[ad_1] One option to your expected output is to create a indexed array with the associative array below it. This will create this kind of array: array(4) { [0]=> array(1) { [203]=> int(1) } [1]=> array(1) { [204]=> int(2) } [2]=> array(1) { [203]=> int(2) } [3]=> array(1) { [205]=> int(1) } } This is … Read more
[ad_1] var goal = 4; array.sort(function(a, b) { if (a.Seats == goal) { return -1; } else if (b.Seats == goal) { return 1; } else { // If neither matches the goal, order them by Seats return a.Seats – b.Seats; } }); 0 [ad_2] solved Sort Array in Javascript [duplicate]
[ad_1] I suggest finding the index of the first mask ocurrence: private static int MaskIndex(byte[] source, byte[] mask) { if (mask.Length == 0) return 0; // or source.Length; or throw exception for (int i = 0; i < source.Length – mask.Length + 1; ++i) { bool found = true; for (int j = 0; j … Read more
[ad_1] A loop or two will do. We need to compare all pairs to all others. let arrayFrom = [‘1’, ‘6.1’, ’10’, ’31’, ‘6.2’, 3]; let arrayTo = [‘2’, ‘9.9’, ’30’, ‘401’, ‘7’, 5]; function is_pair_included(a, in_b) { return (a[0] > in_b[0] && a[1] < in_b[1]) } var to_remove = []; for (var i = … Read more
[ad_1] this should work: const all = [ { “a”: “Content A”, “b”: { “1”: “Content 1”, “2”: “Content 2” } }, { “y”: “Content Y”, “x”: { “3”: “Content 3”, “4”: “Content 4” } }, ]; console.log(all.reduce((prev, el) =>{ let curr = Object.entries(el); let k1 = curr[0][0]; let k2 = curr[1][0]; Object.entries(curr[1][1]).forEach((o => { … Read more
[ad_1] Yes: let array = [“ABC”, “BCA”, “CDA”, “APA”] let array2 = [“ABC”, “APA”] const output = array.map((item) => { let isFound = array2.includes(item); return { value: item, matched: isFound }; }) console.log(output) 1 [ad_2] solved 2 arrays matchings , transformation in arrays of objects [closed]
[ad_1] Instead of the standard for loop to go through an array, for example: for($i = 0; $i < count($array); $i++) { echo $array[$i] . ” “; } Outputs “one two three four five”. Just set $i equal to 1 instead, like so: for($i = 1; $i < count($array); $i++) { echo $array[$i] . ” … Read more
[ad_1] Look at the index values of the list. alist=[[1],[2]] Here alist[1] is [2], so alist[1][0] is 2. 4%1 is 0. [ad_2] solved Why are lists giving different value?
[ad_1] if i ask to print array 1, 1 then i would get back “d”? No that would return an error, as arrays start at 0. So you must return words[0][0], that would get you back “d”. [ad_2] solved Multi-Dimensional String Arrays
[ad_1] You can easily group by name to get what you need. Below is a quite general function to group any array of objects to a Map, by specified property. Then simply make it an array and you’re done function group(arr, propertyToGroupBy) { return arr.reduce(function(a, b) { return a.set(b[propertyToGroupBy], (a.get(b[propertyToGroupBy]) || []).concat(b)); }, new Map); … Read more
[ad_1] how can I declare an array of 10 pointers to objects of the circle class? Like this: circle* myArray[10]; It is then your responsibility to assign those pointers to point at valid circle objects. But how you do that exactly is outside the scope of your question, as you did not explain how you … Read more
[ad_1] Do you mean for leader in leaders where !leader.isEmpty { alert.addAction(UIAlertAction(title: leader[0].name, style: .default, handler:{(_) -> Void in })) } 0 [ad_2] solved how to increase the array index in loop by +1 using swift [closed]
[ad_1] To solve the array issue you just need to make a few small changes. int stop=0,d,*t; // Declare an uninitialized pointer to int float f(float x) { int loop,loopa; float add=0.0,sum=0.0; for(;stop==0;) { int p ; cout << “Enter the degree of the poly. eq. (+ve integer)” << endl; cin >> d ; t … Read more
[ad_1] Assuming that class CatchesItem has time field defined as LocalDateTime, the check for today’s timestamp may be implemented as follows: import java.time.*; // … LocalDate today = LocalDate.now(); for (CatchesItem item : items) { if (today.equals(item.getTime().toLocalDate())) { item.setAmount(item.getAmount() + 5); // add some value to amount } } items.forEach(System.out::println); Output (for the input data … Read more
[ad_1] You have a few errors, can you try this: void load(char *name, float *share, float *buyprice, float *currprice) { printf(“Enter stock name”); gets(name); printf(“Enter share, buyprice, currprice”); scanf(“%f %f %f”, share, buyprice, currprice); // changes: removed &* } void calc(float share, float buyprice, float currprice, float *buytotal, float *currtotal, float *profit) { *buytotal = … Read more