使用numpy在矩阵中查找哪些行的所有元素都为零 [英] Finding which rows have all elements as zeros in a matrix with numpy

问题描述

我有一个大的numpy矩阵M.矩阵的某些行的所有元素均为零，我需要获取这些行的索引.我正在考虑的天真的方法是遍历矩阵中的每一行，然后检查每个元素.但是，我认为有一种更好，更快的方法可以使用numpy来完成此任务.希望您能提供帮助！

I have a large numpy matrix M. Some of the rows of the matrix have all of their elements as zero and I need to get the indices of those rows. The naive approach I'm considering is to loop through each row in the matrix and then check each elements. However I think there's a better and a faster approach to accomplish this using numpy. I hope you can help!

推荐答案

这是一种方法.我假设已经使用import numpy as np导入了numpy.

Here's one way. I assume numpy has been imported using import numpy as np.

In [20]: a
Out[20]: 
array([[0, 1, 0],
       [1, 0, 1],
       [0, 0, 0],
       [1, 1, 0],
       [0, 0, 0]])

In [21]: np.where(~a.any(axis=1))[0]
Out[21]: array([2, 4])

此答案略有不同:这是怎么回事:

如果数组中的任何值为"truthy"，则any方法将返回True.非零数字被认为是True，而0被认为是False.通过使用参数axis=1，该方法将应用于每一行.对于示例a，我们有:

The any method returns True if any value in the array is "truthy". Nonzero numbers are considered True, and 0 is considered False. By using the argument axis=1, the method is applied to each row. For the example a, we have:

In [32]: a.any(axis=1)
Out[32]: array([ True,  True, False,  True, False], dtype=bool)

因此，每个值指示相应的行是否包含非零值. ~运算符是二进制"not"或补码:

So each value indicates whether the corresponding row contains a nonzero value. The ~ operator is the binary "not" or complement:

In [33]: ~a.any(axis=1)
Out[33]: array([False, False,  True, False,  True], dtype=bool)

(给出相同结果的替代表达式是(a == 0).all(axis=1).)

(An alternative expression that gives the same result is (a == 0).all(axis=1).)

要获取行索引，我们使用where函数.它返回参数为True的索引:

To get the row indices, we use the where function. It returns the indices where its argument is True:

In [34]: np.where(~a.any(axis=1))
Out[34]: (array([2, 4]),)

请注意，where返回了一个包含单个数组的元组. where适用于n维数组，因此它总是返回一个元组.我们想要该元组中的单个数组.

Note that where returned a tuple containing a single array. where works for n-dimensional arrays, so it always returns a tuple. We want the single array in that tuple.

In [35]: np.where(~a.any(axis=1))[0]
Out[35]: array([2, 4])

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