David Richmond - 24 days ago 8

Python Question

so here is my matrix:

`A = matrix([[1,1,2,2,2],`

[3,1,4,3,3],

[0,4,2,4,0],

[0,0,2,0,0],

[0,0,2,0,0]])

I would like to insert a 0 row that displays the number of column entries.

This was my original code:

`A = matrix([[1,1,2,2,2],`

[3,1,4,3,3],

[0,4,2,4,0],

[0,0,2,0,0],

[0,0,2,0,0]])

puzzle = A.insert_row(0, sum(A))

Whats wrong with this code is that it gives me the sum of each column, when I just need the count of each column.

What I have

`[ 4 6 12 9 5]`

[ 1 1 2 2 2]

[ 3 1 4 3 3]

[ 0 4 2 4 0]

[ 0 0 2 0 0]

[ 0 0 2 0 0]

Desired

`[ 2 3 5 3 2]`

[ 1 1 2 2 2]

[ 3 1 4 3 3]

[ 0 4 2 4 0]

[ 0 0 2 0 0]

[ 0 0 2 0 0]

Best,

David

Answer

Try the `nonzero_positions_in_column`

method:

```
sage: A = matrix([[1,1,2,2,2],
....: [3,1,4,3,3],
....: [0,4,2,4,0],
....: [0,0,2,0,0],
....: [0,0,2,0,0]])
sage: A.column(1)
(1, 1, 4, 0, 0)
sage: A.nonzero_positions_in_column(1)
[0, 1, 2]
sage: v = vector([len(A.nonzero_positions_in_column(i)) for i in range(A.ncols())])
sage: A.insert_row(0, v)
[2 3 5 3 2]
[1 1 2 2 2]
[3 1 4 3 3]
[0 4 2 4 0]
[0 0 2 0 0]
[0 0 2 0 0]
```

Source (Stackoverflow)

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