I am new to Swift Lang, have seen lots of tutorials, but it's not clear – my question is what's the main difference between the Array, Set and Dictionary collection type?
Here are the practical differences between the different types:
Arrays are effectively ordered lists and are used to store lists of information in cases where order is important.
For example, posts in a social network app being displayed in a tableView may be stored in an array.
Sets are different in the sense that order does not matter and these will be used in cases where order does not matter.
Sets are especially useful when you need to ensure that an item only appears once in the set.
Dictionaries are used to store key, value pairs and are used when you want to easily find a value using a key, just like in a dictionary.
For example, you could store a list of items and links to more information about these items in a dictionary.
Hope this helps :)
(For more information and to find Apple's own definitions, check out Apple's guides at https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/CollectionTypes.html)
Detailed documentation can be found here on Apple's guide. Below are some quick definations extracted from there:
Array
An array stores values of the same type in an ordered list. The same value can appear in an array multiple times at different positions.
Set
A set stores distinct values of the same type in a collection with no defined ordering. You can use a set instead of an array when the order of items is not important, or when you need to ensure that an item only appears once.
Dictionary
A dictionary stores associations between keys of the same type and values of the same type in a collection with no defined ordering. Each value is associated with a unique key, which acts as an identifier for that value within the dictionary. Unlike items in an array, items in a dictionary do not have a specified order. You use a dictionary when you need to look up values based on their identifier, in much the same way that a real-world dictionary is used to look up the definition for a particular word.
Old thread yet worth to talk about performance.
With given N element inside an array or a dictionary it worth to consider the performance when you try to access elements or to add or to remove objects.
Arrays
To access a random element will cost you the same as accessing the first or last, as elements follow sequentially each other so they are accessed directly. They will cost you 1 cycle.
Inserting an element is costly. If you add to the beginning it will cost you 1 cycle. Inserting to the middle, the remainder needs to be shifted. It can cost you as much as N cycle in worst case (average N/2 cycles). If you append to the end and you have enough room in the array it will cost you 1 cycle. Otherwise the whole array will be copied which will cost you N cycle. This is why it is important to assign enough space to the array at the beginning of the operation.
Deleting from the beginning or the end it will cost you 1. From the middle shift operation is required. In average it is N/2.
Finding element with a given property will cost you N/2 cycle.
So be very cautious with huge arrays.
Dictionaries
While Dictionaries are disordered they can bring you some benefits here. As keys are hashed and stored in a hash table any given operation will cost you 1 cycle. Only exception can be finding an element with a given property. It can cost you N/2 cycle in the worst case. With clever design however you can assign property values as dictionary keys so the lookup will cost you 1 cycle only no matter how many elements are inside.
Swift Collections - Array, Dictionary, Set
Every collection is dynamic that is why it has some extra steps for expanding and collapsing. Array should allocate more memory and copy an old date into new one, Dictionary additionally should recalculate basket indexes for every object inside
Big O (O) notation describes a performance of some function
Array - ArrayList - a dynamic array of objects. It is based on usual array. It is used for task where you very often should have an access by index
get by index - O(1)
find element - O(n) - you try to find the latest element
insert/delete - O(n) - every time a tail of array is copied/pasted
Dictionary - HashTable, HashMap - saving key/value pairs. It contains a buckets/baskets(array structure, access by index) where each of them contains another structure(array list, linked list, tree). Collisions are solved by Separate chaining. The main idea is:
calculate key's hash code[About] (Hashable) and based on this hash code the index of bucket is calculated(for example by using modulo(mod)).
Since Hashable function returns Int it can not guarantees that two different objects will have different hash codes. More over count of basket is not equals Int.max. When we have two different objects with the same hash codes, or situation when two objects which have different hash codes are located into the same basket - it is a collision. Than is why when we know the index of basket we should check if anybody there is the same as our key, and Equatable is to the rescue. If two objects are equal the key/value object will be replaces, otherwise - new key/value object will be added inside
find element - O(1) to O(n)
insert/delete - O(1) to O(n)
O(n) - in case when hash code for every object is the same, that is why we have only one bucket. So hash function should evenly distributes the elements
As you see HashMap doesn't support access by index but in other cases it has better performance
Set - hash Set. Is based on HashTable without value
*Also you are able to implement a kind of Java TreeMap/TreeSet which is sorted structure but with O(log(n)) complexity to access an element
[Java Thread safe Collections]
Related
You might have come across someplace where it is mentioned that it is faster to find elements in hashmap/dictionary/table than list/array. My question is WHY?
