blocks|key|298767|text|filter(lambda+v:+v==v,+x)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|298768|同时适用于列表和数值数组，因为v!=v仅适用于NaN|unstyled|298769|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>filter(lambda v: v==v, x)
</code></pre>

works both for lists and numpy array
since v!=v only for NaN

blocks|key|515328|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|515329|import+math
print+[value+for+value+in+x+if+not+math.isnan(value)]|code-block|syntax|javascript|515330|有关更多信息，请在List+Comprehensions上阅读。|offset|length|515331|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/tutorial/datastructures.html#list-comprehensions^0|0|0|9|J|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Try this:

<pre><code>import math
print [value for value in x if not math.isnan(value)]
</code></pre>

For more, read on <a href="http://docs.python.org/tutorial/datastructures.html#list-comprehensions">List Comprehensions</a>.

blocks|key|515501|text|对我来说，@jmetz的答案不起作用，但是使用pandas+isnull()可以。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|515502|x+=+x[~pd.isnull(x)]|code-block|syntax|javascript|515503|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 me the answer by @jmetz didn't work, however using pandas isnull() did.

<pre><code>x = x[~pd.isnull(x)]
</code></pre>

blocks|key|298998|text|正如其他人所展示的那样|type|unstyled|depth|inlineStyleRanges|entityRanges|data|298999|x[~numpy.isnan(x)]|code-block|syntax|javascript|299000|很管用。但是，如果numpy数据类型不是本机数据类型，例如，如果它是object，则会抛出错误。在这种情况下，你可以使用熊猫。|299001|x[~pandas.isna(x)]+or+x[~pandas.isnull(x)]|299002|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|$]]

As shown by others 

<pre><code>x[~numpy.isnan(x)]
</code></pre>

works. But it will throw an error if the numpy dtype is not a native data type, for example if it is object. In that case you can use pandas.

<pre><code>x[~pandas.isna(x)] or x[~pandas.isnull(x)]
</code></pre>

blocks|key|515712|text|@jmetz's+answer可能是大多数人需要的；但是它会产生一维数组，例如，使其无法删除矩阵中的整行或整列。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|515713|为此，应将逻辑数组缩减为一维，然后对目标数组进行索引。例如，以下代码将删除至少具有一个NaN值的行：|515714|x+=+x[~numpy.isnan(x).any(axis=1)]|code-block|syntax|javascript|515715|请参阅更多详细信息here。|515716|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/11620982/3645374|1|https://www.w3resource.com/python-exercises/numpy/python-numpy-exercise-91.php^0|0|F|0|0|0|0|9|4|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|6|7|10|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|11|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|M|3|-4|5|6|7|16|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

<a href="https://stackoverflow.com/a/11620982/3645374">@jmetz's answer</a> is probably the one most people need; however it yields a one-dimensional array, e.g. making it unusable to remove entire rows or columns in matrices.

To do so, one should reduce the logical array to one dimension, then index the target array. For instance, the following will remove rows which have at least one NaN value:

<pre class="lang-py prettyprint-override"><code>x = x[~numpy.isnan(x).any(axis=1)]
</code></pre>

See more detail <a href="https://www.w3resource.com/python-exercises/numpy/python-numpy-exercise-91.php" rel="noreferrer">here</a>.

blocks|key|298840|text|执行上述操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|298841|x+=+x[~numpy.isnan(x)]|code-block|syntax|javascript|298842|或|298843|x+=+x[numpy.logical_not(numpy.isnan(x))]|298844|我发现，重置为相同的变量(x)并不会删除实际的NaN值，而必须使用不同的变量。将其设置为不同的变量删除了nans。例如：|298845|y+=+x[~numpy.isnan(x)]|298846|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|$]]

Doing the above :

<pre><code>x = x[~numpy.isnan(x)]
</code></pre>

or

<pre><code>x = x[numpy.logical_not(numpy.isnan(x))]
</code></pre>

I found that resetting to the same variable (x) did not remove the actual nan values and had to use a different variable. Setting it to a different variable removed the nans.
e.g. 

