multidimensional-array

average numpy array but retain shape

不问归期 提交于 2020-01-03 08:08:10
问题 I have a Numpy 3 axis array whose elements are 3 dimensional. I'd like to average them and return the same shape of the array. The normal average function removes the 3 dimensions and replace it with the average (as expected): a = np.array([[[0.1, 0.2, 0.3], [0.2, 0.3, 0.4]], [[0.4, 0.4, 0.4], [0.7, 0.6, 0.8]]], np.float32) b = np.average(a, axis=2) # b = [[0.2, 0.3], # [0.4, 0.7]] Result required : # b = [[[0.2, 0.2, 0.2], [0.3, 0.3, 0.3]], # [[0.4, 0.4, 0.4], [0.7, 0.7, 0.7]]] Can you do

How to pass a method the height and width of a 2d array as parameters?

可紊 提交于 2020-01-03 06:40:15
问题 Say I have an array: array[2][4] , and inside the main method, I have a call to a function blackandwhite . How can I pass this method the length and width of the array as arguments? 回答1: This is a possible solution : void blackandwhite(int* array, int height, int width) { // Array-processing done here. // array is pointer to int, // initially points to element myarray[0][0]. // variable height = 2; // variable width = 4; } int main() { int myarray[2][4]; blackandwhite(&myarray[0][0], 2, 4); }

Is there a cleaner way to use a 2 dimensional array?

痞子三分冷 提交于 2020-01-03 05:47:10
问题 I'm trying to make a 2D array class, and ran into a problem. The best way I could figure out to do it was to pass get/setitem a tuple of the indices, and have it unpacked in the function. Unfortunately though, the implementation looks really messy: class DDArray: data = [9,8,7,6,5,4,3,2,1,0] def __getitem__ (self, index): return (self.data [index [0]], self.data [index [1]]) def __setitem__ (self, index, value): self.data [index [0]] = value self.data [index [1]] = value test = DDArray ()

new matrix[N][N] failure [duplicate]

前提是你 提交于 2020-01-03 05:35:25
问题 This question already has answers here : How do I declare a 2d array in C++ using new? (23 answers) Closed 6 years ago . I'm having a stack overflow allocating a huge matrix on the stack (and I agree with that: it's stupid to allocate it there) and I'm writing the following code since I want to access the matrix's elements with the subscripts indices mat[x][y] double (*mul1)[N][N]; mul1 = new double[N][N]; I'm receiving an error: error C2440: '=' : cannot convert from 'double (*)[1000]' to

new matrix[N][N] failure [duplicate]

六眼飞鱼酱① 提交于 2020-01-03 05:35:05
问题 This question already has answers here : How do I declare a 2d array in C++ using new? (23 answers) Closed 6 years ago . I'm having a stack overflow allocating a huge matrix on the stack (and I agree with that: it's stupid to allocate it there) and I'm writing the following code since I want to access the matrix's elements with the subscripts indices mat[x][y] double (*mul1)[N][N]; mul1 = new double[N][N]; I'm receiving an error: error C2440: '=' : cannot convert from 'double (*)[1000]' to

Java - how to return in a method multidimensional array without aliasing

元气小坏坏 提交于 2020-01-03 05:33:16
问题 I'm aware that arrays are objects and in java objects are transfered by reference which could cause aliasing so objects should be returned with in this form to not cause aliasing: return new (object(parameters)); So this is what I'm trying to do with multidimensional arrays, however for some reason compiler says I have an error : "array dimension missing". public int[][] Testing(int[][]arr) { int[][]newArr=new int[arr.length][arr[0].length]; for(int i=0;i<arr.length;i++) { for(int j=0;j<arr[0

Print Specific Row in 2D Array

我的未来我决定 提交于 2020-01-03 04:40:10
问题 How would I get just the second row to print in this 4x4 array? double [][] table = new double[4][4]; for(int i = 0; i < table.length; i++){ for(int j = 0; j < table[i].length; j++) table[i][j] = (Math.random() * 10); } 回答1: Use this for(int j = 0; j < table[1].length; j++) System.out.print(table[1][j]+" - "); 回答2: Use this // table[0] == 1st row // table[1] == 2nd row // etc.. for(int i = 0; i < table[1].length; i++) System.out.println(table[1][i]); // Print each item of the 2nd row 回答3:

Building and printing a multidimensional list in Perl without looping

让人想犯罪 __ 提交于 2020-01-03 04:35:42
问题 The top answer in this post: How can I create a multidimensional array in Perl? suggests building a multi-dimensional array as follows: my @array = (); foreach my $i ( 0 .. 10 ) { foreach my $j ( 0 .. 10 ) { push @{ $array[$i] }, $j; } } I am wondering if there is a way of building the array more compactly and avoiding the nested loop, e.g. using something like: my @array = (); my @other_array = (0 ... 10); foreach my $i ( 0 .. 10 ) { $array[$i] = @other_array; # This does not work in Perl }

Right Way of Passing a 2D vector as an argument to a function in C++

走远了吗. 提交于 2020-01-03 03:39:11
问题 I'm trying to write a function that takes a 2D vector as an argument. This compiles just fine, but I get a segmentation fault whilst execution. //http://www.codechef.com/problems/SUMTRIAN #include<iostream> #include<algorithm> #include<vector> using namespace std; int max_sum_path(int maximum_rows,vector<vector<int> >& matrix,int row_index,int colm_index); int main() { vector<vector<int> > Triangle; int num_test_cases; cin>>num_test_cases; //to iterate over the test cases for(int i=0;i<num

string to associative array conversion

▼魔方 西西 提交于 2020-01-03 03:25:13
问题 I've been struggling with this for a few days and wanted to throw it out there and see if someone has any ideas. Basically I have a string e.g 1) "/0/bar" 2) "/build/0/foo/1" and need to convert this into a multidimensional array 1) $result[0][bar] 2) $result[build][0][foo][1] So far I've tried: $query = "/build/0/foo/1"; $queryAr = []; $current = &$queryAr; $keys = explode("/", $query); foreach($keys as $key) { @$current = &$current[$key]; } $current = $value; quieting the output is a pretty