基于Numba的解决方案明显比NumPy解决方案更快，并且，对于非常大的输入，并行方法比朴素方法更快。|2712516|2712517|完整的代码可用here。|2712518|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/1cb9624c27ac7fb821ed52b0546e67f5.png|imageAlt|1|https://ask.qcloudimg.com/http-save/yehe-900000/181621799fa8e5e292d7f3b6217b1216.png|2|LINK|MUTABLE|url|https://colab.research.google.com/drive/1A0LKo-Qdd9k8okVjmCiQurx6LRZuIpUs^0|1W|D|2D|3|0|0|0|0|4|I|N|B|0|0|0|0|0|0|0|0|1|0|0|0|1|1|0|0|0|0|7|4|2|0^^$0|@$1|2|3|4|5|6|7|1S|8|@$9|1T|A|1U|B|C]|$9|1V|A|1W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|20|8|@$9|21|A|22|B|C]|$9|23|A|24|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|25|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|H|7|26|8|@]|D|@]|E|$I|J]]|$1|U|3|V|5|6|7|27|8|@]|D|@]|E|$]]|$1|W|3|X|5|H|7|28|8|@]|D|@]|E|$I|J]]|$1|Y|3|Z|5|6|7|29|8|@]|D|@]|E|$]]|$1|10|3|11|5|6|7|2A|8|@]|D|@]|E|$]]|$1|12|3|13|5|14|7|2B|8|@]|D|@$9|2C|A|2D|1|2E]]|E|$]]|$1|15|3|13|5|14|7|2F|8|@]|D|@$9|2G|A|2H|1|2I]]|E|$]]|$1|16|3|11|5|6|7|2J|8|@]|D|@]|E|$]]|$1|17|3|18|5|6|7|2K|8|@]|D|@]|E|$]]|$1|19|3|-4|5|6|7|2L|8|@]|D|@]|E|$]]|$1|1A|3|1B|5|6|7|2M|8|@]|D|@$9|2N|A|2O|1|2P]]|E|$]]|$1|1C|3|-4|5|6|7|2Q|8|@]|D|@]|E|$]]]|1D|$1E|$5|1F|1G|1H|E|$1I|1J|1K|-4]]|1L|$5|1F|1G|1H|E|$1I|1M|1K|-4]]|1N|$5|1O|1G|1P|E|$1Q|1R]]]]

If you are interested in the fastest execution, you know in advance which value(s) to look for, and your array is 1D, or you are otherwise interested in the result on the flattened array (in which case the input of the function should be <code>np.ravel(arr)</code> rather than just <code>arr</code>), then Numba is your friend:
<pre><code>import numba as nb


@nb.jit
def count_nb(arr, value):
 result = 0
 for x in arr:
 if x == value:
 result += 1
 return result
</code></pre>
or, for very large arrays where parallelization may be beneficial:
<pre><code>@nb.jit(parallel=True)
def count_nbp(arr, value):
 result = 0
 for i in nb.prange(arr.size):
 if arr[i] == value:
 result += 1
 return result
</code></pre>
Benchmarking these against <code>np.count_nonzero()</code> (which also has a problem of creating a temporary array which may be avoided) and <code>np.unique()</code>-based solution
<pre><code>import numpy as np


def count_np(arr, value):
 return np.count_nonzero(arr == value)
</code></pre>
<pre><code>import numpy as np


def count_np2(arr, value):
 uniques, counts = np.unique(a, return_counts=True)
 counter = dict(zip(uniques, counts))
 return counter[value] if value in counter else 0 
</code></pre>
for input generated with:
<pre><code>def gen_input(n, a=0, b=100):
 return np.random.randint(a, b, n)
</code></pre>
the following plots are obtained (the second row of plots is a zoom on the faster approach):
<a href="https://i.stack.imgur.com/noMlo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/noMlo.png" alt="bm_full" /></a>
<a href="https://i.stack.imgur.com/taLwx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/taLwx.png" alt="bm_zoom" /></a>
Showing that Numba-based solution are noticeably faster than the NumPy counterparts, and, for very large inputs, the parallel approach is faster than the naive one.
<hr />
Full code available <a href="https://colab.research.google.com/drive/1A0LKo-Qdd9k8okVjmCiQurx6LRZuIpUs" rel="nofollow noreferrer">here</a>.

