# Forming a frame of zeros around a matrix in python

## Question:

I am trying to pad a matrix with zeros, but am not really sure how to do it. Basically I need to surround a matrix with an n amount of zeros. The input matrix is huge (it represents an image)

Example:

```
Input:
1 2 3 4
5 6 7 8
4 3 2 1
n = 2
Output:
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 1 2 3 4 0 0
0 0 5 6 7 8 0 0
0 0 4 3 2 1 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
```

The problem is that I get **"k" is not accessed** and **"l" is not accessed**.

Code:

```
import numpy as np
n = 2
matrix = [[1, 2, 3, 4],
[5, 6, 7, 8],
[4, 3, 2, 1]]
modified_matrix = np.zeros(shape=((len(matrix) + n), (len(matrix[0]) + n)), dtype=int)
k = n
l = n
modified_matrix = [[l] for l in range(len(matrix[k])] for k in range(len(matrix))]
```

## Answers:

You can use NumPy’s slice notation.

```
import numpy as np
#input matrix
A = np.array([[1,2,3,4],
[3,4,5,6]])
#get matrix shape
x,y=A.shape
#set amount of zeros
n=2
#create zero's matrix
B=np.zeros((x+2*n,y+2*n),dtype=int)
# insert & slice
B[n:x+n, n:y+n] = A
#show result
for row in B:
print(row)
```

**Output**:

```
[0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0]
[0 0 1 2 3 4 0 0]
[0 0 3 4 5 6 0 0]
[0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0]
```