Hello, need to solve a problem like such:

Question

If the number in the line is already wroten, I must print 0 and if it's not, I must print 1. Got ValueError and tried to catch it, but it fails.

a=int(input())
b=input()
n='1 '
c=b.split()
for i in range(a-1):
 l=c[0]
 c.remove(l)
 d=c.index(l)
 try:
  a=a
 except ValueError as ve:
  if d > 0:
   c = c.replace(l,'zero')
  else:
   c=c.replace(l,'one')
for amogus in range(a):
 a=a.replace('one',1)
 a=a.replace('zero',0)
print(a)

The error tracebacks like so:

Traceback (most recent call last):
  File "program.pys3", line 8, in <module>
    d=c.index(l)
ValueError: '1' is not in list

Answer 1

The first part of your loop is like this:

for i in range(a-1):
    l = c[0]
    c.remove(l)
    d = c.index(l)

l is set to the first item in the list c. You then remove that item. The last line above then tries to find the index of that item again. This will work if there is another one, but if not will throw the error you report.

Did you mean to put the last line inside the try:?

for i in range(a-1):
    l = c[0]
    c.remove(l)
    try:
        d = c.index(l)
    except ValueError as ve:
        if d > 0:
            c = c.replace(l,'zero')
        else:
            c = c.replace(l,'one')

Now a friendly warning: if this fixes the reported error you will encounter another one.