blocks|key|2712114|text|另一种简单的解决方案可能是使用numpy.count_nonzero()|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2712115|import+numpy+as+np
y+=+np.array([0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1])
y_nonzero_num+=+np.count_nonzero(y==1)
y_zero_num+=+np.count_nonzero(y==0)
y_nonzero_num
4
y_zero_num
8|code-block|syntax|javascript|2712116|不要让名称误导您，如果您像示例中那样将其与布尔值一起使用，它将会起到作用。|2712117|entityMap^0|F|L|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

Yet another simple solution might be to use numpy.count_nonzero():

<pre><code>import numpy as np
y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
y_nonzero_num = np.count_nonzero(y==1)
y_zero_num = np.count_nonzero(y==0)
y_nonzero_num
4
y_zero_num
8
</code></pre>

Don't let the name mislead you, if you use it with the boolean just like in the example, it will do the trick.

blocks|key|139890|text|要计算出现的次数，可以使用np.unique(array,+return_counts=True)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|139891|In+[75]:+boo+=+np.array([0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1])
+
#+use+bool+value+`True`+or+equivalently+`1`
In+[77]:+uniq,+cnts+=+np.unique(boo,+return_counts=1)
In+[81]:+uniq
Out[81]:+array([0,+1])+++#unique+elements+in+input+array+are:+0,+1

In+[82]:+cnts
Out[82]:+array([8,+4])+++#+0+occurs+8+times,+1+occurs+4+times|code-block|syntax|javascript|139892|entityMap|0|LINK|mutability|MUTABLE|url|https://numpy.org/doc/stable/reference/generated/numpy.unique.html^0|D|10|D|10|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

To count the number of occurrences, you can use <a href="https://numpy.org/doc/stable/reference/generated/numpy.unique.html" rel="nofollow noreferrer"><code>np.unique(array, return_counts=True)</code></a>:
<pre><code>In [75]: boo = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
 
# use bool value `True` or equivalently `1`
In [77]: uniq, cnts = np.unique(boo, return_counts=1)
In [81]: uniq
Out[81]: array([0, 1]) #unique elements in input array are: 0, 1

In [82]: cnts
Out[82]: array([8, 4]) # 0 occurs 8 times, 1 occurs 4 times
</code></pre>

blocks|key|2712380|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2712381|a+=+np.array([0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1])
list(a).count(1)|code-block|syntax|javascript|2712382|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 this:

<pre><code>a = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
list(a).count(1)
</code></pre>

blocks|key|2712358|text|充分利用本系列提供的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2712359|>>>+import+pandas+as+pd
>>>+y+=+[0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1]
>>>+pd.Series(y).value_counts()
0++++8
1++++4
dtype:+int64|code-block|syntax|javascript|2712360|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|$]]

take advantage of the methods offered by a Series:

<pre><code>&gt;&gt;&gt; import pandas as pd
&gt;&gt;&gt; y = [0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1]
&gt;&gt;&gt; pd.Series(y).value_counts()
0 8
1 4
dtype: int64
</code></pre>

blocks|key|2924879|text|您可以使用字典理解来创建整洁的一行程序。有关字典理解can+be+found+here+的更多信息|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2924880|>>>counts+=+{int(value):+list(y).count(value)+for+value+in+set(y)}
>>>print(counts)
{0:+8,+1:+4}|code-block|syntax|javascript|2924881|这将创建一个字典，将ndarray中的值作为键，并将值的计数分别作为键的值。|2924882|当您想要计算此格式的数组中某个值的出现次数时，此方法将起作用。|2924883|entityMap|0|LINK|mutability|MUTABLE|url|https://www.python.org/dev/peps/pep-0274/^0|Q|I|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

You can use dictionary comprehension to create a neat one-liner. More about dictionary comprehension <a href="https://www.python.org/dev/peps/pep-0274/" rel="nofollow noreferrer">can be found here </a>

<pre><code>&gt;&gt;&gt;counts = {int(value): list(y).count(value) for value in set(y)}
&gt;&gt;&gt;print(counts)
{0: 8, 1: 4}
</code></pre>

This will create a dictionary with the values in your ndarray as keys, and the counts of the values as the values for the keys respectively.

