Ultimately it probably doesn't have a safe .get method because a dict is an associative collection (values are associated with names) where it is inefficient to check if a key is present (and return its value) without throwing an exception, while it is super trivial to avoid exceptions accessing list elements (as the len method is very fast). The .get method allows you to query the value associated with a name, not directly access the 37th item in the dictionary (which would be more like what you're asking of your list).

Of course, you can easily implement this yourself:

def safe_list_get (l, idx, default):
  try:
    return l[idx]
  except IndexError:
    return default

You could even monkeypatch it onto the __builtins__.list constructor in __main__, but that would be a less pervasive change since most code doesn't use it. If you just wanted to use this with lists created by your own code you could simply subclass list and add the get method.

Answer from Nick Bastin on Stack Overflow
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I'm coding a space shooter game with pygame python 3.9.13 (if it can help) and i can't go around it because i need a list with my class 'Bulet' and i can't found a solution, code: from pygame import * ... ... I am trying to create a random numpy array using np.random() but for some reason, instead of taking it as a function, google collab is taking it as a module, I checked the documentation, but I would ... ... Upcoming initiatives on Stack Overflow ... More on stackoverflow.com
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January 21, 2025 - After going over the top questions on Pandas and Python dictionaries, in this article, we will go through the 10 most frequently asked Python list questions on Stack Overflow. List **** is a built-in data structure in Python. It is represented as a collection of data points in square brackets and can be used for storing data of different types. I searched for the questions using "python" and "list" tags and sorted them by score. Let’s start. This question can be generalized to other iterables such as tuples and strings. It’s basically asking how to get the index of an item along with the item itself.
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I'm coding a space shooter game with pygame python 3.9.13 (if it can help) and i can't go around it because i need a list with my class 'Bulet' and i can't found a solution, code: from pygame import * ... ... I am trying to create a random numpy array using np.random() but for some reason, instead of taking it as a function, google collab is taking it as a module, I checked the documentation, but I would ... ... Upcoming initiatives on Stack Overflow ...
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3. Using queue.LifoQueue: LifoQueue from the queue module implements a stack using the LIFO principle. It is thread-safe which makes it useful in multi-threaded programs. However, it is generally slower than lists and deque. Example: Here, elements are inserted using put() and removed using get().
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August 29, 2023 - To access an item in a dictionary, ... avoid this, you can use .get and pass in a default value to return instead: d.get("some_key", "default value"). Lists don’t have such a method for safely indexing....
Top answer
1 of 7
55

Lists are a mutable type - in order to create a copy (rather than just passing the same list around), you need to do so explicitly:

listoflists.append((list[:], list[0]))

However, list is already the name of a Python built-in - it'd be better not to use that name for your variable. Here's a version that doesn't use list as a variable name, and makes a copy:

listoflists = []
a_list = []
for i in range(0,10):
    a_list.append(i)
    if len(a_list)>3:
        a_list.remove(a_list[0])
        listoflists.append((list(a_list), a_list[0]))
print listoflists

Note that I demonstrated two different ways to make a copy of a list above: [:] and list().

The first, [:], is creating a slice (normally often used for getting just part of a list), which happens to contain the entire list, and thus is effectively a copy of the list.

The second, list(), is using the actual list type constructor to create a new list which has contents equal to the first list. (I didn't use it in the first example because you were overwriting that name in your code - which is a good example of why you don't want to do that!)

2 of 7
25

I came here because I'm new with python and lazy so I was searching an example to create a list of 2 lists, after a while a realized the topic here could be wrong... This is a code to create a list of lists:

listoflists = []
for i in range(0,2):
    sublist = []
    for j in range(0,10)
        sublist.append((i,j))
    listoflists.append(sublist)
print listoflists

this is the output:

[
    [(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9)],
    [(1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9)]
]

The problem with your code seems to be you are creating a tuple with your list and you get the reference to the list instead of a copy. That I guess should fall under a tuple topic...

