multidimensional-array

问题 How to remove leading / trailing zeros from a NumPy array? Trim_zeros works only for 1D. 回答1: The following function works for any dimension: def trim_zeros(arr, margin=0): ''' Trim the leading and trailing zeros from a N-D array. :param arr: numpy array :param margin: how many zeros to leave as a margin :returns: trimmed array :returns: slice object ''' s = [] for dim in range(arr.ndim): start = 0 end = -1 slice_ = [slice(None)]*arr.ndim go = True while go: slice_[dim] = start go = not np

问题 Mongo DB Array Structure: { "time" :[ [2018-12-18T20:16:28.800Z , 2018-12-18T20:16:28.800Z, 2016-11-14T10:16:32.700Z], [2016-11-14T10:16:32.700Z, 2017-09-17T11:16:54.500Z, 2018-12-18T20:16:28.800Z], [2017-09-17T11:16:54.500Z, 2018-12-18T20:16:28.800Z, 2016-11-14T10:16:32.700Z] ], "bar": [ [2,5,9,8], [6,3,7,None], [7,8,None] ] } Expected output with One array without None values: "time" :[ 2018-12-18T20:16:28.800Z, 2018-12-18T20:16:28.800Z, 2016-11-14T10:16:32.700Z, 2016-11-14T10:16:32.700Z,

问题 Mongo DB Array Structure: { "time" :[ [2018-12-18T20:16:28.800Z , 2018-12-18T20:16:28.800Z, 2016-11-14T10:16:32.700Z], [2016-11-14T10:16:32.700Z, 2017-09-17T11:16:54.500Z, 2018-12-18T20:16:28.800Z], [2017-09-17T11:16:54.500Z, 2018-12-18T20:16:28.800Z, 2016-11-14T10:16:32.700Z] ], "bar": [ [2,5,9,8], [6,3,7,None], [7,8,None] ] } Expected output with One array without None values: "time" :[ 2018-12-18T20:16:28.800Z, 2018-12-18T20:16:28.800Z, 2016-11-14T10:16:32.700Z, 2016-11-14T10:16:32.700Z,

问题 Could anyone help with an an explanation of what axis is in TensorFlow 's one_hot function? According to the documentation: axis: The axis to fill (default: -1, a new inner-most axis) Closest I came to an answer on SO was an explanation relevant to Pandas: Not sure if the context is just as applicable. 回答1: Here's an example: x = tf.constant([0, 1, 2]) ... is the input tensor and N=4 (each index is transformed into 4D vector). axis=-1 Computing one_hot_1 = tf.one_hot(x, 4).eval() yields a (3,

问题 I have a 2D array where I need to filter the rows having date field (3d column) var data = [ ['1','a','12-12-2019','A'], ['2','b','','A'], ['3','c','12-1-2019','A'], ['4','d','','A'], ]; The expected result is result = [ ['1','a','12-12-2019','A'], ['3','c','12-1-2019','A'], ]; Using for loop for comprisons is time intensive, Is there a fastest way to retrieve? 回答1: It seems pretty straightforward to just use Array.filter, checking to see if the date field is empty or not. var data = [ ['1',

问题 So I have this dummy 2d array: int mat[][] = { {10, 20, 30, 40, 50, 60, 70, 80, 90}, {15, 25, 35, 45}, {27, 29, 37, 48}, {32, 33, 39, 50, 51, 89}}; I want to add up all the values by columns so it would add 10 + 15 + 27 + 32 and return 84 and so on. I have this so far: public void sum(int[][] array) { int count = 0; for (int rows = 0; rows < array.length; rows++) { for (int columns = 0; columns < array[rows].length; columns++) { System.out.print(array[rows][columns] + "\t"); count += array[0]

问题 So I have this dummy 2d array: int mat[][] = { {10, 20, 30, 40, 50, 60, 70, 80, 90}, {15, 25, 35, 45}, {27, 29, 37, 48}, {32, 33, 39, 50, 51, 89}}; I want to add up all the values by columns so it would add 10 + 15 + 27 + 32 and return 84 and so on. I have this so far: public void sum(int[][] array) { int count = 0; for (int rows = 0; rows < array.length; rows++) { for (int columns = 0; columns < array[rows].length; columns++) { System.out.print(array[rows][columns] + "\t"); count += array[0]

问题 This question already has answers here : Rotating a NxN matrix in Java (11 answers) Closed 19 hours ago . I want a 2d matrix to rotate to the right, it compiles fine but when I try to the run it says that the array index is out of bounds exception . For example, I want {{10,20,30},{40,50,60}} to rotate into {{40,10},{50,20},{60,30}} : public static int[][] rotate(int[][] m) { int[][] rotateM = new int[m[0].length][m.length]; for (int i = 0; i < m.length; i++) { for (int j = 0; j < m[0].length

问题 This question already has answers here : Rotating a NxN matrix in Java (11 answers) Closed 19 hours ago . I want a 2d matrix to rotate to the right, it compiles fine but when I try to the run it says that the array index is out of bounds exception . For example, I want {{10,20,30},{40,50,60}} to rotate into {{40,10},{50,20},{60,30}} : public static int[][] rotate(int[][] m) { int[][] rotateM = new int[m[0].length][m.length]; for (int i = 0; i < m.length; i++) { for (int j = 0; j < m[0].length

问题 This question already has answers here : Rotating a NxN matrix in Java (11 answers) Closed 19 hours ago . I want a 2d matrix to rotate to the right, it compiles fine but when I try to the run it says that the array index is out of bounds exception . For example, I want {{10,20,30},{40,50,60}} to rotate into {{40,10},{50,20},{60,30}} : public static int[][] rotate(int[][] m) { int[][] rotateM = new int[m[0].length][m.length]; for (int i = 0; i < m.length; i++) { for (int j = 0; j < m[0].length