(inference so far I made: Why should it be faster, as far I see, in both data structure, it has to travel throughout till it reaches the required element)
Let’s reason by analogy. Suppose you want to find a specific shirt to put on in the morning. I assume that, in doing so, you don’t have to look at literally every item of clothing you have. Rather, you probably do something like checking a specific drawer in your dresser or a specific section of your closet and only look there. After all, you’re not (I hope) going to find your shirt in your sock drawer.
Hash tables are faster to search than lists because they employ a similar strategy - they organize data according to the principle that every item has a place it “should” be, then search for the item by just looking in that place. Contrast this with a list, where items are organized based on the order in which they were added and where there isn’t a a particular pattern as to why each item is where it is.
More specifically: one common way to implement a hash table is with a strategy called chained hashing. The idea goes something like this: we maintain an array of buckets. We then come up with a rule that assigns each object a bucket number. When we add something to the table, we determine which bucket number it should go to, then jump to that bucket and then put the item there. To search for an item, we determine the bucket number, then jump there and only look at the items in that bucket. Assuming that the strategy we use to distribute items ends up distributing the items more or less evenly across the buckets, this means that we won’t have to look at most of the items in the hash table when doing a search, which is why the hash table tends to be much faster to search than a list.
For more details on this, check out these lecture slides on hash tables, which fills in more of the details about how this is done.
Hope this helps!
To understand this you can think of how the elements are stored in these Data structures.
HashMap/Dictionary as you know it is a key-value data structure. To store the element, you first find the Hash value (A function which always gives a unique value to a key. For example, a simple hash function can be made by doing the modulo operation.). Then you basically put the value against this hashed key.
In List, you basically keep appending the element to the end. The order of the element insertion would matter in this data structure. The memory allocated to this data structure is not contiguous.
In Array, you can think of it as similar to List. But In this case, the memory allocated is contiguous in nature. So, if you know the value of the address for the first index, you can find the address of the nth element.
Now think of the retrieval of the element from these Data structures:
From HashMap/Dictionary: When you are searching for an element, the first thing that you would do is find the hash value for the key. Once you have that, you go to the map for the hashed value and obtain the value. In this approach, the amount of action performed is always constant. In Asymptotic notation, this can be called as O(1).
From List: You literally need to iterate through each element and check if the element is the one that you are looking for. In the worst case, your desired element might be present at the end of the list. So, the amount of action performed varies, and in the worst case, you might have to iterate the whole list. In Asymptotic notation, this can be called as O(n). where n is the number of elements in the list.
From array: To find the element in the array, what you need to know is the address value of the first element. For any other element, you can do the Math of how relative this element is present from the first index.
For example, Let's say the address value of the first element is 100. Each element takes 4 bytes of memory. The element that you are looking for is present at 3rd position. Then you know the address value for this element would be 108. Math used is
Addresses of first element + (position of element -1 )* memory used for each element.
That is 100 + (3 - 1)*4 = 108.
In this case also as you can observe the action performed is always constant to find an element. In Asymptotic notation, this can be called as O(1).
Now to compare, O(1) will always be faster than O(n). And hence retrieval of elements from HashMap/Dictionary or array would always be faster than List.
I hope this helps.
Is there a function in Fortran that deletes a specific element in an array, such that the array upon deletion shrinks its length by the number of elements deleted?
Background:
I'm currently working on a project which contain sets of populations with corresponding descriptions to the individuals (i.e, age, death-age, and so on).
A method I use is to loop through the array, find which elements I need, place it in another array, and deallocate the previous array and before the next time step, this array is moved back to the array before going through the subroutines to find once again the elements not needed.
