blocks|key|1224218|text|如果添加圆括号，则表达式有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1224219|>>>+y[(1+<+x)+&+(x+<+5)]
array(['o',+'o',+'a'],+
++++++dtype='%7CS1')|code-block|syntax|javascript|1224220|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|$]]

Your expression works if you add parentheses:

<pre><code>&gt;&gt;&gt; y[(1 &lt; x) &amp; (x &lt; 5)]
array(['o', 'o', 'a'], 
 dtype='|S1')
</code></pre>

blocks|key|1224300|text|IMO+OP实际上并不想要np.bitwise_and()+(aka+&)，但实际上想要np.logical_and()，因为它们比较的是逻辑值，例如True和False+-请参阅logical+vs.+bitwise上的这篇文章以了解其中的区别。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1224301|>>>+x+=+array([5,+2,+3,+1,+4,+5])
>>>+y+=+array(['f','o','o','b','a','r'])
>>>+output+=+y[np.logical_and(x+>+1,+x+<+5)]+#+desired+output+is+['o','o','a']
>>>+output
array(['o',+'o',+'a'],
++++++dtype='%7CS1')|code-block|syntax|javascript|1224302|实现这一点的等效方法是通过适当设置axis参数来使用np.all()。|1224303|>>>+output+=+y[np.all([x+>+1,+x+<+5],+axis=0)]+#+desired+output+is+['o','o','a']
>>>+output
array(['o',+'o',+'a'],
++++++dtype='%7CS1')|1224304|从数字上看：|1224305|>>>+%25timeit+(a+<+b)+&+(b+<+c)
The+slowest+run+took+32.97+times+longer+than+the+fastest.+This+could+mean+that+an+intermediate+result+is+being+cached.
100000+loops,+best+of+3:+1.15+µs+per+loop

>>>+%25timeit+np.logical_and(a+<+b,+b+<+c)
The+slowest+run+took+32.59+times+longer+than+the+fastest.+This+could+mean+that+an+intermediate+result+is+being+cached.
1000000+loops,+best+of+3:+1.17+µs+per+loop

>>>+%25timeit+np.all([a+<+b,+b+<+c],+0)
The+slowest+run+took+67.47+times+longer+than+the+fastest.+This+could+mean+that+an+intermediate+result+is+being+cached.
100000+loops,+best+of+3:+5.06+µs+per+loop|1224306|所以使用np.all()比较慢，但是&和logical_and差不多。|1224307|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.bitwise_and.html|1|http://docs.scipy.org/doc/numpy/reference/generated/numpy.logical_and.html|2|https://stackoverflow.com/questions/8418295/logical-vs-bitwise|3|http://docs.scipy.org/doc/numpy/reference/generated/numpy.all.html^0|D|G|Z|1|18|G|23|4|28|5|D|O|0|18|G|1|2I|J|2|0|0|H|4|Q|8|Q|8|3|0|0|0|0|4|8|I|1|K|B|0^^$0|@$1|2|3|4|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]]|D|@$9|1J|A|1K|1|1L]|$9|1M|A|1N|1|1O]|$9|1P|A|1Q|1|1R]]|E|$]]|$1|F|3|G|5|H|7|1S|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1T|8|@$9|1U|A|1V|B|C]|$9|1W|A|1X|B|C]]|D|@$9|1Y|A|1Z|1|20]]|E|$]]|$1|M|3|N|5|H|7|21|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|22|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|23|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|24|8|@$9|25|A|26|B|C]|$9|27|A|28|B|C]|$9|29|A|2A|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|2B|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]|12|$5|X|Y|Z|E|$10|13]]|14|$5|X|Y|Z|E|$10|15]]|16|$5|X|Y|Z|E|$10|17]]]]

IMO OP does not actually want <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.bitwise_and.html" rel="noreferrer"><code>np.bitwise_and()</code> (aka <code>&amp;</code>)</a> but actually wants <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.logical_and.html" rel="noreferrer"><code>np.logical_and()</code></a> because they are comparing logical values such as <code>True</code> and <code>False</code> - see this SO post on <a href="https://stackoverflow.com/questions/8418295/logical-vs-bitwise">logical vs. bitwise</a> to see the difference.

