CODEWITHSUNDEEP
python-3.xloopsconditional-statements
Shakib
Shakib

Reputation: 27

Printing patterns using loop in python

My program read an integer number N, that correspond to the order of a Bidimentional array of integers, and build the Array according to the below example. I want to fill the middle elements like my expected output.

My code:

n = int(input())
for row in range(1, n+1):
    for colum in range(1, n+1):
        print(row, end="   ")
    print()


Input:

5


My output:

1   1   1   1   1   
2   2   2   2   2
3   3   3   3   3   
4   4   4   4   4
5   5   5   5   5


The output I want: 

1   1   1   1   1
1   2   2   2   1
1   2   3   2   1
1   2   2   2   1
1   1   1   1   1

I want to fill the middle elements like this. The height number at the middle then the second height number and so on..

Upvotes: 0

Views: 373

Answers (3)

gautham
gautham

Reputation: 1

enter image description here

`for i in range(n):   
````print((n-i)*" ",end=" ")
````print((i+1)*"* ")

Upvotes: 0

user13747387
user13747387

Reputation:

I hope this is not a homework question, but i will help you.
This can be done a lot more easily with lists!:

def cell_value(i, j, n_rows):
    return min(
        abs(i - -1),
        abs(i - n_rows),
        abs(j - -1),
        abs(j - n_rows),
    )

rows=int(input("Enter the number of rows:"))
row2 = [
    [
        cell_value(i, j, rows)
        for j in range(rows)
    ]
    for i in range(rows)
]

for r in row2:
    print(*r)

Or it can be done even more easily like this below:

numberOfRows = int(input("Enter the number of rows:"))
listOut = [[1]*numberOfRows] * numberOfRows #grid of 1s of appropriate size
for j in range(int((numberOfRows+1)/2)): #symmetrical, so only look to the middle
    if j > 0:
        listOut[j] = list(listOut[j-1]) #copy previous row
    for i in range(int((numberOfRows+1)/2)):   
        if i>=j:
            listOut[j][i] = j+1
            listOut[j][numberOfRows-(i+1)] = j+1
    #copy current row to appropriate distance from the end
    listOut[numberOfRows-(j+1)] = list(listOut[j])
for row in listOut:
    print(row)

Both of the above programs give the SAME result

Enter the number of rows:5
1 1 1 1 1
1 2 2 2 1
1 2 3 2 1
1 2 2 2 1
1 1 1 1 1

Note:this is only possible for odd numbers! Let me know if you have any doubts...
Cheers!

Upvotes: 0

jsbueno
jsbueno

Reputation: 110746

for the "1-2-3-2-1" sequence, you can get it as the "minimum between row and n + 1 - row" - - min(row, n + 1 - row). (And the symmetrical for column) - and then you print the min of this calculation for row and cols:

n = int(input())
for row in range(1, n+1):
    for column in range(1, n+1):
        mrow = min(row, n + 1 - row)
        mcol = min(column, n + 1 - column) 
        print(min(mrow, mcol), end="   ")
    print()

Upvotes: 1

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