CODEWITHSUNDEEP
pythonarraysnumpymultidimensional-array
user137828
user137828

Reputation: 17

making multi-dimensional nulls or ones from numpy array

I want to create an array filled with ones or zeros based on this array

testArray = np.array([7,5,3])

so the final result should look like

[[1,1,1,1,1,1,1],
 [1,1,1,1,1],
 [1,1,1]]

Upvotes: 2

Views: 155

Answers (4)

Ryan Saxe
Ryan Saxe

Reputation: 17829

just write a quick list comprehension:

>>> holder = [np.ones((testArray[i])) for i in range(len(testArray))]
>>> holder
[array([[ 1.,  1.,  1.,  1.,  1.,  1.,  1.]]), array([[ 1.,  1.,  1.,  1.,  1.]]), array([[ 1.,  1.,  1.]])]

if you want it to be in the write format you can always reshape it:

>>> np.array(holder).reshape(3,1)
array([[array([ 1.,  1.,  1.,  1.,  1.,  1.,  1.])],
       [array([ 1.,  1.,  1.,  1.,  1.])],
       [array([ 1.,  1.,  1.])]], dtype=object)

problem solved!

Upvotes: 1

askewchan
askewchan

Reputation: 46530

Every row (and column, etc) in a numpy array must have the same length. You can achieve what @ChrisWilson4 did, and fill the empty parts with 0 or np.nan. Create an empty array with number of rows equal to length of lengths, and number of columns equal to the largest row:

fill = 1    # or `0` or `np.nan`
background = 0 # or `np.nan`
lengths = np.array([7,5,3])

a = np.ones((lengths.size, lengths.max()))*background

and fill it up with your fill value:

for row, length in enumerate(lengths):
    a[row,:length] = fill

a
#array([[ 1.,  1.,  1.,  1.,  1.,  1.,  1.],
#       [ 1.,  1.,  1.,  1.,  1.,  0.,  0.],
#       [ 1.,  1.,  1.,  0.,  0.,  0.,  0.]])

Or, for fill = 0 and background = np.nan:

array([[  0.,   0.,   0.,   0.,   0.,   0.,   0.],
       [  0.,   0.,   0.,   0.,   0.,  nan,  nan],
       [  0.,   0.,   0.,  nan,  nan,  nan,  nan]])

Or, you can make a list of lists, in the pure python way (without using numpy) like so:

fill = 1
lengths = [7,5,3]
a = [ [fill]*length for length in lengths ]

a
#[[1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1], [1, 1, 1]]

Upvotes: 1

tiago
tiago

Reputation: 23492

This will give you a ragged array of object dtype:

>>> result = np.array([np.ones(a) for a in testArray])
>>> print result
[[ 1.  1.  1.  1.  1.  1.  1.] [ 1.  1.  1.  1.  1.] [ 1.  1.  1.]]

For zeros, just use np.zeros.

Upvotes: 1

ChrisWilson4
ChrisWilson4

Reputation: 195

Like @JoshAdel said in comments, the int array cannot be jagged, meaning the rows can't be different lengths. Is this what you were looking for?

public class soArray {
public static void main(String[] args) {

    int[][] testArray = soArray.array(7,5,3);
    
    for (int i = 0; i < testArray.length; i++){
        for (int j = 0; j < testArray[0].length; j++){
            System.out.print(testArray[i][j]);
        }
        System.out.println();
    }
}
public static int[][] array(int a, int b, int c){

    int max;
    if(a > b && a > c)
        max = a;
    else if(b > a && b > c)
        max = b;
    else
        max = c;
        
    int[][] out = new int[3][max];
    
    for (int i = 0; i < max; i++){
        if(i < a)
            out[0][i] = 1;
        else
            out[0][i] = 0;
    }
    for(int i = 0; i< b; i++){
        if(i < b)
            out[1][i] = 1;
        else
            out[1][i] = 0;
    }
    for(int i = 0; i < c; i++){
        if(i < c)
            out[2][i] = 1;
        else
            out[2][i] = 0;
    }
    
    return out;
}

}

It will print out :

1111111

1111100

1110000

Upvotes: -1

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