<pre><code>&gt;&gt;&gt; x = array([5, 2, 3, 1, 4, 5])
&gt;&gt;&gt; y = array(['f','o','o','b','a','r'])
&gt;&gt;&gt; output = y[np.logical_and(x &gt; 1, x &lt; 5)] # desired output is ['o','o','a']
&gt;&gt;&gt; output
array(['o', 'o', 'a'],
 dtype='|S1')
</code></pre>

And equivalent way to do this is with <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.all.html" rel="noreferrer"><code>np.all()</code></a> by setting the <code>axis</code> argument appropriately.

<pre><code>&gt;&gt;&gt; output = y[np.all([x &gt; 1, x &lt; 5], axis=0)] # desired output is ['o','o','a']
&gt;&gt;&gt; output
array(['o', 'o', 'a'],
 dtype='|S1')
</code></pre>

by the numbers:

<pre><code>&gt;&gt;&gt; %timeit (a &lt; b) &amp; (b &lt; c)
The slowest run took 32.97 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 1.15 µs per loop

&gt;&gt;&gt; %timeit np.logical_and(a &lt; b, b &lt; c)
The slowest run took 32.59 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 1.17 µs per loop

&gt;&gt;&gt; %timeit np.all([a &lt; b, b &lt; c], 0)
The slowest run took 67.47 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 5.06 µs per loop
</code></pre>

so using <code>np.all()</code> is slower, but <code>&amp;</code> and <code>logical_and</code> are about the same.

blocks|key|1009898|text|在@J.F.+Sebastian和@Mark+Mikofski的回答中添加一个细节：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1009899|如果想要获得相应的索引(而不是数组的实际值)，可以使用以下代码：|1009900|要满足多个(全部)条件：|1009901|select_indices+=+np.where(+np.logical_and(+x+>+1,+x+<+5)+)[0]+#+++1+<+x+<5|code-block|syntax|javascript|1009902|用于满足多个(或)条件：|1009903|select_indices+=+np.where(+np.logical_or(+x+<+1,+x+>+5+)+)[0]+#+x+<1+or+x+>5|1009904|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|T|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Add one detail to @J.F. Sebastian's and @Mark Mikofski's answers: 
If one wants to get the corresponding indices (rather than the actual values of array), the following code will do:

For satisfying multiple (all) conditions:

<pre><code>select_indices = np.where( np.logical_and( x &gt; 1, x &lt; 5) )[0] # 1 &lt; x &lt;5
</code></pre>

For satisfying multiple (or) conditions:

<pre><code>select_indices = np.where( np.logical_or( x &lt; 1, x &gt; 5 ) )[0] # x &lt;1 or x &gt;5
</code></pre>

blocks|key|1224395|text|实际上，我会这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1224396|L1是满足条件1的元素的索引列表；(也许您可以使用somelist.index(condition1)或np.where(condition1)来获取L1。)|offset|length|style|CODE|1224397|类似地，您将获得L2，这是一个满足条件2的元素列表；|1224398|然后使用intersect(L1,L2)查找交叉点。|1224399|如果你有多个条件需要满足，你也可以找到多个列表的交集。|1224400|然后，您可以在任何其他数组中应用索引，例如x。|1224401|entityMap^0|0|P|Q|1G|K|0|0|4|G|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@$D|T|E|U|F|G]|$D|V|E|W|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|Y|8|@$D|Z|E|10|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|11|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|12|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|Q|$]]

Actually I would do it this way:

L1 is the index list of elements satisfying condition 1;(maybe you can use <code>somelist.index(condition1)</code> or <code>np.where(condition1)</code> to get L1.)

