Python - If Statement with two-dimensional array

Question

I'm trying to generate a Sudoku. I thought about using an array for this. THis is how I create the array (ignore the fact that these are not correct Sudoku numbers):

row_array = [[1, 99],[2],[3], [4], [5], [6], [7], [8], [9]]

It does work fine. The problem is with an if statement. If I do:

print(row_array[0][1])

or:

r = 1
print(row_array[0][r])

to get the "99" it does work fine! But if I use this variable method in an if statement like this:

if y not in row_array[1] and y not in row_array[0][r]:
    row_array[1].append(y)

I get the following ERROR:

TypeError: argument of type 'int' is not iterable

I had the Sudoku working, but it took to much code, so I want to set up a counter to replace the rows and columns with an it. I'm aware of you guys probably already having a much better code for a sudoku, but I want to teach myself, and make the code better and shorter over time. I don't learn to code real good, by just copy someone else's idea or way to code.

Hello! Can you elaborate on how you expect the row_array to look if the append call succeeds? — Artyom Neustroev
– Artyom Neustroev, Commented Sep 23, 2018 at 14:10
That is because row_array[0][r] is no more a sublist but a scalar — Sheldore
– Sheldore, Commented Sep 23, 2018 at 14:20

bnaecker · Accepted Answer · 2018-09-23 14:15:23Z

1

The problem is here:

if y not in row_array[0][r]

The in operator checks if the left-hand side is in the sequence or iterable on the right hand side. But type(row_array[0][r]) will return int, which is not an iterable.

edited Sep 23, 2018 at 14:15

answered Sep 23, 2018 at 14:13

bnaecker

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3 Comments

N Chauhan Over a year ago

Typo: if y not int -> if y not in

Marie S Over a year ago

Thank you. But how do i fix this? I have no clue how to check otherwise then that.

bnaecker Over a year ago

Use if y == row_array[0][r].

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