Python error: AttributeError: 'int' object has no attribute 'append'

Question

So I've looked through similar questions, and I'm still getting the same problem and can't figure it out. For this programming assignment, I'm creating a simplified version of lexical analysis for a small subset of the Clite lexicon. I'm extracting tokens from an input file, outputting the results of my analysis. I'm creating a symbol table for the identifiers founds using a dictionary. When I find the same identifiers on different lines, I need to append to the symbol table the line that it was found on. For example, I find the identifier "number18" on line 2 and again on line 7. So the symbol table would need to go from {number18: 2} to {number18: 2,7}

The problem comes when I try to append the new line number onto a current dictionary entry. I get the error as I posted in the question title. Here's my code thus far

y = 0
s2 = ()
stable = dict()

for line in open("Sample.txt","r"):
    x1 = ''
    for char in line:
    if char.isalpha():
        x1 = x1 + char
    elif char.isdigit():
        x1 = x1 + char
    elif char == '.':
        x1 = x1 + char
    elif x1 != '':
        break

    #print (x1)    
    if (x1 == "for" or x1 == "bool" or x1 == "char" or x1 == "else" or x1 == "false" or x1 == "float" or x1 == "if" or x1 == "int" or x1 == "main" or x1 == "true" or x1 == "while"):
        s2=(y,"Keyword",x1)
    elif x1.isidentifier():
        s2=(y,"Identifier",x1)
    if x1 in stable.keys():
        stable[x1].append(y)
    else:
        stable[x1]=y


    elif x1.isdigit():
        s2=(y,"Int",x1)
    else:
        s2=(y,"Float",x1)
    print (s2)
    y=y+1

print (stable)

Answer 1

You first set your dict values to be an int:

    stable[x1]=y

but then you later on you try to treat it as if it is a list:

    stable[x1].append(y)

Start out with a list containing your first int instead:

    stable[x1]=[y]

and the .append() will work.

Alternatively, you could use a defaultdict:

stable = defaultdict(list)

and then append at will without needing to test if the key is already there:

    stable[x1].append(y)  # No need to do `if x1 in stable`.

Answer 2

elif x1.isidentifier():
    s2=(y,"Identifier",x1)
    if x1 in stable.keys():
        stable[x1].append(y)
    else:
        stable[x1]=y

In your else part above, you are adding integer first time. So when you use append next time, you would get that error.

Rather wrap your integer y in a list [y], when you add your value first time to your dict

    else:
        stable[x1]=[y]

Well, you already know from @Martijn's answer that using a defaultdict will be better option here, as then you don't need to check for containment of key.

But, still relating to the way you are checking for the key in dict: -

if x1 in stable.keys():

You don't need to use stable.keys(), just use stable, that will check in keys only.

if x1 in stable: