blocks|key|903251|text|您可以创建一个接受可调用参数的函数，该参数将在列表理解的条件部分中使用。然后您可以使用lambda或其他函数对象来传递您的任意条件：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|903252|def+indices(a,+func):
++++return+[i+for+(i,+val)+in+enumerate(a)+if+func(val)]

a+=+[1,+2,+3,+1,+2,+3,+1,+2,+3]

inds+=+indices(a,+lambda+x:+x+>+2)

>>>+inds
[2,+5,+8]|code-block|syntax|javascript|903253|它更接近于您的Matlab示例，而不必加载所有numpy。|903254|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/tutorial/controlflow.html#lambda-forms^0|17|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

You can make a function that takes a callable parameter which will be used in the condition part of your list comprehension. Then you can use a <a href="http://docs.python.org/tutorial/controlflow.html#lambda-forms" rel="noreferrer">lambda</a> or other function object to pass your arbitrary condition:

<pre><code>def indices(a, func):
 return [i for (i, val) in enumerate(a) if func(val)]

a = [1, 2, 3, 1, 2, 3, 1, 2, 3]

inds = indices(a, lambda x: x &gt; 2)

&gt;&gt;&gt; inds
[2, 5, 8]
</code></pre>

It's a little closer to your Matlab example, without having to load up all of numpy.

blocks|key|903167|text|在numpy中有where：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|903168|>>+import+numpy+as+np
>>+x+=+np.random.randint(0,+20,+10)
>>+x
array([14,+13,++1,+15,++8,++0,+17,+11,+19,+13])
>>+np.where(x+>+10)
(array([0,+1,+3,+6,+7,+8,+9],+dtype=int64),)|code-block|syntax|javascript|903169|entityMap^0|8|5|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|$]]

in numpy you have <code>where</code> :

<pre><code>&gt;&gt; import numpy as np
&gt;&gt; x = np.random.randint(0, 20, 10)
&gt;&gt; x
array([14, 13, 1, 15, 8, 0, 17, 11, 19, 13])
&gt;&gt; np.where(x &gt; 10)
(array([0, 1, 3, 6, 7, 8, 9], dtype=int64),)
</code></pre>

blocks|key|903437|text|或者使用numpy的非零函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|903438|import+numpy+as+np
a++++=+np.array([1,2,3,4,5])
inds+=+np.nonzero(a>2)
a[inds]+
array([3,+4,+5])|code-block|syntax|javascript|903439|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|$]]

Or use numpy's nonzero function:

<pre><code>import numpy as np
a = np.array([1,2,3,4,5])
inds = np.nonzero(a&gt;2)
a[inds] 
array([3, 4, 5])
</code></pre>

blocks|key|686925|text|为什么不使用这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|686926|[i+for+i+in+range(len(a))+if+a[i]+>+2]|code-block|syntax|javascript|686927|或者，对于任意条件，为您的条件定义一个函数f并执行以下操作：|offset|length|style|CODE|686928|[i+for+i+in+range(len(a))+if+f(a[i])]|686929|entityMap^0|0|0|L|1|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]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|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|$]]

Why not just use this:

<pre><code>[i for i in range(len(a)) if a[i] &gt; 2]
</code></pre>

or for arbitrary conditions, define a function <code>f</code> for your condition and do:

<pre><code>[i for i in range(len(a)) if f(a[i])]
</code></pre>

blocks|key|903517|text|此应用程序更常用的numpy例程是numpy.where()；不过，我认为它的工作方式与numpy.nonzero()相同。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|903518|import+numpy
a++++=+numpy.array([1,2,3,4,5])
inds+=+numpy.where(a>2)|code-block|syntax|javascript|903519|要获取这些值，您可以存储索引并对其进行切片：|903520|a[inds]|903521|或者，您可以将数组作为可选参数进行传递：|903522|numpy.where(a>2,+a)|903523|或多个数组：|903524|b+=+numpy.array([11,22,33,44,55])
numpy.where(a>2,+a,+b)|903525|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.where.html|1|http://docs.scipy.org/doc/numpy/reference/generated/numpy.nonzero.html^0|9|5|H|D|18|F|H|D|0|18|F|1|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@$9|1D|A|1E|1|1F]|$9|1G|A|1H|1|1I]]|E|$]]|$1|F|3|G|5|H|7|1J|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1K|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1L|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1M|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1N|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1O|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1P|8|@]|D|@]|E|$I|J]]|$1|W|3|-4|5|6|7|1Q|8|@]|D|@]|E|$]]]|X|$Y|$5|Z|10|11|E|$12|13]]|14|$5|Z|10|11|E|$12|15]]]]

The <code>numpy</code> routine more commonly used for this application is <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.where.html" rel="nofollow"><code>numpy.where()</code></a>; though, I believe it works the same as <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.nonzero.html" rel="nofollow"><code>numpy.nonzero()</code></a>.

<pre><code>import numpy
a = numpy.array([1,2,3,4,5])
inds = numpy.where(a&gt;2)
</code></pre>

To get the values, you can either store the indices and slice withe them:

<pre><code>a[inds]
</code></pre>

or you can pass the array as an optional parameter:

<pre><code>numpy.where(a&gt;2, a)
</code></pre>

or multiple arrays:

