Is there a more 'mathematical' way to do the following:
1.2738 * [list_of_items]
So for what I'm doing is:
[1.2738 * item for item in list_of_items]
asked Mar 2, 2015 at 23:40
David542David542
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The mathematical equivalent of what you're describing is the operation of multiplication by a scalar for a vector. Thus, my suggestion would be to convert your list of elements into a "vector" and then multiply that by the scalar.
A standard way of doing that would be using numpy.
Instead of
1.2738 * [list_of_items]
You can use
import numpy
1.2738 * numpy.array[list_of_items]
Sample Output:
In [8]: list_of_items
Out[8]: [1, 2, 4, 5]
In [9]: import numpy
In [10]: 1.2738 * numpy.array[list_of_items]
Out[10]: array[[ 1.2738, 2.5476, 5.0952, 6.369 ]]
answered Mar 2, 2015 at 23:42
Another approach
map[lambda x:x*1.2738,list_of_items]
answered Mar 2, 2015 at 23:43
levilevi
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Given a matrix and a scalar element k, our task is to find out the scalar product of that matrix.
Examples:
Input : mat[][] = {{2, 3}
{5, 4}}
k = 5
Output : 10 15
25 20
We multiply 5 with every element.
Input : 1 2 3
4 5 6
7 8 9
k = 4
Output : 4 8 12
16 20 24
28 32 36 The scalar multiplication of a number k[scalar], multiply it on every entry in the matrix. and a matrix A is the matrix kA.
C++
#include
using namespace std;
#define N 3
void scalarProductMat[int mat[][N], int k]
{
for [int i = 0; i < N; i++]
for [int j = 0; j < N; j++]
mat[i][j] = mat[i][j] * k;
}
int
main[]
{
int mat[N][N] = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
int k = 4;
scalarProductMat[mat, k];
printf["Scalar Product Matrix is : \n"];
for [int i = 0; i < N; i++] {
for [int j = 0; j < N; j++]
printf["%d ", mat[i][j]];
printf["\n"];
}
return 0;
}
Java
import java.io.*;
class GFG {
static final int N = 3;
static void scalarProductMat[int mat[][],
int k]
{
for [int i = 0; i < N; i++]
for [int j = 0; j < N; j++]
mat[i][j] = mat[i][j] * k;
}
public static void main [String[] args]
{
int
mat[][] = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
int k = 4;
scalarProductMat[mat, k];
System.out.println["Scalar Product Matrix is : "];
for [int i = 0; i < N; i++]
{
for [int j = 0; j < N; j++]
System.out.print[mat[i][j] + " "];
System.out.println[];
}
}
}
Python 3
N = 3
def scalarProductMat[ mat, k]:
for i in range[ N]:
for j in range[ N]:
mat[i][j] = mat[i][j] * k
if __name__ == "__main__":
mat =
[[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]]
k = 4
scalarProductMat[mat, k]
print["Scalar Product Matrix is : "]
for i in range[N]:
for j in range[N]:
print[mat[i][j], end =
" "]
print[]
C#
using System;
class GFG{
static int N = 3;
static void scalarProductMat[int[,] mat,
int k]
{
for [int i = 0; i < N; i++]
for [int j = 0; j < N; j++]
mat[i,j] = mat[i, j] * k;
}
static public void Main []
{
int[,] mat = {{1, 2, 3},
{4, 5, 6},
{7, 8, 9}};
int k = 4;
scalarProductMat[mat, k];
Console.WriteLine["Scalar Product Matrix is : "];
for [int i = 0; i < N; i++] {
for [int j = 0; j < N; j++]
Console.Write[mat[i, j] + " "];
Console.WriteLine[];
}
}
}