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how to covert 1/4 elements of a matrix to zero


how to covert 1/4 elements of a matrix to zero

By : n3bu
Date : November 28 2020, 08:01 AM
I hope this helps you . If you want the points chosen randomly:
If you have a 512 x 512 array:
code :


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how can I covert this to a matrix

how can I covert this to a matrix


By : Karl
Date : March 29 2020, 07:55 AM
it should still fix some issue I'd suggest replacing your for-loop with something like the following:
code :
A = np.loadtxt("C:\Users\KEMAL\Desktop\piton\test.txt").T

s = A.max(axis=1).cumsum()
data = np.zeros((int(s[-1]), A.shape[1]))

A[1:] += s[:2, None]
i = (A - 1).astype(int)
j = np.tile(np.arange(A.shape[1]), (A.shape[0], 1))

data[i, j] = 1
array([[ 0.,  1.,  0.,  1.,  0.,  0.,  0.,  0.],
       [ 1.,  0.,  0.,  0.,  0.,  1.,  0.,  1.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  1.,  0.],
       [ 0.,  0.,  1.,  0.,  1.,  0.,  0.,  0.],
       [ 0.,  1.,  1.,  1.,  1.,  0.,  0.,  1.],
       [ 1.,  0.,  0.,  0.,  0.,  1.,  1.,  0.],
       [ 0.,  0.,  1.,  1.,  0.,  1.,  0.,  1.],
       [ 0.,  0.,  0.,  0.,  1.,  0.,  0.,  0.],
       [ 0.,  1.,  0.,  0.,  0.,  0.,  1.,  0.],
       [ 1.,  0.,  0.,  0.,  0.,  0.,  0.,  0.]])
A=np.loadtxt("C:\Users\KEMAL\Desktop\piton\test.txt")
n=3
arrs = []
for i in range(0,n):
    b=[row[i] for row in A]
    D=np.array(b)
    size=[len(D),np.max(D)]
    B=np.zeros(size)
    for i in range(len(D)):
        idx=D[i]-1
        B[i,idx]=1
        U=B.transpose(1, 0)

    arrs.append(U)

np.vstack(arrs)
How to covert array to matrix for Google Sheets

How to covert array to matrix for Google Sheets


By : Hyun Min
Date : March 29 2020, 07:55 AM
Hope that helps How about this modification?
Modification points : In your script, "[\"" + arr[i] + "\"]" becomes string. You can use [arr[i]]. You can put array = toMatrix(array); to outside of for loop. Modified script :
code :
// function pastValues should past 6 columns each comprised of a,b,c,d,e,f. Each letter is in its own cell.
function pasteValues() {
  var sheet1 = SpreadsheetApp.getActiveSpreadsheet().getActiveSheet();

  var array = ["a","b","c","d","e","f"];
  array = toMatrix(array);
  for(var i=1; i < array.length; i++) {
    sheet1.getRange(2, i, 6, 1).setValues(array);
  }
}

// convert array [a,b,c,d,e,f] into matrix [[a],[b],[c],[d],[e],[f]] so it can be set into newShceduleSheet.
function toMatrix(arr){
  var newMatrix = [];
  for (var i=0; i<arr.length; i++) {
    newMatrix.push([arr[i]]);
  }
  return newMatrix;
}
function pasteValues() {
  var sheet1 = SpreadsheetApp.getActiveSpreadsheet().getActiveSheet();
  var array = ["a","b","c","d","e","f"];
  array = array.map(function(e){return [e]});
  for(var i=1; i < array.length; i++) {
    sheet1.getRange(2, i, 6, 1).setValues(array);
  }
}
function pasteValues() {
  var sheet1 = SpreadsheetApp.getActiveSpreadsheet().getActiveSheet();
  var array = ["a","b","c","d","e","f"];
  array = array.map(function(e){return [e]});

  var n = array.length; // Please input the number of columns that you want to copy.
  for (var i = 1; i <= n; i++) {
    sheet1.getRange(2, i, array.length, array[0].length).setValues(array);
  }
}
How to covert 1d array to Logical matrix

How to covert 1d array to Logical matrix


By : Helgi Viggosson
Date : March 29 2020, 07:55 AM
help you fix your problem You can create an identity matrix and then use the indices to create a new re-ordered matrix:
code :
>>> a = np.eye(4)
[Out]: array([[1., 0., 0., 0.],
              [0., 1., 0., 0.],
              [0., 0., 1., 0.],
              [0., 0., 0., 1.]])

>>> indices = [1, 3, 1, 2]
>>> a[indices]
[Out]: array([[0., 1., 0., 0.],
              [0., 0., 0., 1.],
              [0., 1., 0., 0.],
              [0., 0., 1., 0.]])
Matlab: covert m x n matrix into 1 x n cell array

Matlab: covert m x n matrix into 1 x n cell array


By : uhligfd
Date : March 29 2020, 07:55 AM
around this issue You can use mat2cell that converts array to cell array.
For example:
code :
A=randi(10,4,3)
A =

     7     3     7
     2     1     4
     8     1    10
     1     9     1

C = mat2cell(A, size(A,1), ones(1, size(A,2)))

C = 

    [4x1 double]    [4x1 double]    [4x1 double] 
Covert a dataframe into a matrix form

Covert a dataframe into a matrix form


By : Qinhong.Jiang
Date : January 02 2021, 06:48 AM
I wish this helpful for you You can use df.iloc[]:
code :
df.iloc[:,1:].to_numpy()
array([[1, 2, 5],
   [1, 2, 1],
   [1, 2, 4],
   [1, 5, 1],
   [1, 4, 4]], dtype=int64)
df.astype(str).iloc[:,1:].to_numpy()
array([['1', '2', '5'],
   ['1', '2', '1'],
   ['1', '2', '4'],
   ['1', '5', '1'],
   ['1', '4', '4']], dtype=object)
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