check if a number already exist in a list in python

175,610

Solution 1

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

Solution 2

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

Solution 3

If you want your numbers in ascending order you can add them into a set and then sort the set into an ascending list.

s = set()
if number1 not in s:
  s.add(number1)
if number2 not in s:
  s.add(number2)
...
s = sorted(s)  #Now a list in ascending order

Solution 4

You could probably use a set object instead. Just add numbers to the set. They inherently do not replicate.

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PhoonOne
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PhoonOne

Updated on December 23, 2021

Comments

  • PhoonOne
    PhoonOne over 2 years

    I am writing a python program where I will be appending numbers into a list, but I don't want the numbers in the list to repeat. So how do I check if a number is already in the list before I do list.append()?