python: How come list element lookup is O(1)

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Today in class, we learned that retrieving an element from a list is O(1) in Python. Why is this the case? Suppose I have a list of 4 items ie:



li = ["perry", 1, 23.5, "s"]


These items have different sizes in memory. And so it is not possible to take the memory location of li[0] and add 3 times the size of each element to get the memory location of li[3]. So how does the interpreter know where li[3] is without having to traverse the list in order to retrieve the element?










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  • 6




    What makes you think that arrays are linearly allocated rather than a list of pointers. - [me confused by your profile description]
    – Morrison Chang
    3 hours ago







  • 9




    Don't confuse item access, which is O(1), with item lookup / search, which is O(n).
    – Mike Scotty
    3 hours ago







  • 1




    Lists don't physically contain other objects. It's almost impossible to have one object physically contain another object in Python.
    – user2357112
    3 hours ago






  • 2




    Some relevant reading material: Python list implementation, How is Python's List Implemented?, What is the underlying data structure for Python lists?.
    – Warren Weckesser
    3 hours ago










  • Python list objects are implemented as array-lists, not a linked-list structure. They are essentially arrays of PyObject pointers, in CPython.
    – juanpa.arrivillaga
    2 hours ago














up vote
6
down vote

favorite
2












Today in class, we learned that retrieving an element from a list is O(1) in Python. Why is this the case? Suppose I have a list of 4 items ie:



li = ["perry", 1, 23.5, "s"]


These items have different sizes in memory. And so it is not possible to take the memory location of li[0] and add 3 times the size of each element to get the memory location of li[3]. So how does the interpreter know where li[3] is without having to traverse the list in order to retrieve the element?










share|improve this question



















  • 6




    What makes you think that arrays are linearly allocated rather than a list of pointers. - [me confused by your profile description]
    – Morrison Chang
    3 hours ago







  • 9




    Don't confuse item access, which is O(1), with item lookup / search, which is O(n).
    – Mike Scotty
    3 hours ago







  • 1




    Lists don't physically contain other objects. It's almost impossible to have one object physically contain another object in Python.
    – user2357112
    3 hours ago






  • 2




    Some relevant reading material: Python list implementation, How is Python's List Implemented?, What is the underlying data structure for Python lists?.
    – Warren Weckesser
    3 hours ago










  • Python list objects are implemented as array-lists, not a linked-list structure. They are essentially arrays of PyObject pointers, in CPython.
    – juanpa.arrivillaga
    2 hours ago












up vote
6
down vote

favorite
2









up vote
6
down vote

favorite
2






2





Today in class, we learned that retrieving an element from a list is O(1) in Python. Why is this the case? Suppose I have a list of 4 items ie:



li = ["perry", 1, 23.5, "s"]


These items have different sizes in memory. And so it is not possible to take the memory location of li[0] and add 3 times the size of each element to get the memory location of li[3]. So how does the interpreter know where li[3] is without having to traverse the list in order to retrieve the element?










share|improve this question















Today in class, we learned that retrieving an element from a list is O(1) in Python. Why is this the case? Suppose I have a list of 4 items ie:



li = ["perry", 1, 23.5, "s"]


These items have different sizes in memory. And so it is not possible to take the memory location of li[0] and add 3 times the size of each element to get the memory location of li[3]. So how does the interpreter know where li[3] is without having to traverse the list in order to retrieve the element?







python arrays list






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edited 27 mins ago

























asked 3 hours ago









Teererai Marange

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  • 6




    What makes you think that arrays are linearly allocated rather than a list of pointers. - [me confused by your profile description]
    – Morrison Chang
    3 hours ago







  • 9




    Don't confuse item access, which is O(1), with item lookup / search, which is O(n).
    – Mike Scotty
    3 hours ago







  • 1




    Lists don't physically contain other objects. It's almost impossible to have one object physically contain another object in Python.
    – user2357112
    3 hours ago






  • 2




    Some relevant reading material: Python list implementation, How is Python's List Implemented?, What is the underlying data structure for Python lists?.
    – Warren Weckesser
    3 hours ago










  • Python list objects are implemented as array-lists, not a linked-list structure. They are essentially arrays of PyObject pointers, in CPython.
    – juanpa.arrivillaga
    2 hours ago












  • 6




    What makes you think that arrays are linearly allocated rather than a list of pointers. - [me confused by your profile description]
    – Morrison Chang
    3 hours ago







  • 9




    Don't confuse item access, which is O(1), with item lookup / search, which is O(n).
    – Mike Scotty
    3 hours ago







  • 1




    Lists don't physically contain other objects. It's almost impossible to have one object physically contain another object in Python.
    – user2357112
    3 hours ago






  • 2




    Some relevant reading material: Python list implementation, How is Python's List Implemented?, What is the underlying data structure for Python lists?.
    – Warren Weckesser
    3 hours ago










  • Python list objects are implemented as array-lists, not a linked-list structure. They are essentially arrays of PyObject pointers, in CPython.
    – juanpa.arrivillaga
    2 hours ago







