问题：在NumPy数组中查找等于零的元素的索引
NumPy具有有效的功能/方法nonzero()来标识ndarray对象中非零元素的索引。什么是最有效的方式来获得该元素的索引做具有零值？
NumPy has the efficient function/method nonzero() to identify the indices of non-zero elements in an ndarray object. What is the most efficient way to obtain the indices of the elements that do have a value of zero?
回答 0
numpy.where（）是我的最爱。
>>> x = numpy.array([1,0,2,0,3,0,4,5,6,7,8])
>>> numpy.where(x == 0)[0]
array([1, 3, 5])
numpy.where() is my favorite.
>>> x = numpy.array([1,0,2,0,3,0,4,5,6,7,8])
>>> numpy.where(x == 0)[0]
array([1, 3, 5])
回答 1
有np.argwhere，
import numpy as np
arr = np.array([[1,2,3], [0, 1, 0], [7, 0, 2]])
np.argwhere(arr == 0)
返回所有找到的索引作为行：
array([[1, 0], # Indices of the first zero
[1, 2], # Indices of the second zero
[2, 1]], # Indices of the third zero
dtype=int64)
There is np.argwhere,
import numpy as np
arr = np.array([[1,2,3], [0, 1, 0], [7, 0, 2]])
np.argwhere(arr == 0)
which returns all found indices as rows:
array([[1, 0], # Indices of the first zero
[1, 2], # Indices of the second zero
[2, 1]], # Indices of the third zero
dtype=int64)
回答 2
您可以使用以下方法搜索任何标量条件：
>>> a = np.asarray([0,1,2,3,4])
>>> a == 0 # or whatver
array([ True, False, False, False, False], dtype=bool)
这将把数组作为条件的布尔掩码返回。
You can search for any scalar condition with:
>>> a = np.asarray([0,1,2,3,4])
>>> a == 0 # or whatver
array([ True, False, False, False, False], dtype=bool)
Which will give back the array as an boolean mask of the condition.
回答 3
您也可以nonzero()在条件的布尔掩码中使用它，因为False它也是零。
>>> x = numpy.array([1,0,2,0,3,0,4,5,6,7,8])
>>> x==0
array([False, True, False, True, False, True, False, False, False, False, False], dtype=bool)
>>> numpy.nonzero(x==0)[0]
array([1, 3, 5])
它的mtrw方法与s的方法完全相同，但是与问题更相关；）
You can also use nonzero() by using it on a boolean mask of the condition, because False is also a kind of zero.
>>> x = numpy.array([1,0,2,0,3,0,4,5,6,7,8])
>>> x==0
array([False, True, False, True, False, True, False, False, False, False, False], dtype=bool)
>>> numpy.nonzero(x==0)[0]
array([1, 3, 5])
It’s doing exactly the same as mtrw‘s way, but it is more related to the question ;)
回答 4
您可以使用numpy.nonzero查找零。
>>> import numpy as np
>>> x = np.array([1,0,2,0,3,0,0,4,0,5,0,6]).reshape(4, 3)
>>> np.nonzero(x==0) # this is what you want
(array([0, 1, 1, 2, 2, 3]), array([1, 0, 2, 0, 2, 1]))
>>> np.nonzero(x)
(array([0, 0, 1, 2, 3, 3]), array([0, 2, 1, 1, 0, 2]))
You can use numpy.nonzero to find zero.
>>> import numpy as np
>>> x = np.array([1,0,2,0,3,0,0,4,0,5,0,6]).reshape(4, 3)
>>> np.nonzero(x==0) # this is what you want
(array([0, 1, 1, 2, 2, 3]), array([1, 0, 2, 0, 2, 1]))
>>> np.nonzero(x)
(array([0, 0, 1, 2, 3, 3]), array([0, 2, 1, 1, 0, 2]))
回答 5
如果使用一维数组，则有一个语法糖：
>>> x = numpy.array([1,0,2,0,3,0,4,5,6,7,8])
>>> numpy.flatnonzero(x == 0)
array([1, 3, 5])
If you are working with a one-dimensional array there is a syntactic sugar:
>>> x = numpy.array([1,0,2,0,3,0,4,5,6,7,8])
>>> numpy.flatnonzero(x == 0)
array([1, 3, 5])
回答 6
import numpy as np
x = np.array([1,0,2,3,6])
non_zero_arr = np.extract(x>0,x)
min_index = np.amin(non_zero_arr)
min_value = np.argmin(non_zero_arr)
import numpy as np
x = np.array([1,0,2,3,6])
non_zero_arr = np.extract(x>0,x)
min_index = np.amin(non_zero_arr)
min_value = np.argmin(non_zero_arr)
回答 7
我将通过以下方式进行操作：
>>> x = np.array([[1,0,0], [0,2,0], [1,1,0]])
>>> x
array([[1, 0, 0],
[0, 2, 0],
[1, 1, 0]])
>>> np.nonzero(x)
(array([0, 1, 2, 2]), array([0, 1, 0, 1]))
# if you want it in coordinates
>>> x[np.nonzero(x)]
array([1, 2, 1, 1])
>>> np.transpose(np.nonzero(x))
array([[0, 0],
[1, 1],
[2, 0],
[2, 1])
I would do it the following way:
>>> x = np.array([[1,0,0], [0,2,0], [1,1,0]])
>>> x
array([[1, 0, 0],
[0, 2, 0],
[1, 1, 0]])
>>> np.nonzero(x)
(array([0, 1, 2, 2]), array([0, 1, 0, 1]))
# if you want it in coordinates
>>> x[np.nonzero(x)]
array([1, 2, 1, 1])
>>> np.transpose(np.nonzero(x))
array([[0, 0],
[1, 1],
[2, 0],
[2, 1])