This will work whenever you want to count occurences of a value in arrays of this format.

blocks|key|141985|text|它涉及到更多的步骤，但更灵活的解决方案也适用于二维数组和更复杂的过滤器是创建一个布尔掩码，然后在掩码上使用.sum()。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|141986|>>>>y+=+np.array([0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1])
>>>>mask+=+y+==+0
>>>>mask.sum()
8|code-block|syntax|javascript|141987|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|$]]

It involves one more step, but a more flexible solution which would also work for 2d arrays and more complicated filters is to create a boolean mask and then use .sum() on the mask.

<pre><code>&gt;&gt;&gt;&gt;y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
&gt;&gt;&gt;&gt;mask = y == 0
&gt;&gt;&gt;&gt;mask.sum()
8
</code></pre>

blocks|key|2712155|text|一个普遍而简单的答案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2712156|numpy.sum(MyArray==x)+++#+sum+of+a+binary+list+of+the+occurence+of+x+(=0+or+1)+in+MyArray|code-block|syntax|javascript|2712157|这将导致生成以下完整代码作为示例|2712158|import+numpy
MyArray=numpy.array([0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1])++#+array+we+want+to+search+in
x=0+++#+the+value+I+want+to+count+(can+be+iterator,+in+a+list,+etc.)
numpy.sum(MyArray==0)+++#+sum+of+a+binary+list+of+the+occurence+of+x+in+MyArray|2712159|现在，如果MyArray在多维中，并且您想要计算行中值分布的出现次数(=下文中的模式)|offset|length|style|BOLD|2712160|MyArray=numpy.array([[6,+1],[4,+5],[0,+7],[5,+1],[2,+5],[1,+2],[3,+2],[0,+2],[2,+5],[5,+1],[3,+0]])
x=numpy.array([5,1])+++#+the+value+I+want+to+count+(can+be+iterator,+in+a+list,+etc.)
temp+=+numpy.ascontiguousarray(MyArray).view(numpy.dtype((numpy.void,+MyArray.dtype.itemsize+*+MyArray.shape[1])))++#+convert+the+2d-array+into+an+array+of+analyzable+patterns
xt=numpy.ascontiguousarray(x).view(numpy.dtype((numpy.void,+x.dtype.itemsize+*+x.shape[0])))++#+convert+what+you+search+into+one+analyzable+pattern
numpy.sum(temp==xt)++#+count+of+the+searched+pattern+in+the+list+of+patterns|2712161|entityMap^0|0|0|0|0|D|2|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

A general and simple answer would be:

<pre><code>numpy.sum(MyArray==x) # sum of a binary list of the occurence of x (=0 or 1) in MyArray
</code></pre>

which would result into this full code as exemple

<pre><code>import numpy
MyArray=numpy.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1]) # array we want to search in
x=0 # the value I want to count (can be iterator, in a list, etc.)
numpy.sum(MyArray==0) # sum of a binary list of the occurence of x in MyArray
</code></pre>

Now if MyArray is in multiple dimensions and you want to count the occurence of a distribution of values in line (= pattern hereafter)

