Introduction Splitting a dictionary by multiple values and one of the keys is a useful way to organize data. It allows you to quickly and easily access specific information from a large set of data. This technique can be used to create subsets of data based on certain criteria, such as a specific key or … Read more
Extract only the students with grade 80 or higher with Array.filter. Map the remaining students into a new format using Array.map. Result: let students = [{ fname: “Jane”, lname: “Brazier”, snum: “100366942”, agrade: 67.59127376966494, tgrade: 64.86530868914188, egrade: 70.52944558104066 }, { fname: “Ricardo”, lname: “Allen”, snum: “100345641”, agrade: 65.80370345301014, tgrade: 75.40211705841241, egrade: 55.39348896202821 }, { fname: … Read more
Introduction Javascript is a powerful programming language that can be used to manipulate data in a variety of ways. In this tutorial, we will be looking at how to print all names in an array containing multiple objects. We will be using the forEach() method to iterate through the array and print out each name. … Read more
I think you want var test = dictionary Remove the quotations 3 solved Swift: access to the content of a dictionary [duplicate]
There could be gundreds of way to do that. Here is one: template<class T> void dumpVector( const std::vector<T>& vect ) { for ( std::vector<T>::const_iterator iter = vect.begin(); iter != vect.end(); ++iter ) { std::cout << ” ” << *iter; } } for ( std::map<vector<double>,vector<int>>::const_iterator mapIter = path.begin(); mapIter != path.end(); ++mapIter ) { std::cout << … Read more
If you want only print the resulting dictionaries, uncomment the print statement (and comment the following 2). d1 = [ {‘index’:’1′,’color’:’red’}, {‘index’:’2′,’color’:’blue’}, {‘index’:’3′,’color’:’green’} ] d2 = [ {‘device’:’1′,’name’:’x’}, {‘device’:’2′,’name’:’y’}, {‘device’:’3′,’name’:’z’} ] result_list = [] for dict1 in d1: merged_dict = dict1.copy() for dict2 in d2: merged_dict.update(dict2) # print(merged_dict) result_list.append(merged_dict.copy()) print(result_list) The result: [{‘name’: ‘x’, ‘device’: … Read more
If you want a more pythonic way: from itertools import groupby from pprint import pprint from collections import ChainMap a = [{‘a’:0,’b’:23}, {‘a’:3,’b’:77}, {‘a’:1,’b’:99}] b = [{‘a’:1,’c’:666}, {‘a’:4,’c’:546}] c = [{‘d’:33,’a’:3}, {‘d’:1111,’a’:4}, {‘d’:76,’a’:1}, {‘d’:775,’a’:0}] d = [{‘a’:2,’e’:12}, {‘a’:4,’e’:76}] dict_list = a + b + c + d # You just need to specify the key … Read more
I believe you want a dictionary with a single key and a list: l = [1,2,3,4,5] d = {l[0]:l[1:]} Output: {1: [2, 3, 4, 5]} 0 solved Turning a list into a dictionary?
The simple way to get a count of unique words is to use a set. I put your text into a file called ‘qdata.txt’. The file is very small, so there’s no need to read it line by line: just read the whole thing into a single string, then split that string on whitespace and … Read more
You can use a defaultdict as follow: class Account: def __init__(self, x, y): self.x = x self.y = y d_in = {0: Account(13, 1), 1: Account(15, 5), 2: Account(55, 1), 3: Account(33 ,1)} d_out = collections.defaultdict(list) for index, k in enumerate(d_in.keys()): d_out[d_in[k].y].append(index) print d_out Giving the following output: defaultdict(<type ‘list’>, {1: [0, 2, 3], 5: … Read more
Here’s a straightforward solution with a simple loop: dict_x = {} for value in list_x: if isinstance(value, str): dict_x[value] = current_list = [] else: current_list.append(value) Basically, if the value is a string then a new empty list is added to the dict, and if it’s a list, it’s appended to the previous list. solved Convert … Read more
Here is an example with marshal and unmarshal using your object definition package main import ( “encoding/json” “fmt” ) type MyObject struct { ID string `json:”id”` Name string `json:”name”` Planets map[string]int `json:”planets”` } func main() { aa := &MyObject{ ID: “123”, Name: “pepe”, Planets: map[string]int{ “EARTH”: 3, “MARS”: 4, }, } // Marshal out, err … Read more
If what you want is to split the string at the first space, then you can use the string.split(s, maxsplit=n) where s is the string to split by and maxsplit=n is the number of splits to stop at. If you do not give any value for s that is only call the function as string.split(maxsplit=n) … Read more
from collections import defaultdict d = defaultdict(list) for i,j in enumerate(x): d[j].append(i) Using defaultdict with the default being empty list and enumerate to provide the index solved Python: build dictionary from list
If you want to compare keys of Dictionary, then: var dict1 = new Dictionary<string, List<string>>(); var dict2 = new Dictionary<string, List<string>>(); // something.. if (dict1.Keys.SequenceEqual(dict2.Keys)) { // your code } If you want to compare values of Dictionary, then: var dict1 = new Dictionary<string, List<string>>(); var dict2 = new Dictionary<string, List<string>>(); // something.. var d1Keys … Read more