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.

Answer 2

Based on the OP's comment It should print "t" if there is a 0 in a row and there is not a 1 in the row.

change if 1 not in c to if 1 not in row

for c, row in enumerate(matrix):
    if 0 in row:
        print("Found 0 on row,", c, "index", row.index(0))
        if 1 not in row: #change here
            print ("t")

Further clarification: The row variable holds a single row itself, ie [0, 5, 0, 0, 0, 3, 0, 0, 0] . The c variable holds the index of which row it is. ie, if row holds the 3rd row in the matrix, c = 2 . Remember that c is zero-based, ie the first row is at index 0, second row at index 1 etc.

Answer 3

c is the row number, so it's an int . So numbers can't be in other numbers.

Answer 4

You're trying to iterate over 'c' which is just an integer, holding your row number.

It should print "t" if there is a 0 in a row

Then just replace the c with the row so it says:

if 1 not in row:

Answer 5

Well, if we closely look at the error, it says that we are looking to iterate over an object which is not iterable. Basically, what I mean is if we write 'x' in 1 , it would throw error .And if we write 'x' in [1] it would return False

>>> 'x' in 1
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: argument of type 'int' is not iterable

>>> 'x' in [1]
False

So all we need to do it make the item iterable, in case of encountering this error.In this question, we can just make c by a list [c] to resolve the error. if 1 not in [c]:

Answer 6

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!