I need to figure out how I can find all the index of a value in a 2d numpy array.
我需要弄清楚如何在2d numpy数组中找到值的所有索引。
For example, I have the following 2d array:
例如，我有以下2d数组：
([[1 1 0 0],
[0 0 1 1],
[0 0 0 0]])
I need to find the index of all the 1's and 0's.
我需要找到所有1和0的索引。
1: [(0, 0), (0, 1), (1, 2), (1, 3)]
0: [(0, 2), (0, 3), (1, 0), (1, 1), (the entire all row)]
I tried this but it doesn't give me all the indexes:
我试过这个，但它没有给我所有的索引：
t = [(index, row.index(1)) for index, row in enumerate(x) if 1 in row]
Basically, it gives me only one of the index in each row [(0, 0), (1, 2)].
基本上，它只给出了每行[（0,0），（1,2）]中的一个索引。
2 个解决方案
#1
18
You can use np.where to return a tuple of arrays of x and y indices where a given condition holds in an array.
您可以使用np.where返回一个x和y索引数组的元组，其中给定条件保存在数组中。
If a is the name of your array:
如果a是数组的名称：
>>> np.where(a == 1)
(array([0, 0, 1, 1]), array([0, 1, 2, 3]))
If you want a list of (x, y) pairs, you could zip the two arrays:
如果你想要一个（x，y）对的列表，你可以压缩两个数组：
>>> zip(*np.where(a == 1))
[(0, 0), (0, 1), (1, 2), (1, 3)]
Or, even better, @jme points out that np.asarray(x).T can be a more efficient way to generate the pairs.
或者，甚至更好，@ jme指出np.asarray（x）.T可以是生成对的更有效方式。
#2
6
The problem with the list comprehension you provided is that it only goes one level deep, you need a nested list comprehension:
你提供的列表理解的问题是它只有一个深度，你需要一个嵌套的列表理解：
a = [[1,0,1],[0,0,1], [1,1,0]]
>>> [(ix,iy) for ix, row in enumerate(a) for iy, i in enumerate(row) if i == 0]
[(0, 1), (1, 0), (1, 1), (2, 2)]
That being said, if you are working with a numpy array, it's better to use the built in functions as suggested by ajcr.
话虽这么说，如果你正在使用numpy数组，最好使用ajcr建议的内置函数。
#1
18
You can use np.where to return a tuple of arrays of x and y indices where a given condition holds in an array.
您可以使用np.where返回一个x和y索引数组的元组，其中给定条件保存在数组中。
If a is the name of your array:
如果a是数组的名称：
>>> np.where(a == 1)
(array([0, 0, 1, 1]), array([0, 1, 2, 3]))
If you want a list of (x, y) pairs, you could zip the two arrays:
如果你想要一个（x，y）对的列表，你可以压缩两个数组：
>>> zip(*np.where(a == 1))
[(0, 0), (0, 1), (1, 2), (1, 3)]
Or, even better, @jme points out that np.asarray(x).T can be a more efficient way to generate the pairs.
或者，甚至更好，@ jme指出np.asarray（x）.T可以是生成对的更有效方式。
#2
6
The problem with the list comprehension you provided is that it only goes one level deep, you need a nested list comprehension:
你提供的列表理解的问题是它只有一个深度，你需要一个嵌套的列表理解：
a = [[1,0,1],[0,0,1], [1,1,0]]
>>> [(ix,iy) for ix, row in enumerate(a) for iy, i in enumerate(row) if i == 0]
[(0, 1), (1, 0), (1, 1), (2, 2)]
That being said, if you are working with a numpy array, it's better to use the built in functions as suggested by ajcr.
话虽这么说，如果你正在使用numpy数组，最好使用ajcr建议的内置函数。