Top answer
1 of 14
1828

As for your first question: "if item is in my_list:" is perfectly fine and should work if item equals one of the elements inside my_list. The item must exactly match an item in the list. For instance, "abc" and "ABC" do not match. Floating point values in particular may suffer from inaccuracy. For instance, 1 - 1/3 != 2/3.

As for your second question: There's actually several possible ways if "finding" things in lists.

Checking if something is inside

This is the use case you describe: Checking whether something is inside a list or not. As you know, you can use the in operator for that:

3 in [1, 2, 3] # => True

Filtering a collection

That is, finding all elements in a sequence that meet a certain condition. You can use list comprehension or generator expressions for that:

matches = [x for x in lst if fulfills_some_condition(x)]
matches = (x for x in lst if x > 6)

The latter will return a generator which you can imagine as a sort of lazy list that will only be built as soon as you iterate through it. By the way, the first one is exactly equivalent to

matches = filter(fulfills_some_condition, lst)

in Python 2. Here you can see higher-order functions at work. In Python 3, filter doesn't return a list, but a generator-like object.

Finding the first occurrence

If you only want the first thing that matches a condition (but you don't know what it is yet), it's fine to use a for loop (possibly using the else clause as well, which is not really well-known). You can also use

next(x for x in lst if ...)

which will return the first match or raise a StopIteration if none is found. Alternatively, you can use

next((x for x in lst if ...), [default value])

Finding the location of an item

For lists, there's also the index method that can sometimes be useful if you want to know where a certain element is in the list:

[1,2,3].index(2) # => 1
[1,2,3].index(4) # => ValueError

However, note that if you have duplicates, .index always returns the lowest index:......

[1,2,3,2].index(2) # => 1

If there are duplicates and you want all the indexes then you can use enumerate() instead:

[i for i,x in enumerate([1,2,3,2]) if x==2] # => [1, 3]
2 of 14
281

If you want to find one element or None use default in next, it won't raise StopIteration if the item was not found in the list:

first_or_default = next((x for x in lst if ...), None)
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Top answer
1 of 11
2992

The len() function can be used with several different types in Python - both built-in types and library types. For example:

>>> len([1, 2, 3])
3
2 of 11
323

How do I get the length of a list?

To find the number of elements in a list, use the builtin function len:

items = []
items.append("apple")
items.append("orange")
items.append("banana")

And now:

len(items)

returns 3.

Explanation

Everything in Python is an object, including lists. All objects have a header of some sort in the C implementation.

Lists and other similar builtin objects with a "size" in Python, in particular, have an attribute called ob_size, where the number of elements in the object is cached. So checking the number of objects in a list is very fast.

But if you're checking if list size is zero or not, don't use len - instead, put the list in a boolean context - it is treated as False if empty, and True if non-empty.

From the docs

len(s)

Return the length (the number of items) of an object. The argument may be a sequence (such as a string, bytes, tuple, list, or range) or a collection (such as a dictionary, set, or frozen set).

len is implemented with __len__, from the data model docs:

object.__len__(self)

Called to implement the built-in function len(). Should return the length of the object, an integer >= 0. Also, an object that doesn’t define a __nonzero__() [in Python 2 or __bool__() in Python 3] method and whose __len__() method returns zero is considered to be false in a Boolean context.

And we can also see that __len__ is a method of lists:

items.__len__()

returns 3.

Builtin types you can get the len (length) of

And in fact we see we can get this information for all of the described types:

>>> all(hasattr(cls, '__len__') for cls in (str, bytes, tuple, list, 
                                            range, dict, set, frozenset))
True

Do not use len to test for an empty or nonempty list

To test for a specific length, of course, simply test for equality:

if len(items) == required_length:
    ...

But there's a special case for testing for a zero length list or the inverse. In that case, do not test for equality.

Also, do not do:

if len(items): 
    ...

Instead, simply do:

if items:     # Then we have some items, not empty!
    ...

or

if not items: # Then we have an empty list!
    ...

I explain why here but in short, if items or if not items is more readable and performant than other alternatives.

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