blocks|key|329125|text|这会更快一点(看起来也更好)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|329126|np.argmax(aa>5)|code-block|syntax|javascript|329127|由于argmax将在第一个True处停止(“如果多次出现最大值，则返回与第一个出现的索引对应的索引。”)并且不保存另一个列表。|offset|length|style|CODE|329128|In+[2]:+N+=+10000

In+[3]:+aa+=+np.arange(-N,N)

In+[4]:+timeit+np.argmax(aa>N/2)
100000+loops,+best+of+3:+52.3+us+per+loop

In+[5]:+timeit+np.where(aa>N/2)[0][0]
10000+loops,+best+of+3:+141+us+per+loop

In+[6]:+timeit+np.nonzero(aa>N/2)[0][0]
10000+loops,+best+of+3:+142+us+per+loop|329129|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.argmax.html^0|0|0|2|6|D|4|2|6|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@$I|13|J|14|1|15]]|A|$]]|$1|M|3|N|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

This is a little faster (and looks nicer)

<pre><code>np.argmax(aa&gt;5)
</code></pre>

Since <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.argmax.html" rel="noreferrer"><code>argmax</code></a> will stop at the first <code>True</code> ("In case of multiple occurrences of the maximum values, the indices corresponding to the first occurrence are returned.") and doesn't save another list.

<pre><code>In [2]: N = 10000

In [3]: aa = np.arange(-N,N)

In [4]: timeit np.argmax(aa&gt;N/2)
100000 loops, best of 3: 52.3 us per loop

In [5]: timeit np.where(aa&gt;N/2)[0][0]
10000 loops, best of 3: 141 us per loop

In [6]: timeit np.nonzero(aa&gt;N/2)[0][0]
10000 loops, best of 3: 142 us per loop
</code></pre>

blocks|key|329139|text|In+[34]:+a=np.arange(-10,10)

In+[35]:+a
Out[35]:
array([-10,++-9,++-8,++-7,++-6,++-5,++-4,++-3,++-2,++-1,+++0,+++1,+++2,
+++++++++3,+++4,+++5,+++6,+++7,+++8,+++9])

In+[36]:+np.where(a>5)
Out[36]:+(array([16,+17,+18,+19]),)

In+[37]:+np.where(a>5)[0][0]
Out[37]:+16|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|329140|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>In [34]: a=np.arange(-10,10)

In [35]: a
Out[35]:
array([-10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2,
 3, 4, 5, 6, 7, 8, 9])

In [36]: np.where(a&gt;5)
Out[36]: (array([16, 17, 18, 19]),)

In [37]: np.where(a&gt;5)[0][0]
Out[37]: 16
</code></pre>

blocks|key|111773|text|对于已排序的数组内容，还有一个更快的方法：searchsorted。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|111774|import+time
N+=+10000
aa+=+np.arange(-N,N)
%25timeit+np.searchsorted(aa,+N/2)%2B1
%25timeit+np.argmax(aa>N/2)
%25timeit+np.where(aa>N/2)[0][0]
%25timeit+np.nonzero(aa>N/2)[0][0]

#+Output
100000+loops,+best+of+3:+5.97+µs+per+loop
10000+loops,+best+of+3:+46.3+µs+per+loop
10000+loops,+best+of+3:+154+µs+per+loop
10000+loops,+best+of+3:+154+µs+per+loop|code-block|syntax|javascript|111775|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.searchsorted.html^0|L|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]]]]

given the sorted content of your array, there is an even faster method: <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.searchsorted.html">searchsorted</a>.

<pre><code>import time
N = 10000
aa = np.arange(-N,N)
%timeit np.searchsorted(aa, N/2)+1
%timeit np.argmax(aa&gt;N/2)
%timeit np.where(aa&gt;N/2)[0][0]
%timeit np.nonzero(aa&gt;N/2)[0][0]

# Output
100000 loops, best of 3: 5.97 µs per loop
10000 loops, best of 3: 46.3 µs per loop
10000 loops, best of 3: 154 µs per loop
10000 loops, best of 3: 154 µs per loop
</code></pre>

blocks|key|111819|text|如果是我的话|type|unstyled|depth|inlineStyleRanges|entityRanges|data|111820|i+=+np.min(np.where(V+>=+x))|code-block|syntax|javascript|111821|其中V是向量(一维数组)，x是值，i是结果索引。|offset|length|style|CODE|111822|entityMap^0|0|0|2|1|D|1|H|1|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|N|$]]

I would go with

<pre><code>i = np.min(np.where(V &gt;= x))
</code></pre>

where <code>V</code> is vector (1d array), <code>x</code> is the value and <code>i</code> is the resulting index.

blocks|key|329570|text|我想向你求婚|type|unstyled|depth|inlineStyleRanges|entityRanges|data|329571|np.min(np.append(np.where(aa>5)[0],np.inf))|code-block|syntax|javascript|329572|这将返回满足条件的最小索引，如果从未满足条件，则返回无穷大(+where返回一个空数组)。|offset|length|style|CODE|329573|entityMap^0|0|0|V|5|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I'd like to propose 

<pre><code>np.min(np.append(np.where(aa&gt;5)[0],np.inf))
</code></pre>

This will return the smallest index where the condition is met, while returning infinity if the condition is never met (and <code>where</code> returns an empty array).

I have a 1D array in numpy and I want to find the position of the index where a value exceeds the value in numpy array.

E.g.

<pre><code>aa = range(-10,10)
</code></pre>

Find position in <code>aa</code> where, the value <code>5</code> gets exceeded.

Numpy first occurrence of value greater than existing value

我在numpy中有一个一维数组，我想找出一个值超过numpy数组中的值的索引的位置。例如。aa = range(-10,10)在超过值5的aa中查找位置。

问Numpy第一次出现大于现有值的值
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