Similarly, you get L2, a list of elements satisfying condition 2;

Then you find intersection using <code>intersect(L1,L2)</code>.

You can also find intersection of multiple lists if you get multiple conditions to satisfy.

Then you can apply index in any other array, for example, x.

blocks|key|1224459|text|我喜欢使用np.vectorize来完成这样的任务。请考虑以下几点：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1224460|>>>+#+Arrays
>>>+x+=+np.array([5,+2,+3,+1,+4,+5])
>>>+y+=+np.array(['f','o','o','b','a','r'])

>>>+#+Function+containing+the+constraints
>>>+func+=+np.vectorize(lambda+t:+t>1+and+t<5)

>>>+#+Call+function+on+x
>>>+y[func(x)]
>>>+array(['o',+'o',+'a'],+dtype='<U1')|code-block|syntax|javascript|1224461|优点是您可以在矢量化函数中添加更多类型的约束。|1224462|希望能有所帮助。|1224463|entityMap^0|5|C|0|0|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]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|V|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

I like to use <code>np.vectorize</code> for such tasks. Consider the following:

<pre><code>&gt;&gt;&gt; # Arrays
&gt;&gt;&gt; x = np.array([5, 2, 3, 1, 4, 5])
&gt;&gt;&gt; y = np.array(['f','o','o','b','a','r'])

&gt;&gt;&gt; # Function containing the constraints
&gt;&gt;&gt; func = np.vectorize(lambda t: t&gt;1 and t&lt;5)

&gt;&gt;&gt; # Call function on x
&gt;&gt;&gt; y[func(x)]
&gt;&gt;&gt; array(['o', 'o', 'a'], dtype='&lt;U1')
</code></pre>

The advantage is you can add many more types of constraints in the vectorized function.

Hope it helps.

blocks|key|1224559|text|对于二维数组，您可以这样做。使用条件创建二维蒙版。根据数组的不同，将条件掩码类型转换为int或float，并将其与原始数组相乘。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1224560|In+[8]:+arr
Out[8]:+
array([[+1.,++2.,++3.,++4.,++5.],
+++++++[+6.,++7.,++8.,++9.,+10.]])

In+[9]:+arr*(arr+%25+2+==+0).astype(np.int)+
Out[9]:+
array([[+0.,++2.,++0.,++4.,++0.],
+++++++[+6.,++0.,++8.,++0.,+10.]])|code-block|syntax|javascript|1224561|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|$]]

For 2D arrays, you can do this. Create a 2D mask using the condition. Typecast the condition mask to int or float, depending on the array, and multiply it with the original array.

<pre><code>In [8]: arr
Out[8]: 
array([[ 1., 2., 3., 4., 5.],
 [ 6., 7., 8., 9., 10.]])

In [9]: arr*(arr % 2 == 0).astype(np.int) 
Out[9]: 
array([[ 0., 2., 0., 4., 0.],
 [ 6., 0., 8., 0., 10.]])
</code></pre>

Suppose I have a numpy array <code>x = [5, 2, 3, 1, 4, 5]</code>, <code>y = ['f', 'o', 'o', 'b', 'a', 'r']</code>. I want to select the elements in <code>y</code> corresponding to elements in <code>x</code> that are greater than 1 and less than 5.

I tried

<pre><code>x = array([5, 2, 3, 1, 4, 5])
y = array(['f','o','o','b','a','r'])
output = y[x &gt; 1 &amp; x &lt; 5] # desired output is ['o','o','a']
</code></pre>

but this doesn't work. How would I do this?

How do I select elements of an array given condition?

假设我有一个数值数组x = [5, 2, 3, 1, 4, 5]，y = ['f', 'o', 'o', 'b', 'a', 'r']。我想选择与x中大于1和小于5的元素相对应的y中的元素。我试过了x = array([5, 2, 3, 1, 4, 5])y = array(['f','o','o','b','a','...

问如何在给定条件下选择数组中的元素？
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问如何在给定条件下选择数组中的元素？
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