unordered

来源： https://stackoverflow.com/questions/39398330/nesting-a-ul-inside-an-ol

来源： https://stackoverflow.com/questions/39398330/nesting-a-ul-inside-an-ol

问题 The following list has duplicated sublists. However, they are in different order: l1 = [['The', 'quick', 'brown', 'fox'], ['hi', 'there'], ['jumps', 'over', 'the', 'lazy', 'dog'], ['there', 'hi'], ['jumps', 'dog', 'over','lazy', 'the']] How can I remove them in order to get: l1 = [['The', 'quick', 'brown', 'fox'], ['hi', 'there'], ['jumps', 'over', 'the', 'lazy', 'dog']] I tried to: [list(i) for i in set(map(tuple, l1))] Nevertheless, I do not know if this is the fastest way of doing it for

问题 I have 2 arrays of objects and I have to compare them, but the order of the objects DOES NOT matter. I can't sort them because I won't have their keys' names because the functions must be generic. The only information that I'll have about the array is that both array's objects have the same amount of keys and those keys have the same name. So the array1 must contain the same objects as the array2. var array1 = [{"key1":"Banana", "key2":"Yammy"}, {"key1":"Broccoli", "key2":"Ew"}]; var array2 =

问题 Please link me to answer if this has already been answered, my problem is i want to get diff of multilevel json which is unordered. x=json.loads('''[{"y":2,"x":1},{"x":3,"y":4}]''') y=json.loads('''[{"x":1,"y":2},{"x":3,"y":4}]''') z=json.loads('''[{"x":3,"y":4},{"x":1,"y":2}]''') import json_tools as jt import json_delta as jd print jt.diff(y,z) print jd.diff(y,z) print y==z print x==y output is [{'prev': 2, 'value': 4, 'replace': u'/0/y'}, {'prev': 1, 'value': 3, 'replace': u'/0/x'}, {'prev

问题 To narrow it down: I'm currently using Boost.Unordered. I see two possible solutions: Define my own Equality Predicates and Hash Functions and to utilize templates (maybe is_pointer ) to distinct between pointers and instances; Simply to extend boost::hash by providing hash_value(Type* const& x) as for hashing; and add == operator overload as free function with (Type* const& x, Type* const& y) parameters as for equality checking. I'm not sure whether both variations are actually possible,

问题 This question already has answers here : How can I access and process nested objects, arrays or JSON? (23 answers) Closed 5 years ago . I need Help regarding the traversal of each and every elements in the following JSON and get both key and value and print it like an unordered list. var dataSource = ({ "Items": ({ "Deserts": ({}), "Veg": ({ "VegPulao": "Veg Pulao", "PalakPaneer": "Palak Paneer", "PaneerButterMasala": "Paneer Butter Masala" }), "Chicken": ({ "Tandoori": "Tandoori special" }),

问题 I'd like to calculate a hash of a set of strings in Java. Yes I can sort the strings and calculate the MD5 hash iterative using digest.update . But I'd prefer to omit the sort and use something like combineUnordered https://github.com/google/guava/wiki/HashingExplained There is a lot of similar question asking the same such as Order-independant Hash Algorithm but non of them provides a simple example showing how to calculate iterative an order independent hash in Java. 回答1: Just XOR each hash

问题 This is for a "Meet Our Staff" page where there is a vertical unordered list of small images associated with a staff member (#staffDirectory). When the user clicks a different staff member than the one displayed, I want the current large listitem (which includes an image and info about the member - and is assigned the class "staffSelected" and is displayed in the div #staffMember) to fadeOut, lose the class, then take the corresponding listitem (the one the user clicked in #staffDirectory),

问题 To put an element into, say, an unordered set we calculate its hash and put it to the corresponding bucket. However we usually have many fewer buckets than the range of values of the hash function. How is the correspondence of buckets and hash values calculated? It seems like some function is used reflecting (0 ... size_t) -> (0 ... size_of_buckets - 1). But using such a function could lead to big number of collisions even for good hash function. 回答1: I'm not sure whether std::unordered_map