<pre><code>y = x[~numpy.isnan(x)]
</code></pre>

blocks|key|299022|text|如果您使用的是numpy|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|299023|#+first+get+the+indices+where+the+values+are+finite
ii+=+np.isfinite(x)

#+second+get+the+values
x+=+x[ii]|code-block|syntax|javascript|299024|entityMap^0|7|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|$]]

If you're using <code>numpy</code> 

<pre><code># first get the indices where the values are finite
ii = np.isfinite(x)

# second get the values
x = x[ii]
</code></pre>

blocks|key|515768|text|如果它有帮助，对于简单的一维阵列：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|515769|x+=+np.array([np.nan,+1,+2,+3,+4])

x[~np.isnan(x)]
>>>+array([1.,+2.,+3.,+4.])|code-block|syntax|javascript|515770|但是，如果您希望扩展到矩阵并保留其形状：|515771|x+=+np.array([
++++[np.nan,+np.nan],
++++[np.nan,+0],
++++[1,+2],
++++[3,+4]
])

x[~np.isnan(x).any(axis=1)]
>>>+array([[1.,+2.],
+++++++++++[3.,+4.]])|515772|我在处理pandas+.shift()功能时遇到了这个问题，我希望避免不惜一切代价使用.apply(...,+axis=1)，因为它的效率很低。|offset|length|style|CODE|515773|entityMap^0|0|0|0|0|B|8|17|J|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

In case it helps, for simple 1d arrays:
<pre class="lang-py prettyprint-override"><code>x = np.array([np.nan, 1, 2, 3, 4])

x[~np.isnan(x)]
&gt;&gt;&gt; array([1., 2., 3., 4.])
</code></pre>
but if you wish to expand to matrices and preserve the shape:
<pre class="lang-py prettyprint-override"><code>x = np.array([
 [np.nan, np.nan],
 [np.nan, 0],
 [1, 2],
 [3, 4]
])

x[~np.isnan(x).any(axis=1)]
&gt;&gt;&gt; array([[1., 2.],
 [3., 4.]])
</code></pre>
I encountered this issue when dealing with pandas <code>.shift()</code> functionality, and I wanted to avoid using <code>.apply(..., axis=1)</code> at all cost due to its inefficiency.

blocks|key|515748|text|只需简单地填充|type|unstyled|depth|inlineStyleRanges|entityRanges|data|515749|+x+=+numpy.array([
+[0.99929941,+0.84724713,+-0.1500044],
+[-0.79709026,+numpy.NaN,+-0.4406645],
+[-0.3599013,+-0.63565744,+-0.70251352]])

x[numpy.isnan(x)]+=+.555

print(x)

#+[[+0.99929941++0.84724713+-0.1500044+]
#++[-0.79709026++0.555++++++-0.4406645+]
#++[-0.3599013++-0.63565744+-0.70251352]]|code-block|syntax|javascript|515750|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|$]]

Simply fill with
<pre><code> x = numpy.array([
 [0.99929941, 0.84724713, -0.1500044],
 [-0.79709026, numpy.NaN, -0.4406645],
 [-0.3599013, -0.63565744, -0.70251352]])

x[numpy.isnan(x)] = .555

print(x)

# [[ 0.99929941 0.84724713 -0.1500044 ]
# [-0.79709026 0.555 -0.4406645 ]
# [-0.3599013 -0.63565744 -0.70251352]]
</code></pre>

blocks|key|298960|text|最简单的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|298961|numpy.nan_to_num(x)|code-block|syntax|javascript|298962|文档：https://docs.scipy.org/doc/numpy/reference/generated/numpy.nan_to_num.html|offset|length|298963|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/numpy/reference/generated/numpy.nan_to_num.html^0|0|0|3|22|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

A simplest way is: 

<pre><code>numpy.nan_to_num(x)
</code></pre>

Documentation: <a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.nan_to_num.html" rel="nofollow noreferrer">https://docs.scipy.org/doc/numpy/reference/generated/numpy.nan_to_num.html</a>

I want to figure out how to remove nan values from my array. My array looks something like this: 

<pre><code>x = [1400, 1500, 1600, nan, nan, nan ,1700] #Not in this exact configuration
</code></pre>

How can I remove the <code>nan</code> values from <code>x</code>?

Removing nan values from an array

我想弄清楚如何从我的数组中删除NaN值。我的数组看起来像这样：x = [1400, 1500, 1600, nan, nan, nan ,1700] #Not in this exact configuration如何从x中删除nan值

问从数组中删除NaN值
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