TypeError: argument of type 'int' is not iterable

Question

I am getting this error when I run my program and I have no idea why. The error is occurring on the line that says "if 1 not in c:"

Code:

matrix = [
    [0, 0, 0, 5, 0, 0, 0, 0, 6],
    [8, 0, 0, 0, 4, 7, 5, 0, 3],
    [0, 5, 0, 0, 0, 3, 0, 0, 0],
    [0, 7, 0, 8, 0, 0, 0, 0, 9],
    [0, 0, 0, 0, 1, 0, 0, 0, 0],
    [9, 0, 0, 0, 0, 4, 0, 2, 0],
    [0, 0, 0, 9, 0, 0, 0, 1, 0],
    [7, 0, 8, 3, 2, 0, 0, 0, 5],
    [3, 0, 0, 0, 0, 8, 0, 0, 0],
    ]
a = 1
while a:
     try:
        for c, row in enumerate(matrix):
            if 0 in row:
                print("Found 0 on row,", c, "index", row.index(0))
                if 1 not in c:
                    print ("t")
    except ValueError:
         break

What I would like to know is how I can fix this error from happening an still have the program run correctly.

Thanks in advance!

Answer 1

Here c is the index not the list that you are searching. Since you cannot iterate through an integer, you are getting that error.

>>> myList = ['a','b','c','d']
>>> for c,element in enumerate(myList):
...     print c,element
... 
0 a
1 b
2 c
3 d

You are attempting to check if 1 is in c, which does not make sense.