Numpy返回非预期值

I'm trying to get the index of elements in a row that is filled.

我正在尝试获取已填充的行中的元素索引。

I tried to use numpy where() function, but it's only returning the index of non-zero elements.

我试图使用numpy where()函数,但它只返回非零元素的索引。

import numpy as np
board = np.array([[0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [0, 2, 0, 0],
                  [0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [2, 2, 2, 2],
                  [0, 0, 0, 0]])
for rows in board:
    if set(rows) == {2}:
        if len(set(rows)) <= 1:
            print(np.where(board == rows))

My desired output is the following:

我想要的输出如下:

(array([9, 9, 9, 9], dtype=int32), array([0, 1, 2, 3], dtype=int32))

that is, row, col.

就是,行,col。

However, I'm getting this instead:

但是,我得到了这个:

(array([2, 9, 9, 9, 9], dtype=int32), array([1, 0, 1, 2, 3], dtype=int32))

As stated above, it should only get the index of elements in a filled row. The board1 is not filled with 2's, yet it's being included.

如上所述,它应该只获得填充行中元素的索引。 board1没有填充2,但它被包含在内。

3 个解决方案

#1

You don't need loops for this at all. For finding the rows that are filled with something other than 0, use .all(axis=1). The axis argument tells which axis to look for matching values across:

你根本不需要循环。要查找用0以外的其他内容填充的行,请使用.all(axis = 1)。 axis参数告诉哪个轴在以下位置查找匹配值:

>>> (board != 0).all(axis=1)
array([False, False, False, False, False, False, False, False, False,
    True, False], dtype=bool)

The booleans are True wherever there is a row filled with non-zero values, and False otherwise. If you take the outer product with a row array full of True values, you'll get True exactly in your matrix locations corresponding to the full row:

如果有一行填充非零值,则布尔值为True,否则为False。如果您使用一个充满True值的行数组来获取外部产品,那么您将在与整行对应的矩阵位置中获得True:

>>> np.outer((board!=0).all(axis=1),np.ones(board.shape[1], dtype=bool))
array([[False, False, False, False],
       [False, False, False, False],
       [False, False, False, False],
       [False, False, False, False],
       [False, False, False, False],
       [False, False, False, False],
       [False, False, False, False],
       [False, False, False, False],
       [False, False, False, False],
       [ True,  True,  True,  True],
       [False, False, False, False]], dtype=bool)

Then you can get the indices of those True values with np.where():

然后,您可以使用np.where()获取这些True值的索引:

>>> np.where(np.outer((board!=0).all(axis=1),np.ones(board.shape[1], dtype=bool)))
(array([9, 9, 9, 9]), array([0, 1, 2, 3]))

Note that np.where() returns np.int64 values. If you want np.int32 then just assign variables and convert them:

请注意,np.where()返回np.int64值。如果你想要np.int32,那么只需分配变量并转换它们:

>>> yind, xind = np.where(np.outer((board!=0).all(axis=1),np.ones(board.shape[1], dtype=bool)))
>>> yind = np.int32(yind)
>>> xind = np.int32(xind)
>>> yind, xind
(array([9, 9, 9, 9], dtype=int32), array([0, 1, 2, 3], dtype=int32))

To instead put all of these indices as tuples, use zip(*) (see here for an explanation if you're not familiar with * unpacking):

要将所有这些索引作为元组,请使用zip(*)(如果您不熟悉*解包,请参阅此处获取解释):

>>> [(y,x) for y,x in zip(*np.where(np.outer((board!=0).all(1), np.ones(board.shape[1]))))]
[(9, 0), (9, 1), (9, 2), (9, 3)]

And again if you need np.int32 just specify that in the comprehension:

如果你需要np.int32,请再次在理解中指定:

>>> [(y.astype(np.int32),x.astype(np.int32)) for y,x in zip(*np.where(np.outer((board == 2).all(1), np.ones(board.shape[1]))))]
[(9, 0), (9, 1), (9, 2), (9, 3)]

#2

If i understood well you are trying to find the row where your numpy array is completely filled ( no 0 values )

如果我理解你正在尝试找到你的numpy数组完全填充的行(没有0值)

so i ask of you to consider this:

所以我请你考虑一下:

import numpy as np
board = np.array([[0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [0, 2, 0, 0],
                  [1, 1, 1, 1],   # added this one just to generalize in 
                  [0, 0, 0, 0],   # case you have numbers != 2
                  [0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [2, 2, 2, 2],
                  [0, 0, 0, 0]])
for i,rows in enumerate(board):
    if not 0 in set(rows): # changed this to work on all non zero numbers
        print(rows)
        if len(set(rows)) <= 1:
            print(i)

output:

[1 1 1 1] # first row filled
3         # its index as int ( if that's what you're looking for )
[2 2 2 2] # seconds row filled
9         # its index as int

all you needed is enumerate built-in function. Happy Coding

你需要的只是枚举内置函数。快乐的编码

#3

If I understood your problem correctly, you want to find the row index of a specific element in the array. So, you need the following print() statement instead of print(np.where(board == rows)).

如果我正确理解了您的问题,您希望找到数组中特定元素的行索引。因此,您需要以下print()语句而不是print(np.where(board == rows))。

print(np.where((board == rows).all(axis=1)))

It outputs:

(array([9], dtype=int64),)

To get the specific row number, you can do the following.

要获取特定的行号,您可以执行以下操作。

loc = np.where((board == rows).all(axis=1))
print(loc[0][0]) # prints 9

Complete program:

import numpy as np
board = np.array([[0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [0, 2, 0, 0],
                  [0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [0, 0, 0, 0],
                  [2, 2, 2, 2],
                  [0, 0, 0, 0]])
for rows in board:
    if set(rows) == {2}:
        print(rows)
        if len(set(rows)) <= 1:
            print(np.where((board == rows).all(axis=1)))

#1