Argument of type 'int' is not iterable

Question

Problem:

Well, I am getting an error from my python program that says, argument of type 'int' is not iterable. 

Could anybody save my day by fixing the error? Find my sample code snippet below

def duplicates(n):
    my_list = n[0]
    for i in range(1,len(n)):
        if n[i] in my_list:
            print "duplicate " + str(n[i])
        else:
            out.append(n[i])    
    return my_list
duplicates([4,5,5,4])

Thanks.

Answer 1

Once your compiler throws back an error, you should read the error message carefully. The error argument of type 'int' is not iterable means, you haven’t declared any list in your program but an integer. Whenever you write

my_list = n[0]

it becomes an integer. Hence my_list is an integer you can not write n[ i ] or my_list.append(n [ i ])

If you want to treat your my_list [ n ]  as a list you need to re-write the code like this:

my_list = [n[0]]

Good day!

Answer 2

Here c is the index not the list that you are looking. Because you cannot iterate through an integer, you are obtaining 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 trying to check if 1 is in c, which does not make sense

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

Alter 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")

More clarification: The row variable retain a single row itself, ie [0, 5, 0, 0, 0, 3, 0, 0, 0]. The c variable apprehend the index of which row it is. ie, in case row contains the 3rd row in the matrix, c = 2. Keep in mind that c is zero-based, for example the first row is at index 0, second row at index 1 etc.

Cast, Cast, Cast.

word = 'stuff'
blur = 12344566
word in str(blur)

This will output false successfully saying you whether stuff can be figured out inside the now string ‘12344566’. And currently for a more objective example.

array = ['Foo','Bar', 1]
for element in array:
    print(element)
    if 'sandwich' in element:
        print('Found the sandwich')

Here we can view all elements of the array are printed since the error happens on the third element. One will never trace a string in an int. You can eliminate checking types and error handling by wrapping element in str() to cast it as a string in case we come across any unruly data types.

array = ['Foo','Bar', 1]
for element in array:
    print(element)
    if 'sandwich' in str(element):
        print('Found the sandwich'