<pre><code>b = numpy.array([11,22,33,44,55])
numpy.where(a&gt;2, a, b)
</code></pre>

blocks|key|686890|text|要获取具有任意条件的值，您可以使用带有lambda函数的filter()：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|686891|>>>+a+=+[1,2,3,1,2,3,1,2,3]
>>>+filter(lambda+x:+x+>+2,+a)
[3,+3,+3]|code-block|syntax|javascript|686892|获取索引的一种可能方法是使用enumerate()构建一个同时包含索引和值的元组，然后对其进行过滤：|686893|>>>+a+=+[1,2,3,1,2,3,1,2,3]
>>>+aind+=+tuple(enumerate(a))
>>>+print+aind
((0,+1),+(1,+2),+(2,+3),+(3,+1),+(4,+2),+(5,+3),+(6,+1),+(7,+2),+(8,+3))
>>>+filter(lambda+x:+x[1]+>+2,+aind)
((2,+3),+(5,+3),+(8,+3))|686894|entityMap^0|S|8|0|0|E|B|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

To get values with arbitrary conditions, you could use <code>filter()</code> with a lambda function:

<pre><code>&gt;&gt;&gt; a = [1,2,3,1,2,3,1,2,3]
&gt;&gt;&gt; filter(lambda x: x &gt; 2, a)
[3, 3, 3]
</code></pre>

One possible way to get the indices would be to use <code>enumerate()</code> to build a tuple with both indices and values, and then filter that:

<pre><code>&gt;&gt;&gt; a = [1,2,3,1,2,3,1,2,3]
&gt;&gt;&gt; aind = tuple(enumerate(a))
&gt;&gt;&gt; print aind
((0, 1), (1, 2), (2, 3), (3, 1), (4, 2), (5, 3), (6, 1), (7, 2), (8, 3))
&gt;&gt;&gt; filter(lambda x: x[1] &gt; 2, aind)
((2, 3), (5, 3), (8, 3))
</code></pre>

blocks|key|903571|text|我想我可能找到了一个快速而简单的替代品。顺便说一句，我觉得np.where()函数不是很令人满意，从某种意义上说，它包含了令人讨厌的零元素行。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|903572|import+matplotlib.mlab+as+mlab
a+=+np.random.randn(1,5)
print+a

>>+[[+1.36406736++1.45217257+-0.06896245++0.98429727+-0.59281957]]

idx+=+mlab.find(a<0)
print+idx
type(idx)

>>+[2+4]
>>+np.ndarray|code-block|syntax|javascript|903573|最好的，达|903574|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|$]]

I think I may have found one quick and simple substitute.
BTW I felt that the np.where() function not very satisfactory, in a sense that somehow it contains an annoying row of zero-element.

<pre><code>import matplotlib.mlab as mlab
a = np.random.randn(1,5)
print a

&gt;&gt; [[ 1.36406736 1.45217257 -0.06896245 0.98429727 -0.59281957]]

idx = mlab.find(a&lt;0)
print idx
type(idx)

&gt;&gt; [2 4]
&gt;&gt; np.ndarray
</code></pre>

Best,
Da

blocks|key|903478|text|Matlab的find代码有两个参数。John的代码说明了第一个参数，但不是第二个参数。例如，如果您想知道索引中满足条件的位置:Mtlab的函数为：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|903479|find(x>2,1)|code-block|syntax|javascript|903480|使用John的代码，您所要做的就是在index函数的末尾添加一个x，其中x是您要查找的索引号。|903481|def+indices(a,+func):
++++return+[i+for+(i,+val)+in+enumerate(a)+if+func(val)]

a+=+[1,+2,+3,+1,+2,+3,+1,+2,+3]

inds+=+indices(a,+lambda+x:+x+>+2)[0]+#[0]+being+the+2nd+matlab+argument|903482|它返回>>>+2，这是第一个超过2的索引。|903483|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|$]]

Matlab's find code has two arguments. John's code accounts for the first argument but not the second. For instance, if you want to know where in the index the condition is satisfied: Mtlab's function would be:

<pre><code>find(x&gt;2,1)
</code></pre>

Using John's code, all you have to do is add a [x] at the end of the indices function, where x is the index number you're looking for.

<pre><code>def indices(a, func):
 return [i for (i, val) in enumerate(a) if func(val)]

a = [1, 2, 3, 1, 2, 3, 1, 2, 3]

inds = indices(a, lambda x: x &gt; 2)[0] #[0] being the 2nd matlab argument
</code></pre>

which returns >>> 2, the first index to exceed 2.

In MATLAB it is easy to find the indices of values that meet a particular condition:

<pre><code>&gt;&gt; a = [1,2,3,1,2,3,1,2,3];
&gt;&gt; find(a &gt; 2) % find the indecies where this condition is true
[3, 6, 9] % (MATLAB uses 1-based indexing)
&gt;&gt; a(find(a &gt; 2)) % get the values at those locations
[3, 3, 3]
</code></pre>

What would be the best way to do this in Python?

So far, I have come up with the following. To just get the values:

<pre><code>&gt;&gt;&gt; a = [1,2,3,1,2,3,1,2,3]
&gt;&gt;&gt; [val for val in a if val &gt; 2]
[3, 3, 3]
</code></pre>

But if I want the index of each of those values it's a bit more complicated:

<pre><code>&gt;&gt;&gt; a = [1,2,3,1,2,3,1,2,3]
&gt;&gt;&gt; inds = [i for (i, val) in enumerate(a) if val &gt; 2]
&gt;&gt;&gt; inds
[2, 5, 8]
&gt;&gt;&gt; [val for (i, val) in enumerate(a) if i in inds]
[3, 3, 3]
</code></pre>

Is there a better way to do this in Python, especially for arbitrary conditions (not just 'val > 2')?

I found functions equivalent to MATLAB 'find' in NumPy but I currently do not have access to those libraries.

MATLAB-style find() function in Python

在MATLAB中，很容易找到满足特定条件的值的索引：>> a = [1,2,3,1,2,3,1,2,3];>> find(a > 2)     % find the indecies where this condition is true[3, 6, 9]          % (MATLAB uses 1-based indexing)>> a(find(a > 2))  % get the 

问Python中的MATLAB风格的find()函数
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