6




6




What makes you think that arrays are linearly allocated rather than a list of pointers. - [me confused by your profile description]
– Morrison Chang
3 hours ago





What makes you think that arrays are linearly allocated rather than a list of pointers. - [me confused by your profile description]
– Morrison Chang
3 hours ago





9




9




Don't confuse item access, which is O(1), with item lookup / search, which is O(n).
– Mike Scotty
3 hours ago





Don't confuse item access, which is O(1), with item lookup / search, which is O(n).
– Mike Scotty
3 hours ago





1




1




Lists don't physically contain other objects. It's almost impossible to have one object physically contain another object in Python.
– user2357112
3 hours ago




Lists don't physically contain other objects. It's almost impossible to have one object physically contain another object in Python.
– user2357112
3 hours ago




2




2




Some relevant reading material: Python list implementation, How is Python's List Implemented?, What is the underlying data structure for Python lists?.
– Warren Weckesser
3 hours ago




Some relevant reading material: Python list implementation, How is Python's List Implemented?, What is the underlying data structure for Python lists?.
– Warren Weckesser
3 hours ago












Python list objects are implemented as array-lists, not a linked-list structure. They are essentially arrays of PyObject pointers, in CPython.
– juanpa.arrivillaga
2 hours ago




Python list objects are implemented as array-lists, not a linked-list structure. They are essentially arrays of PyObject pointers, in CPython.
– juanpa.arrivillaga
2 hours ago












1 Answer
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A list in Python is implemented as an array of pointers1. So, what's really happening when you create the list:



["perry", 1, 23.5, "s"]


is that you are actually creating an array of pointers like so:



[0xa3d25342, 0x635423fa, 0xff243546, 0x2545fade]


Each pointer "points" to the respective objects in memory, so that the string "perry" will be stored at address 0xa3d25342 and the number 1 will be stored at 0x635423fa, etc.



Since all pointers are the same size, the interpreter can in fact add 3 times the size of an element to the address of li[0] to get to the pointer stored at li[3].




1 Get more details from: the horse's mouth (CPython source code on GitHub).






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    9
    down vote



    accepted










    A list in Python is implemented as an array of pointers1. So, what's really happening when you create the list:



    ["perry", 1, 23.5, "s"]


    is that you are actually creating an array of pointers like so:



    [0xa3d25342, 0x635423fa, 0xff243546, 0x2545fade]


    Each pointer "points" to the respective objects in memory, so that the string "perry" will be stored at address 0xa3d25342 and the number 1 will be stored at 0x635423fa, etc.



    Since all pointers are the same size, the interpreter can in fact add 3 times the size of an element to the address of li[0] to get to the pointer stored at li[3].




    1 Get more details from: the horse's mouth (CPython source code on GitHub).






    share|improve this answer


























      up vote
      9
      down vote



      accepted










      A list in Python is implemented as an array of pointers1. So, what's really happening when you create the list:



      ["perry", 1, 23.5, "s"]


      is that you are actually creating an array of pointers like so:



      [0xa3d25342, 0x635423fa, 0xff243546, 0x2545fade]


      Each pointer "points" to the respective objects in memory, so that the string "perry" will be stored at address 0xa3d25342 and the number 1 will be stored at 0x635423fa, etc.



      Since all pointers are the same size, the interpreter can in fact add 3 times the size of an element to the address of li[0] to get to the pointer stored at li[3].




      1 Get more details from: the horse's mouth (CPython source code on GitHub).






      share|improve this answer
























        up vote
        9
        down vote



        accepted







        up vote
        9
        down vote



        accepted






        A list in Python is implemented as an array of pointers1. So, what's really happening when you create the list:



        ["perry", 1, 23.5, "s"]


        is that you are actually creating an array of pointers like so:



        [0xa3d25342, 0x635423fa, 0xff243546, 0x2545fade]


        Each pointer "points" to the respective objects in memory, so that the string "perry" will be stored at address 0xa3d25342 and the number 1 will be stored at 0x635423fa, etc.



        Since all pointers are the same size, the interpreter can in fact add 3 times the size of an element to the address of li[0] to get to the pointer stored at li[3].




        1 Get more details from: the horse's mouth (CPython source code on GitHub).






        share|improve this answer














        A list in Python is implemented as an array of pointers1. So, what's really happening when you create the list:



        ["perry", 1, 23.5, "s"]


        is that you are actually creating an array of pointers like so:



        [0xa3d25342, 0x635423fa, 0xff243546, 0x2545fade]


        Each pointer "points" to the respective objects in memory, so that the string "perry" will be stored at address 0xa3d25342 and the number 1 will be stored at 0x635423fa, etc.



        Since all pointers are the same size, the interpreter can in fact add 3 times the size of an element to the address of li[0] to get to the pointer stored at li[3].




        1 Get more details from: the horse's mouth (CPython source code on GitHub).







        share|improve this answer














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        edited 2 hours ago

























        answered 3 hours ago









        DJG

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