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# confused about resizing array in Python

 P: n/a My confusion comes from the following piece of code: memo = {1:1, 2:1} def fib_memo(n): global memo if not n in memo: memo[n] = fib_memo(n-1) + fib_memo(n-2) return memo[n] I used to think that the time complexity for this code is O(n) due to its use of memoization. However, I was told recently that in Python, dictionary is a special kind of array and to append new element to it or to resize it, it is in fact internally inplemented by creating another array and copying the old one to it and append a new one. Therefore, for "memo[n] = fib_memo(n-1) + fib_memo(n-2)", the time it taks is not at all constant. The larger the n grows, the more time this statement takes. Can anybody here familiar with the internal mechanism of python confirm this? Feb 3 '07 #1
10 Replies

 P: n/a Ruan schreef: My confusion comes from the following piece of code: memo = {1:1, 2:1} def fib_memo(n): global memo if not n in memo: memo[n] = fib_memo(n-1) + fib_memo(n-2) return memo[n] I used to think that the time complexity for this code is O(n) due to its use of memoization. However, I was told recently that in Python, dictionary is a special kind of array and to append new element to it or to resize it, it is in fact internally inplemented by creating another array and copying the old one to it and append a new one. That's not correct. Python dictionaries are highly optimized and I believe the time complexity is amortized constant (i.e. O(1)) for both insertions and lookups. -- If I have been able to see further, it was only because I stood on the shoulders of giants. -- Isaac Newton Roel Schroeven Feb 3 '07 #2

 P: n/a Then how about Python's list? What is done exactly when list.append is executed? For list, is there another larger list initialized and the contents from the old list is copied to it together with the new appended list? "Roel Schroeven"

 P: n/a On Feb 4, 7:41 am, "Ruan"

 P: n/a Ruan schreef: "Roel Schroeven" Ruan schreef: >>My confusion comes from the following piece of code: memo = {1:1, 2:1}def fib_memo(n):global memoif not n in memo:memo[n] = fib_memo(n-1) + fib_memo(n-2)return memo[n]I used to think that the time complexity for this code is O(n) due toits use of memoization.However, I was told recently that in Python, dictionary is a specialkind of array and to append new element to it or to resize it, it is in factinternally inplemented by creating another array and copying the old one toit and append a new one. >That's not correct. Python dictionaries are highly optimized and Ibelieve the time complexity is amortized constant (i.e. O(1)) for bothinsertions and lookups. Then how about Python's list? What is done exactly when list.append is executed? For list, is there another larger list initialized and the contents from the old list is copied to it together with the new appended list? I'm not sure, but I think each time the list needs to grow, it doubles in size. That leaves room to add a number of elements before the allocated space needs to grow again. It's a frequently used approach, since it is quite efficient and the memory needed is never double the amount of memory strictly needed for the elements of the list. You can always study the source code for all gory details of course. -- If I have been able to see further, it was only because I stood on the shoulders of giants. -- Isaac Newton Roel Schroeven Feb 3 '07 #5

 P: n/a You mentioned "it doubles in size". Are you saying that a new double sized array is allocated and the contents of the old list is copied there? Then the old list is freed from memory? It seems to be what is called amortized constant. Say the list size is 100, before it is fully used, the append takes O(1) time. But for the 101th element, the time will be O(100+1), and then from then on, it is O(1) again. Like John Machin said in the previous post? But on average, it is O(1). I guess this is the amortized constant. Isn't it? "Roel Schroeven" "Roel Schroeven" >Ruan schreef:My confusion comes from the following piece of code: memo = {1:1, 2:1}def fib_memo(n):global memoif not n in memo:memo[n] = fib_memo(n-1) + fib_memo(n-2)return memo[n]I used to think that the time complexity for this code is O(n) due toits use of memoization.However, I was told recently that in Python, dictionary is a specialkind of array and to append new element to it or to resize it, it is infactinternally inplemented by creating another array and copying the oldone toit and append a new one. >>That's not correct. Python dictionaries are highly optimized and Ibelieve the time complexity is amortized constant (i.e. O(1)) for bothinsertions and lookups. >Then how about Python's list?What is done exactly when list.append is executed?For list, is there another larger list initialized and the contents fromtheold list is copied to it together with the new appended list? I'm not sure, but I think each time the list needs to grow, it doubles in size. That leaves room to add a number of elements before the allocated space needs to grow again. It's a frequently used approach, since it is quite efficient and the memory needed is never double the amount of memory strictly needed for the elements of the list. You can always study the source code for all gory details of course. -- If I have been able to see further, it was only because I stood on the shoulders of giants. -- Isaac Newton Roel Schroeven Feb 3 '07 #6