You can use the PACK intrinsic function and intrinsic assignment to create an array value that is comprised of selected elements from another array. Assuming array is allocatable, and the elements to be removed are nominated by a logical mask logical_mask that is the same size as the original value of array:
array = PACK(array, .NOT. logical_mask)
Succinct syntax for a single element nominated by its index is:
array = [array(:index-1), array(index+1:)]
Depending on your Fortran processor, the above statements may result in the compiler creating temporaries that may impact performance. If this is problematic then you will need to use the subroutine approach that you describe.
Maybe you want to look into linked lists. You can insert and remove items and the list automatically resizes. This resource is pretty good.
http://www.iag.uni-stuttgart.de/IAG/institut/abteilungen/numerik/images/4/4c/Pointer_Introduction.pdf
To continue the discussion, the solution you might want to implement depends on the number of delete operation and access you do, where you insert/delete the elements (the first, the last, randomly in the set?), how do you access the data (from the first to the last, randomly in the set?), what are your efficiency requirements in terms of CPU and memory.
Then you might want to go for linked list or for static or dynamic vectors (other types of data structures might also fit better your needs).
For example:
a static vector can be used when you want to access a lot of elements randomly and know the maximum number nmax of elements in the vector. Simply use an array of nmax elements with an associated length variable that will track the last element. A deletion can simply and quickly be done my exchanging the last element with the deleted one and reducing the length.
a dynamic vector can be implemented when you don't know the maximum number of elements. In order to avoid systematic array allocation+copy+unallocation at for each deletion/insertion, you fix the maximum number of elements (as above) and only augment its size (eg. nmax becomes 10*nmax, then reallocate and copy) when reaching the limit (the reverse system can also be implemented to reduce the number of elements).
I was given the following question in a technical interview:
How do i remove duplicates from an unsorted array?
One option I was thinking of:
Create a hash map with the frequency of each number in the array
Go through the array and do a O(1) lookup in the hash map. If the frequency > 0, remove the number from the array.
Is there a more efficient way?
Another option
Sort the array O(nlog n) using quick sort or merge sort
Then iterate through the array and remove duplicates
Why is option 1 better than option 2?
I cannot use any functions that already do the work like array_unique.
Instead of removing the object from the array if the hash map says there is a duplicate, why don't you build a new array for each item in the hash map, and only add it to the array if there isn't a duplicate? The idea is to save the extra step of having 2 arrays with equal overhead at the start. PHP sucks at garbage collection so if you start with a massive array, even though you unset its value, it might still be hanging around in memory.
For the first option, time complexity is O(n); because creating as hash map O(n) and iterating through the array O(n), so in total O(n).
For the second option, time complexity is O(log(n)); because sort O(log(n)) and iterating O(n), so in total O(log(n)).
Clearly first option is better. Hope this helps:)
If you have no constraints on creating another data structure to track state but must mutate the array in-place and only remove duplicates without sorting, then a variant of your first option may be best.
I propose you make a hashmap as you iterate the array, use the array values as keys and any garbage (boolean set to TRUE perhaps) as the value. As you encounter each item in the array (which is O(n)), check the map. If it exists, delete the item from the array, if not add they key-value pair. No need to track count, you only need to track what has been encountered.
Many languages have a built-in set abstract data type which basically perform this operation on a construction or add all operation. If you can provide a separate data structure with duplicates removed, just create a new set with the array's items and let that data structure remove duplicates.
This is probably a common question that arises in search/store situations and there is a standard answer. I'm trying to do this from intuition and am somewhat out of my comfort zone.
I'm attempting to generate all of a certain kind of combinatorial object. Each object of size n can be generated from an object of size n-1, usually in multiple ways. From the single object of size 2, my search generates 6 objects of size 3, about 140 objects of size 4, and about 29,000 objects of size 5. As I generate the objects, I store them in a globally declared array. Before storing each object, I have to check all the previous ones stored for that size, to make sure I didn't generate it already from an earlier (n-1)-object. I currently do this in a naive way, which is just that I go through all the objects currently sitting in the array and compare them to the one currently being generated. Only if it's different from every single one there do I add it to the array and increment the number of objects currently in there. The new object is just added as the most recent object in the array, it is not sorted, and so this is obviously inefficient, and I can't hope to generate the objects of size 6 in this way.
