问题描述
类似于这个问题我想从二维numpy数组中删除一些NAN.但是,不是要删除具有NAN的整个行,而是要从数组的每一行中删除相应的元素.例如(为简单起见,使用列表格式)
Similar to this question I would like to remove some NAN's from a 2-D numpy array. However, instead of removing an entire row that has NAN's I want to remove the corresponding element from each row of the array. For example (using list format for simplicity)
x=[ [1,2,3,4],
[2,4,nan,8],
[3,6,9,0] ]
将成为
x=[ [1,2,4],
[2,4,8],
[3,6,0] ]
我可以想象使用numpy.where
找出NAN在每行中出现的位置,然后使用一些循环和逻辑语句从旧数组中创建一个新数组(跳过NAN和另一个数组中的对应元素)行),但对我而言,这似乎并不是一种非常简化的处理方式.还有其他想法吗?
I can imagine using a numpy.where
to figure out where in each row the NAN's appear and then use some loops and logic statements to make a new array from the old array (skipping over the NAN's and the corresponding elements in the other rows) but that to me doesn't seem to be a very streamlined way to do things. Any other ideas?
推荐答案
您可以使用布尔索引来选择仅不包含nan
的那些列:
You could use boolean indexing to select only those columns which do not contain nan
:
>>> x[:, ~np.isnan(x).any(axis=0)]
array([[ 1., 2., 4.],
[ 2., 4., 8.],
[ 3., 6., 0.]])
(这几乎与您链接的答案相同;只有轴已切换.)
(This is nearly identical to the answer you've linked to; only the axes have been switched.)
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