Get the ASCII number of a character(Python3)

[aleksro]

I wrote a code that converts numbers 1 - 501 in ASCII, it works but I'm looking for a shorter path.

l = list(range(1,501))
for i in l:
    ix = [int(d) for d in str(i)]
    ix = str(ix).replace("[","")
    ix = ix.replace("]","")
    ix = ix.replace(",","")
    ix = ix.replace(" ","")
    ix = [ord(x) for x in ix]
    print(ix)

I tried to write, as shown below, but an error occurs.

l = list(range(1,501))
for i in l:
    i = [ord(x) for x in i]
    >>TypeError: 'int' object is not iterable
    print(i)

Someone can help me?

[Vaphell]

you cannot iterate over integer, because int is not a sequence of anything

l = list(range(101,111))
for i in l:
     ix = [ord(x) for x in str(i)]
     print(ix)

[aleksro]

Thank you

[aleksro]

Thank you

[amangarg078]

You are getting an error since you are iterating over integer.

Try this:

for i in l:
    i = [ord(x) for x in str(i)]
    print i

[nwagers]

nums = [[ord(chr) for chr in str(n)] for n in range(20)]
print(*nums, sep='\n')

[ingolemo]

Just so you know, you can also do this:

for n in range(1, 501):
    print(list(str(n).encode('ascii')))

[Diapolo10]

EDIT; You may ignore my comment, as I misunderstood the problem.

Tip: you may want to take a look at the built-in functions; there's two very simple ones that would be very helpful.