使用np.where在2D数组中查找匹配的行 [英] Using np.where to find matching row in 2D array
问题描述
我想知道如何将 np.where
与2D数组一起使用
I would like to know how I use np.where
with 2D array
我有以下数组:
arr1 = np.array([[ 3., 0.],
[ 3., 1.],
[ 3., 2.],
[ 3., 3.],
[ 3., 6.],
[ 3., 5.]])
我想找到这个数组:
arr2 = np.array([3.,0.])
但是当我使用 np.where()
时:
np.where(arr1 == arr2)
它返回:
(array([0, 0, 1, 2, 3, 4, 5]), array([0, 1, 0, 0, 0, 0, 0]))
我不明白这是什么意思.有人可以帮我解释一下吗?
I can't understand what it means. Can someone explain this for me?
推荐答案
您可能想要所有与 arr2
相等的行:
You probably wanted all rows that are equal to your arr2
:
>>> np.where(np.all(arr1 == arr2, axis=1))
(array([0], dtype=int64),)
这意味着第一行(零索引)匹配.
Which means that the first row (zeroth index) matched.
The problem with your approach is that numpy broadcasts the arrays (visualized with np.broadcast_arrays
):
>>> arr1_tmp, arr2_tmp = np.broadcast_arrays(arr1, arr2)
>>> arr2_tmp
array([[ 3., 0.],
[ 3., 0.],
[ 3., 0.],
[ 3., 0.],
[ 3., 0.],
[ 3., 0.]])
然后进行逐元素比较:
>>> arr1 == arr2
array([[ True, True],
[ True, False],
[ True, False],
[ True, False],
[ True, False],
[ True, False]], dtype=bool)
然后
和 np.where
会为您提供每个 True
的坐标:
and np.where
then gives you the coordinates of every True
:
>>> np.where(arr1 == arr2)
(array([0, 0, 1, 2, 3, 4, 5], dtype=int64),
array([0, 1, 0, 0, 0, 0, 0], dtype=int64))
# ^---- first match (0, 0)
# ^--- second match (0, 1)
# ^--- third match (1, 0)
# ...
这意味着(0,0)
(第一行的左边项目)是第一个 True
,然后是 0,1
(第一行的右边)项目,然后 1、0
(第二行,左侧项目),....
Which means (0, 0)
(first row left item) is the first True
, then 0, 1
(first row right item), then 1, 0
(second row, left item), ....
如果沿第一个轴使用 np.all
,则会得到所有完全相等的行:
If you use np.all
along the first axis you get all rows that are completly equal:
>>> np.all(arr1 == arr2, axis=1)
array([ True, False, False, False, False, False], dtype=bool)
如果保持尺寸不变,则可以更好地可视化:
Can be better visualized if one keeps the dimensions:
>>> np.all(arr1 == arr2, axis=1, keepdims=True)
array([[ True],
[False],
[False],
[False],
[False],
[False]], dtype=bool)
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