 P: n/a Dongsheng Ruan schreef: "Roel Schroeven" Ruan schreef: >>Then how about Python's list?What is done exactly when list.append is executed?For list, is there another larger list initialized and the contents fromthe old list is copied to it together with the new appended list? >I'm not sure, but I think each time the list needs to grow, it doubles insize. That leaves room to add a number of elements before the allocatedspace needs to grow again. It's a frequently used approach, since it isquite efficient and the memory needed is never double the amount of memorystrictly needed for the elements of the list. You mentioned "it doubles in size". Are you saying that a new double sized array is allocated and the contents of the old list is copied there? Then the old list is freed from memory? It seems to be what is called amortized constant. Say the list size is 100, before it is fully used, the append takes O(1) time. But for the 101th element, the time will be O(100+1), and then from then on, it is O(1) again. Like John Machin said in the previous post? But on average, it is O(1). I guess this is the amortized constant. Isn't it? I think so, more or less, but as I said I'm not entirely sure about how Python handles lists. One thing to keep in mind is that the list (like any other Python data structure) doesn't store the objects themselves; it only stores references to the objects. If the list needs to be copied, only the references are copied; the objects themselves can stay where they are. For small objects this doesn't make much difference, but if the objects grow larger it gets much more efficient if you only have to move the references around. -- If I have been able to see further, it was only because I stood on the shoulders of giants. -- Isaac Newton Roel Schroeven Feb 4 '07 #7

 P: n/a This seems to be clever to use reference for list. Is it unique to Python? How about the traditional programming languages like C, Pascal or C++? "Roel Schroeven" "Roel Schroeven" >Ruan schreef:Then how about Python's list?What is done exactly when list.append is executed?For list, is there another larger list initialized and the contentsfrom the old list is copied to it together with the new appended list? >>I'm not sure, but I think each time the list needs to grow, it doublesin size. That leaves room to add a number of elements before theallocated space needs to grow again. It's a frequently used approach,since it is quite efficient and the memory needed is never double theamount of memory strictly needed for the elements of the list. You mentioned "it doubles in size". Are you saying that a new double sized array is allocated and the contents of the old list is copied there? Then the old list is freed from memory? It seems to be what is called amortized constant. Say the list size is 100, before it is fully used, the append takes O(1) time. But for the 101th element, the time will be O(100+1), and then from then on, it is O(1) again. Like John Machin said in the previous post? But on average, it is O(1). I guess this is the amortized constant. Isn't it? I think so, more or less, but as I said I'm not entirely sure about how Python handles lists. One thing to keep in mind is that the list (like any other Python data structure) doesn't store the objects themselves; it only stores references to the objects. If the list needs to be copied, only the references are copied; the objects themselves can stay where they are. For small objects this doesn't make much difference, but if the objects grow larger it gets much more efficient if you only have to move the references around. -- If I have been able to see further, it was only because I stood on the shoulders of giants. -- Isaac Newton Roel Schroeven Feb 4 '07 #8

 P: n/a In , Dongsheng Ruan wrote: This seems to be clever to use reference for list. Is it unique to Python? No of course not. Java is very similar in only passing references around for objects. And `ArrayList` and `Vector` behave similar to Python lists. How about the traditional programming languages like C, Pascal or C++? For a start they don't have a built in list type. C and Pascal don't even have one in the standard library. C++ has STL vectors and if you, the programmer, decide to store pointers in it instead of structures or objects then you have something like Python's list type. Ciao, Marc 'BlackJack' Rintsch Feb 4 '07 #9

 P: n/a On 2007-02-04, Marc 'BlackJack' Rintsch How about the traditional programming languages like C, Pascalor C++? For a start they don't have a built in list type. C and Pascal don't even have one in the standard library. C++ has STL vectors and if you, the programmer, decide to store pointers in it instead of structures or objects then you have something like Python's list type. You need to store some form of smart pointer (rather than a bare pointer) in C++ standard containers in order to avoid heart, head and stomach aches. A reference counted pointer type will come fairly close to Python semantics. -- Neil Cerutti Eddie Robinson is about one word: winning and losing. --Eddie Robinson's agent Paul Collier Feb 5 '07 #10

 P: n/a En Sat, 03 Feb 2007 21:34:19 -0300, Dongsheng Ruan escribió: This seems to be clever to use reference for list. Is it unique to Python? How about the traditional programming languages like C, Pascal or C++? Python is written in C - so obviously it can be done in plain C. Delphi (Pascal) has a similar thing; lists hold only a reference to the object, and grow in discrete steps when needed. And in C++ you have several container variants in the STL to choose from. In all cases, there is a library behind, and a fairly good amount of code. The good news is that it's already done for python: you get a lot of data structures ready to use in Python. -- Gabriel Genellina Feb 6 '07 #11

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