How exactly does Python check through a list?

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I was doing one of the course exercises on codeacademy for python and I had a few questions I couldn't seem to find an answer to:

For this block of code, how exactly does python check whether something is "in" or "not in" a list? Does it run through each item in the list to check or does it use a quicker process?

Also, how would this code be affected if it were running with a massive list of numbers (thousands or millions)? Would it slow down as the list size increases, and are there better alternatives?

numbers = [1, 1, 2, 3, 5, 8, 13]

def remove_duplicates(list):
  new_list = []
  for i in list: 
    if i not in new_list:
      new_list.append(i)
  return new_list

remove_duplicates(numbers)

Thanks!

P.S. Why does this code not function the same?

numbers = [1, 1, 2, 3, 5, 8, 13]

def remove_duplicates(list):
  new_list = []
  new_list.append(i for i in list if i not in new_list)
  return new_list

In order to execute i not in new_list Python has to do a linear scan of the list. The scanning loop breaks as soon as the result of the test is known, but if i is actually not in the list the whole list must be scanned to determine that. It does that at C speed, so it's faster than doing a Python loop to explicitly check each item. Doing the occasional in some_list test is ok, but if you need to do a lot of such membership tests it's much better to use a set.

On average, with random data, testing membership has to scan through half the list items, and in general the time taken to perform the scan is proportional to the length of the list. In the usual notation the size of the list is denoted by n, and the time complexity of this task is written as O(n).

In contrast, determining membership of a set (or a dict) can be done (on average) in constant time, so its time complexity is O(1). Please see TimeComplexity in the Python Wiki for further details on this topic. Thanks, Serge, for that link.

Of course, if your using a set then you get de-duplication for free, since it's impossible to add duplicate items to a set.

One problem with sets is that they generally don't preserve order. But you can use a set as an auxilliary collection to speed up de-duping. Here is an illustration of one common technique to de-dupe a list, or other ordered collection, which does preserve order. I'll use a string as the data source because I'm too lazy to type out a list. ;)

new_list = []
seen = set()
for c in "this is a test":
    if c not in seen:
        new_list.append(c)
        seen.add(c)
print(new_list)

output

['t', 'h', 'i', 's', ' ', 'a', 'e']

Please see How do you remove duplicates from a list whilst preserving order? for more examples. Thanks, Jean-François Fabre, for the link.


As for your PS, that code appends a single generator object to new_list, it doesn't append what the generate would produce.

I assume you alreay tried to do it with a list comprehension:

new_list = [i for i in list if i not in new_list]

That doesn't work, because the new_list doesn't exist until the list comp finishes running, so doing in new_list would raise a NameError. And even if you did new_list = [] before the list comp, it won't be modified by the list comp, and the result of the list comp would simply replace that empty list object with a new one.


BTW, please don't use list as a variable name (even in example code) since that shadows the built-in list type, which can lead to mysterious error messages.

Python List Contains, How do you check if something is in a list in Python? How To Check If a List Is Empty in Python. an empty list or an implicit evaluation of an empty list. What does that mean? to see if our list in question is exactly the same as an empty list.

You are asking multiple questions and one of them asking if you can do this more efficiently. I'll answer that.

Ok let's say you'd have thousands or millions of numbers. From where exactly? Let's say they were stored in some kind of txtfile, then you would probably want to use numpy (if you are sticking with Python that is). Example:

import numpy as np

numbers = np.array([1, 1, 2, 3, 5, 8, 13], dtype=np.int32)
numbers = np.unique(numbers).tolist()

This will be more effective (above all memory-efficient compared) than reading it with python and performing a list(set..)

numbers = [1, 1, 2, 3, 5, 8, 13]
numbers = list(set(numbers))

How to check or find if a value is in a list in Excel?, How do you check if a value is in a list? Let’s fix this inefficiency by turning our list comprehension into a generator expression. Generator expressions. A generator expression is like a list comprehension, but instead of making a list it makes a generator (Python chat on generators). A generator is a lazy iterable: generators don’t compute the items they contain until you loop over them. We’ll see what that means in a moment.

