matrix

问题 We have an n x m matrix whose rows are sorted, we need to print the numbers in the matrix in ascending order. The columns are not necessarly sorted. The solution that came to my mind was simple merging the rows in the matrix considering them as separate lists(basically the merging step in merge sort) which in merge sort takes O(n). I wanted to know what is the complexity of merging n separate arrays as in this case. I thought it would be O(n x m) but I am not sure. Also, what would be a

问题 We have an n x m matrix whose rows are sorted, we need to print the numbers in the matrix in ascending order. The columns are not necessarly sorted. The solution that came to my mind was simple merging the rows in the matrix considering them as separate lists(basically the merging step in merge sort) which in merge sort takes O(n). I wanted to know what is the complexity of merging n separate arrays as in this case. I thought it would be O(n x m) but I am not sure. Also, what would be a

问题 According to the answers from this question and also according to numpy, matrix multiplication of 2-D arrays is best done via a @ b , or numpy.matmul(a,b) as compared to a.dot(b) . If both a and b are 2-D arrays, it is matrix multiplication, but using matmul or a @ b is preferred. I did the following benchmark and found contrary results. Questions: Is there an issue with my benchmark? If not, why does Numpy not recommend a.dot(b) when it is faster than a@b or numpy.matmul(a,b) ? Benchmark

问题 According to the answers from this question and also according to numpy, matrix multiplication of 2-D arrays is best done via a @ b , or numpy.matmul(a,b) as compared to a.dot(b) . If both a and b are 2-D arrays, it is matrix multiplication, but using matmul or a @ b is preferred. I did the following benchmark and found contrary results. Questions: Is there an issue with my benchmark? If not, why does Numpy not recommend a.dot(b) when it is faster than a@b or numpy.matmul(a,b) ? Benchmark

问题 So following is an interview problem. Given two N 2 matrices with entries being 0 or 1 . How can we find out the number of maximum overlapping 1 's possible? Note: You can only move the matrix upward, downward, leftward and rightward, so rotation is not allowed Currently I'm stuck at the most naive O(N^4) method, which being align the top left corner of one matrix to every possible position of the other matrix and count the all the overlap 1s. Example : [0 1 0] [0 0 1] A: [1 0 0] B: [0 0 1]

问题 For tutorial purposes, I'd like to be able to print or display matrices and vectors side-by-side, often to illustrate the result of a matrix equation, like $A x = b$. I could do this using SAS/IML, where the print statement takes an arbitrary collection of (space separated) expressions, evaluates them and prints the result, e.g., print A ' * ' x '=' (A * x) '=' b; A X #TEM1001 B 1 1 -4 * 0.733 = 2 = 2 1 -2 1 -0.33 1 1 1 1 1 -0.4 0 0 Note that quoted strings are printed as is. I've searched,

问题 I'm trying to convert a record into a date and time format using the strptime function. However, I'm not sure why I'm getting the error: number of items to replace is not a multiple of replacement length. I tried to check the length of the record using the length function but both have the same length. data <- DT head(data[6]) # column # 1 2014-12-22 23:53:48 # 2 2014-12-22 23:20:34 # 3 2014-12-22 23:20:30 # 4 2014-12-22 23:20:16 # 5 2014-12-22 23:20:07 # 6 2014-12-22 23:05:49 data[,6] <- as

问题 So, I was constructing a matrix (square) of length N, its element i should between 1 <= i < 2 * N Each element should be between 1 to 2 * N - 1 for each i and j (1 <= i <= N and 1 <= j < 2 * N ) the j element should appear in i th row / i column for any N if the desired matrix is not possible we have print -1; Which means for N = 2 We could get matrix as - 1 2 3 1 As we can see for 1st row and column we can get 1 , 2 , 3. and for 2nd row and column we get 1, 2, 3 Another example : for N = 4 1

问题 I want to compute a "distance" matrix, similarly to scipy.spatial.distance.cdist, but using the intersection over union (IoU) between "bounding boxes" (4-dimensional vectors), instead of a typical distance metric (like the Euclidean distance). For example, let's suppose that we have two collections of bounding boxes, such as import numpy as np A_bboxes = np.array([[0, 0, 10, 10], [5, 5, 15, 15]]) array([[ 0, 0, 10, 10], [ 5, 5, 15, 15]]) B_bboxes = np.array([[1, 1, 11, 11], [4, 4, 13, 13], [9

问题 I want to compute a "distance" matrix, similarly to scipy.spatial.distance.cdist, but using the intersection over union (IoU) between "bounding boxes" (4-dimensional vectors), instead of a typical distance metric (like the Euclidean distance). For example, let's suppose that we have two collections of bounding boxes, such as import numpy as np A_bboxes = np.array([[0, 0, 10, 10], [5, 5, 15, 15]]) array([[ 0, 0, 10, 10], [ 5, 5, 15, 15]]) B_bboxes = np.array([[1, 1, 11, 11], [4, 4, 13, 13], [9