(To give an idea of the problem of the growth of the array: the first couple of 4-objects, from among the 140 or so, give rise to over 2000 new 5-objects in a fraction of a second. By the time I've gotten to the last few 4-objects, with over 25,000 5-objects already stored, each 4-object generates only a handful of previously unseen 5-objects, but takes several seconds for the process for each 4-object. There is very little correlation between the order I generate new objects in, and their eventual position as a consequence of the comparison function I'm using.)
Obviously if I had a sorted array of objects, it would be much more efficient to find out whether I'm looking at a new object: using a binary midpoint search strategy I'd only have to look at roughly log_2(n) of the n objects currently stored, instead of all n of them. But placing the newly generated object at the right place in an array means moving half of the existing ones, on average, to make room for it. (I would implement this with an array of pointers pointing to the unsorted array of object structs, so that I only had to move pointers instead of moving data, but it still seems like a lot of pointers to have to repoint at each insert.)
The other option would be to place the objects in a linked list, as insertion is very cheap in that situation. But then I wouldn't have random access to the elements in the linked list--you can only find the right place to insert the newly generated object (if it's actually new) by traversing the list node by node and comparing. On average you'd have to traverse half the list before finding the right insertion point, which doesn't sound any better than repointing half the pointers.
Is there a third choice I'm missing? I could accomplish this very easily if I had both random access to stored elements so I could find the insertion point quickly (in log_2(n) steps), and I could insert new objects very cheaply, like in a linked list. Am I dreaming?
To summarise: I need to be able to determine whether an object is new or duplicates an existing one, and I need to be able to insert an object at the right place. I don't ever need to delete an object. Thank you.
Say, I have a sorted array of n elements. I want to find 2 different keys k1 and k2 in this array using Binary search.
A basic solution would be to apply Binary search on them separately, like two calls for 2 keys which would maintain the time complexity to 2(logn).
Can we solve this problem using any other approach(es) for different k keys, k < n ?
Each search you complete can be used to subdivide the input to make it more efficient. For example suppose the element corresponding to k1 is at index i1. If k2 > k1 you can restrict the second search to i1..n, otherwise restrict it to 0..i1.
Best case is when your search keys are sorted also, so every new search can begin where the last one was found.
You can reduce the real complexity (although it will still be the same big O) by walking the shared search path once. That is, start the binary search until the element you're at is between the two items you are looking for. At that point, spawn a thread to continue the binary search for one element in the range past the pivot element you're at and spawn a thread to continue the binary search for the other element in the range before the pivot element you're at. Return both results. :-)
EDIT:
As Oli Charlesworth had mentioned in his comment, you did ask for an arbitrary amount of elements. This same logic can be extended to an arbitrary amount of search keys though. Here is an example:
You have an array of search keys like so:
searchKeys = ['findme1', 'findme2', ...]
You have key-value datastructure that maps a search key to the value found:
keyToValue = {'findme1': 'foundme1', 'findme2': 'foundme2', 'findme3': 'NOT_FOUND_VALUE'}
Now, following the same logic as before this EDIT, you can pass a "pruned" searchKeys array on each thread spawn where the keys diverge at the pivot. Each time you find a value for the given key, you update the keyToValue map. When there are no more ranges to search but still values in the searchKeys array, you can assume those keys are not to be found and you can update the mapping to signify that in some way (some null-like value perhaps?). When all threads have been joined (or by use of a counter), you return the mapping. The big win here is that you did not have to repeat the initial search logic that any two keys may share.
Second EDIT:
As Mark has added in his answer, sorting the search keys allows you to only have to look at the first item in the key range.
You can find academic articles calculating the complexity of different schemes for the general case, which is merging two sorted sequences of possibly very different lengths using the minimum number of comparisons. The paper at http://www.math.cmu.edu/~af1p/Texfiles/HL.pdf analyses one of the best known schemes, by Hwang and Lin, and has references to other schemes, and to the original paper by Hwang and Lin.
It looks a lot like a merge which steps through each item of the smaller list, skipping along the larger list with a stepsize that is the ratio of the sizes of the two lists. If it finds out that it has stepped too far along the large list it can use binary search to find a match amongst the values it has stepped over. If it has not stepped far enough, it takes another step.