You are asking for the algorithmic complexity of this function. To find that you need to see what is happening at each step.

You are scanning the list one at a time, which takes 1 unit of work. This is because retrieving something from a list is O(1). If you know the index, it can be retrieved in 1 operation.

The list to which you are going to add it increases at worst case 1 at a time. So at any point in time, the unique items list is going to be of size n.

Now, to add the item you picked to the unique items list is going to take n work in the worst case. Because we have to scan each item to decide that.

So if you sum up the total work in each step, it would be 1 + 2 + 3 + 4 + 5 + ... n which is n (n + 1) / 2. So if you have a million items, you can just find that by applying n = million in the formula.


This is not entirely true because of how list works. But theoretically, it would help to visualize this way.

Python - Check if all elements in a List are same, How do you check if a number is repeated in a list Python? A list is a data structure in Python that is a mutable, or changeable, ordered sequence of elements. Each element or value that is inside of a list is called an item. Each element or value that is inside of a list is called an item.

to answer the question in the title: python has more efficient data types but the list() object is just a plain array, if you want a more efficient way to search values you can use dict() which uses a hash of the object stored to insert it into a tree which i assume is what you were thinking of when you mentioned "a quicker process".

as to the second code snippet: list().append() inserts whatever value you give it to the end of the list, i for i in list if i not in new_list is a generator object and it inserts that generator as an object into the array, list().extend() does what you want: it takes in an iterable and appends all of its elements to the list

Python, Python in is the most conventional way to check if an element exists in list or not. This particular way returns True if element exists in list and False if the element does not exists in list. List need not be sorted to practice this approach of checking. 4 Answers 4. In order to execute i not in new_list Python has to do a linear scan of the list. The scanning loop breaks as soon as the result of the test is known, but if i is actually not in the list the whole list must be scanned to determine that. It does that at C speed, so it's faster than doing a Python loop to explicitly check each item.

Python, Given an object, the task is to check whether the object is list or not. Method #1: Using isinstance. filter_none. edit close. play_arrow. link brightness_4 code  Then using zip(), we combine the full list ([2, 2, 2, 3]) and the list without the first element ([2, 2, 3]) as follows: [(2, 2), (2, 2 ), (2, 3)]. The comparison of elements among themselves will pass to the all() function the set [True, True, False], and as a result we get False, which is the correct answer, since not all elements in the input list are the same.

18 Most Common Python List Questions, The list type is a container that holds a number of other objects, in a given order. And Python never creates a new list if you assign a list to a variable. The in operator can be used to check if an item is present in the list: The time needed to insert an item depends on the size of the list, or more exactly, how many items​  Python method listdir() returns a list containing the names of the entries in the directory given by path. The list is in arbitrary order. The list is in arbitrary order. It does not include the special entries '.' and '..' even if they are present in the directory.

An Introduction to Python Lists, We're creating one iterable from another which is exactly what list comprehensions are good for. Let's copy-paste our way into a list  Similar to the previous post, this article assumes no prior knowledge of statistics, but does require at least a general knowledge of Python and general data science worflows. If you are uncomfortable with for loops and lists, I recommend covering them briefly in our free introductory Python course before progressing.

Comments
  • Your second question is answered by changing append to extend.
  • By the way list(set(numbers)) also removes duplicates
  • BTW, on Stack Exchange sites it's best to ask one question per question. That makes your question more useful to future readers. But I guess it's ok here, since the two questions are related.
  • faster while preserving order: stackoverflow.com/questions/480214/…
  • wiki.python.org/moin/TimeComplexity
  • Small detail: set won't preserve order.
  • you can link to that post then: stackoverflow.com/questions/480214/…
  • Good thinking, @Serge!
  • Thanks for the help! This thoroughly answered my questions!
  • @PM2Ring That was me not realizing that you can only choose one "check-mark"