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Spreading a given value over a given length of an array


By : Payam Amani
Date : October 16 2020, 08:10 PM
I think the issue was by ths following , I guess you can do it in many ways.
As @rahnema1 mentioned you can use a uniform distribution:
code :
repmat(number/Size ,Size ,1)
normcdf(0:Size,1000,100) %choose the parameters yo want


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Spreading undefined in array vs object


By : Jose Mrabet
Date : March 29 2020, 07:55 AM
Hope this helps As noted in the comments, and summarized by @ftor from #687, object spread is equivalent1 to Object.assign() (issues #687, #45), where as spread in array literal context is iterable spread.
Quoting Ecma-262 6.0, Object.assign() is defined as:
code :
const maybeArray = undefined;
const newArray = [ ...(maybeArray || []) ];

How to form a 3d array by spreading a 1d vector


By : imaUser
Date : March 29 2020, 07:55 AM
With these it helps You can chain spread as mentioned by francescalus to obtain a 3D array (see below) The 3 first writes produce all the sequence 1, 2 and 3. However, you might consider to just 3 nested loop which is my opinion is easier to read.
code :
program test_spread
  implicit none
  integer :: a(3) = (/ 1, 2, 3 /)
  integer :: array(3,3,3)
  real    :: arrayR(3,3,3)
  integer :: i,j,k

  array = spread(spread(a, 2, 3),3,3)
  write(6,*) array(:,1,1)

  array = spread(spread(a, 1, 3),3,3)
  write(6,*) array(1,:,1)

  array = spread(spread(a, 1, 3),2,3)
  write(6,*) array(1,1,:)

  arrayR = 2.0**(                           &
     spread(spread(a, 2, 3),3,3) +           &
     spread(spread(a, 1, 3),3,3) +           &
     spread(spread(a, 1, 3),2,3))

  write(6,*) 'arrayR (1)',arrayR
  do k = 1,3
    do j = 1,3
      do i = 1,3
        arrayR(i,j,k) = 2.0**(a(i)+a(j)+a(k))
      end do
    end do
  end do

  write(6,*) 'arrayR (2)',arrayR

 end program test_spread

Spreading array to arguments in typescript


By : user2672443
Date : March 29 2020, 07:55 AM
This might help you You just need to add a type for your args so TypeScript knows that the args variable is an array of tuples with a number and a string.
Then it will work:
code :
const args: [number, string][] = [
    [1, 'a'],
    [2, 'b'],
];

const concatter = (first: number, second: string) => `${first}-${second}`;

const singleTest = concatter(...args[0]);

const test = args.map(a => concatter(...a));

console.log(test);

Laravel - Spreading an eloquent object into an array


By : user3391058
Date : March 29 2020, 07:55 AM
may help you . Consider this code: , Merge the arrays.
code :
$tokenArray = $this->me()->toArray() + [
    'access_token' => $token,
    'token_type' => 'bearer',
    'expires_in' => auth()->factory()->getTTL() * 60,
];

what is the difference between assigning and spreading an array?


By : user3393850
Date : March 29 2020, 07:55 AM
help you fix your problem let x = array; just makes the x variable and the array variable both point to the same array:
code :
array−−−+
        |    +−−−−−−−−−+
        +−−−>| (array) |
        |    +−−−−−−−−−+
x−−−−−−−+    | 0: 1    |
             | 1: 2    |
             | 2: 3    |
             | 3: 4    |
             | 4: 5    |
             +−−−−−−−−−+
             +−−−−−−−−−+
array−−−−−−−>| (array) |
             +−−−−−−−−−+
             | 0: 1    |
             | 1: 2    |
             | 2: 3    |
             | 3: 4    |
             | 4: 5    |
             +−−−−−−−−−+

             +−−−−−−−−−+
y−−−−−−−−−−−>| (array) |
             +−−−−−−−−−+
             | 0: 1    |
             | 1: 2    |
             | 2: 3    |
             | 3: 4    |
             | 4: 5    |
             +−−−−−−−−−+
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