<pre><code>MyArray=numpy.array([[6, 1],[4, 5],[0, 7],[5, 1],[2, 5],[1, 2],[3, 2],[0, 2],[2, 5],[5, 1],[3, 0]])
x=numpy.array([5,1]) # the value I want to count (can be iterator, in a list, etc.)
temp = numpy.ascontiguousarray(MyArray).view(numpy.dtype((numpy.void, MyArray.dtype.itemsize * MyArray.shape[1]))) # convert the 2d-array into an array of analyzable patterns
xt=numpy.ascontiguousarray(x).view(numpy.dtype((numpy.void, x.dtype.itemsize * x.shape[0]))) # convert what you search into one analyzable pattern
numpy.sum(temp==xt) # count of the searched pattern in the list of patterns
</code></pre>

blocks|key|2712324|text|你有一个特殊的数组，这里只有1和0。所以有个技巧就是使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2712325|np.mean(x)|code-block|syntax|javascript|2712326|它给出了数组中1的百分比。或者，使用|2712327|np.sum(x)
np.sum(1-x)|2712328|将给出数组中1和0的绝对数。|2712329|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You have a special array with only 1 and 0 here. So a trick is to use 

<pre><code>np.mean(x)
</code></pre>

which gives you the percentage of 1s in your array. Alternatively, use

<pre><code>np.sum(x)
np.sum(1-x)
</code></pre>

will give you the absolute number of 1 and 0 in your array.

blocks|key|139976|text|dict(zip(*numpy.unique(y,+return_counts=True)))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|139977|我只是在这里复制了赛波·埃纳维的评论，这应该是一个恰当的答案|unstyled|139978|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>dict(zip(*numpy.unique(y, return_counts=True)))
</code></pre>

Just copied Seppo Enarvi's comment here which deserves to be a proper answer

blocks|key|142040|text|这可以通过以下方法轻松完成|type|unstyled|depth|inlineStyleRanges|entityRanges|data|142041|y+=+np.array([0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1])
y.tolist().count(1)|code-block|syntax|javascript|142042|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|$]]

This can be done easily in the following method

<pre><code>y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
y.tolist().count(1)
</code></pre>

blocks|key|138653|text|由于ndarray只包含0和1，因此可以使用sum()获取1的出现次数，使用len()-sum()获取0的出现次数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|138654|num_of_ones+=+sum(array)
num_of_zeros+=+len(array)-sum(array)|code-block|syntax|javascript|138655|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|$]]

Since your ndarray contains only 0 and 1, 
you can use sum() to get the occurrence of 1s 
and len()-sum() to get the occurrence of 0s.

<pre><code>num_of_ones = sum(array)
num_of_zeros = len(array)-sum(array)
</code></pre>

blocks|key|2712280|text|对于通用条目：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2712281|x+=+np.array([11,+2,+3,+5,+3,+2,+16,+10,+10,+3,+11,+4,+5,+16,+3,+11,+4])
n+=+{i:len([j+for+j+in+np.where(x==i)[0]])+for+i+in+set(x)}
ix+=+{i:[j+for+j+in+np.where(x==i)[0]]+for+i+in+set(x)}|code-block|syntax|javascript|2712282|将输出一个计数：|2712283|{2:+2,+3:+4,+4:+2,+5:+2,+10:+2,+11:+3,+16:+2}|2712284|和索引：|2712285|{2:+[1,+5],
3:+[2,+4,+9,+14],
4:+[11,+16],
5:+[3,+12],
10:+[7,+8],
11:+[0,+10,+15],
16:+[6,+13]}|2712286|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]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

For generic entries:

<pre><code>x = np.array([11, 2, 3, 5, 3, 2, 16, 10, 10, 3, 11, 4, 5, 16, 3, 11, 4])
n = {i:len([j for j in np.where(x==i)[0]]) for i in set(x)}
ix = {i:[j for j in np.where(x==i)[0]] for i in set(x)}
</code></pre>

Will output a count:

<pre><code>{2: 2, 3: 4, 4: 2, 5: 2, 10: 2, 11: 3, 16: 2}
</code></pre>

And indices:

<pre><code>{2: [1, 5],
3: [2, 4, 9, 14],
4: [11, 16],
5: [3, 12],
10: [7, 8],
11: [0, 10, 15],
16: [6, 13]}
</code></pre>

blocks|key|2924696|text|如果不想使用numpy或集合模块，可以使用字典：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2924697|d+=+dict()
a+=+[0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1]
for+item+in+a:
++++try:
++++++++d[item]%2B=1
++++except+KeyError:
++++++++d[item]=1|code-block|syntax|javascript|2924698|结果：|2924699|>>>d
{0:+8,+1:+4}|2924700|当然，您也可以使用if/else语句。我认为Counter函数几乎做了同样的事情，但这更透明。|2924701|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If you don't want to use numpy or a collections module you can use a dictionary:

