check if a number already exist in a list in python

You could do

if item not in mylist:
     mylist.append(item)

But you should really use a set, like this :

myset = set()
myset.add(item)

EDIT: If order is important but your list is very big, you should probably use both a list and a set, like so:

mylist = []
myset = set()
for item in ...:
    if item not in myset:
        mylist.append(item)
        myset.add(item)

This way, you get fast lookup for element existence, but you keep your ordering. If you use the naive solution, you will get O(n) performance for the lookup, and that can be bad if your list is big

Or, as @larsman pointed out, you can use OrderedDict to the same effect:

from collections import OrderedDict

mydict = OrderedDict()
for item in ...:
    mydict[item] = True

If you want to have unique elements in your list, then why not use a set, if of course, order does not matter for you: -

>>> s = set()
>>> s.add(2)
>>> s.add(4)
>>> s.add(5)
>>> s.add(2)
>>> s
39: set([2, 4, 5])

If order is a matter of concern, then you can use: -

>>> def addUnique(l, num):
...     if num not in l:
...         l.append(num)
...     
...     return l

You can also find an OrderedSet recipe, which is referred to in Python Documentation