[Solved] React Javascript Array Object [closed]
You need to parse that object with JSON.parse(string) first. 1 solved React Javascript Array Object [closed]
You need to parse that object with JSON.parse(string) first. 1 solved React Javascript Array Object [closed]
You could use Array.FindIndex and do this. var array = new string [] {“windows”,”dual sim”,”32 gb”,”Intel i5″}; string searchString = “android 32 gb”; var index = Array.FindIndex(array, x=> searchString.IndexOf(x) >=0); if you are looking for case insensitive search, use this. var index = Array.FindIndex(array, x=> searchString.IndexOf(x, StringComparison.CurrentCultureIgnoreCase) >=0); Check this Demo 8 solved How to … Read more
The name stands on the image. Mappr. The website is mappr.io. But I’m not sure if the product is still available. The company behind Mappr, Vibrand Data, has been acquired by Slice Technologies. And I don’t find any informations about Mappr there. So, good luck wile searching … Sources: http://mappr.io http://vibrantdata.io http://slice.com http://blog.slice.com 1 solved … Read more
then you can make a loop to make a dictionary that later will be appended to arrayDictionary let array = [[String:Any]]() for item in data { let dictionary = [String:Any]() dictionary[“object1”] = item.id dictionary[“object2”] = item.name dictionary[“object3”] = item.address array.append(dictionary) } 2 solved How to write nested dictionary in Swift 4 [closed]
With the regular expression ^ – start of string (.*) – zero or more characters (Groups[1]) \s – whitespace (\S*) – zero or more non-whitespace characters (Groups[2]) $ – end of string it is possible to split a string into 2 groups, using the last whitespace in the string as the delimiter. Using this regular … Read more
Writing a function to print letters of a string according to their frequency compalsarily using dictionary and tuple [closed] solved Writing a function to print letters of a string according to their frequency compalsarily using dictionary and tuple [closed]
To sum the values, you can loop over the data using .items(), will return the key with its associated value at each iteration: s = {‘A’: [3, 4, 7], ‘B’: [4, 9, 9], ‘C’: [3, 4, 5], ‘D’: [2, 2, 6], ‘E’: [6, 7, 9], ‘F’: [2, 4, 5]} new_dict = {} for a, b … Read more
You cannot use a dictionary of Booleans, because it is limited to only two values. This will inevitably lead to key collisions, because you plan to add more than two items to it. Good news is that you do not need a dictionary anyway, because you use it as a collection of tuples. This will … Read more
there can ba multiple values that match the condition – Dictionary<int, decimal> dict = new Dictionary<int,decimal>(); dict.Add(1, 392.3m); dict.Add(2, 612m); dict.Add(3, 981m); dict.Add(4, 344.23m); List<int> Result = dict.Where(x => x.Value < 400).Select(x => x.Key).ToList(); 0 solved How can you select and filter dictionary keys based on their values using Linq?
Hey bud I only have so much time a day to spend on this but I worked out a little better framwework for you to play with, try using elements in here working with it and modifying it, the skeleton of it should help you get all your goals accomplished with this certain code. I’ll … Read more
Hope this helps. I’m not sure it’s the best way to approach your problem, but it should help you to familiarise yourself with some of the options / see why some of the guys on here are advising against it. If you let us know more about what you’re trying to do we can better … Read more
Yes the dictionary will be re-created if you leave and then re-enter that scope, for example def switch(a): responses = {1: ‘a’, 2: ‘b’, 3: ‘c’} print(id(responses)) return responses[a] Notice that the id of responses keeps changing which illustrates that a new object has been created each time >>> switch(1) 56143240 ‘a’ >>> switch(2) 56143112 … Read more
First, your help_dict is not a dict, it’s a set. Second: you’re comparing a string with a set – which of course won’t never ever compare equal. Testing if a set contains an element is done with the in operator: if something in myset: – and for non-appartenance you simply use not in. IOW, you … Read more
if index==3: indexes=line.split(“\t”) for index in indexes: Old_Values=Old_Values{[key]} # What here? As I understand it, what you want there is simply an empty list to put the corresponding elements from the subsequent lines in: Old_Values[index] = [] Which will give you: {‘WebAddress’: [], ‘Address’: [], ‘Surname’: [], ‘Name’: [], ‘Telephone’: []} I don’t know where … Read more
I would use join to find the matches: Dictionary<int, string> dict = new Dictionary<int, string> { {1, “A”}, {2, “B”}, {3, “c”}, {4, “D”}, {5, “E”}, }; string[] values = new [] {“A”,”D”,”E”}; var query = from kvp in dict join s in values on kvp.Value equals s select new {kvp.Key, Found = true}; You … Read more