<pre><code>d = dict()
a = [0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1]
for item in a:
 try:
 d[item]+=1
 except KeyError:
 d[item]=1
</code></pre>

result:

<pre><code>&gt;&gt;&gt;d
{0: 8, 1: 4}
</code></pre>

Of course you can also use an if/else statement. 
I think the Counter function does almost the same thing but this is more transparant.

blocks|key|141106|text|这里有一些东西，您可以通过它来计算特定数字出现的次数:根据您的代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|141107|count_of_zero=list(yy==0).count(0)|141108|打印(Count_of_zero)|141109|//根据匹配会有布尔值，根据True值返回数字0|141110|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

here I have something, through which you can count the number of occurrence of a particular number:
according to your code

count_of_zero=list(y[y==0]).count(0) 

print(count_of_zero)

// according to the match there will be boolean values and according to True value the number 0 will be return

blocks|key|2712561|text|如果您正在处理非常大的数组，那么使用生成器可能是一种选择。这里的好处是，这种方法对数组和列表都能很好地工作，并且不需要任何额外的包。此外，您没有使用那么多内存。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2712562|my_array+=+np.array([0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1])
sum(1+for+val+in+my_array+if+val==0)
Out:+8|code-block|syntax|javascript|2712563|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 dealing with very large arrays using generators could be an option. The nice thing here it that this approach works fine for both arrays and lists and you dont need any additional package. Additionally, you are not using that much memory. 

<pre><code>my_array = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
sum(1 for val in my_array if val==0)
Out: 8
</code></pre>

blocks|key|2712590|text|此函数返回数组中变量的出现次数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2712591|def+count(array,variable):
++++number+=+0
++++for+i+in+range(array.shape[0]):
++++++++for+j+in+range(array.shape[1]):
++++++++++++if+array[i,j]+==+variable:
++++++++++++++++number+%2B=+1
++++return+number|code-block|syntax|javascript|2712592|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|$]]

This funktion returns the number of occurences of a variable in an array:
<pre><code>def count(array,variable):
 number = 0
 for i in range(array.shape[0]):
 for j in range(array.shape[1]):
 if array[i,j] == variable:
 number += 1
 return number
</code></pre>

blocks|key|2924807|text|Numpy为此提供了一个模块。这只是个小技巧。将您的输入数组放入存储箱。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2924808|numpy.histogram(y,+bins=y)|code-block|syntax|javascript|2924809|输出是2个数组。一个是值本身，另一个是相应的频率。|2924810|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|$]]

Numpy has a module for this. Just a small hack. Put your input array as bins. 

<pre><code>numpy.histogram(y, bins=y)
</code></pre>

The output are 2 arrays. One with the values itself, other with the corresponding frequencies.

blocks|key|2924914|text|using+numpy.count

$+a+=+[0,+0,+0,+1,+0,+1,+1,+0,+0,+0,+0,+1]

$+np.count(a,+1)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2924915|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>using numpy.count

$ a = [0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1]

$ np.count(a, 1)
</code></pre>

In Python, I have an ndarray <code>y</code>
that is printed as <code>array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])</code>

I'm trying to count how many <code>0</code>s and how many <code>1</code>s are there in this array. 

But when I type <code>y.count(0)</code> or <code>y.count(1)</code>, it says 

<blockquote>
 <code>numpy.ndarray</code> object has no attribute <code>count</code>
</blockquote>

What should I do?

How to count the occurrence of certain item in an ndarray?

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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在Python中，我有一个打印为array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])的ndarray y我正在尝试计算这个数组中有多少个0和多少个1。但当我键入y.count(0)或y.count(1)时，它会显示 numpy.ndarray对象没有count属性我该怎么办？

问如何计算某一项在ndarray中的出现次数？
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