I've written a a function which takes score as parameter and should return the letter grade.
There are some conditions to be followed while writing the code. ie: return 'A' if 25 < score <= 30 return 'B' if 20 < score <= 25 and so on. So I wanted to do this by omitting a whole lot of if-else's. As I'm new to javascript this is all I could come up with:
// This function takes Nested arrays and a single number,
// which checks its availability in
// the inside array and then return the index of the array
function my_index(arr, score) {
for (const [index, elem] of arr.entries()) {
if (elem.includes(score)) {
return index;
}
}
}
// function to get letter grade
function getGrade(score) {
let grade;
var gradeDict = {
'A': [26, 27, 28, 29, 30],
'B': [21, 22, 23, 24, 25],
'C': [16, 17, 18, 19, 20],
'D': [11, 12, 13, 14, 15],
'E': [6, 7, 8, 9, 10],
'F': [0, 1, 2, 3, 4, 5]
}
var keys = Object.keys(gradeDict);
var values = [Object.values(gradeDict)]
grade = keys[my_index(values, score)]
return grade;
}
The first function works fine. It returns the index of nested array. But the main function getGrade happens to return 'Undefined'. Can't think of a better solution than this to reduce a bunch of ugly if-elses.
var question = {
'1st': 'Can anybody help me get this done?',
'2nd': 'Is there any better way to do this?'
}
Is there a better way to write this?
I'd do:
function getLetterGrade(score) {
return ['F', 'F', 'E', 'D', 'C', 'B', 'A'][Math.ceil(score / 5)];
}
(F occurs twice because more scores map to F than to other grades)
It may be a bit cryptic, but is easier to tune should the possible score ever change.
Remove the outer [] array of the Object.values. Object.values already returns values in array.
from
var values = [Object.values(gradeDict)];
to
var values = Object.values(gradeDict);
working example:
function my_index(arr, score) {
for (const [index, elem] of arr.entries()) {
if (elem.includes(score)) {
return index;
}
}
}
function getGrade(score) {
let grade;
var gradeDict = {
A: [26, 27, 28, 29, 30],
B: [21, 22, 23, 24, 25],
C: [16, 17, 18, 19, 20],
D: [11, 12, 13, 14, 15],
E: [6, 7, 8, 9, 10],
F: [0, 1, 2, 3, 4, 5],
};
var keys = Object.keys(gradeDict);
var values = Object.values(gradeDict);
grade = keys[my_index(values, score)];
return grade;
}
console.log(getGrade(5));
console.log(getGrade(25));
Alternate solution
function getGrade(score) {
let grade;
var gradeDict = {
A: [26, 27, 28, 29, 30],
B: [21, 22, 23, 24, 25],
C: [16, 17, 18, 19, 20],
D: [11, 12, 13, 14, 15],
E: [6, 7, 8, 9, 10],
F: [0, 1, 2, 3, 4, 5],
};
for (let key in gradeDict) {
if (gradeDict[key].includes(score)) return key;
}
return "Not found";
}
console.log(getGrade(5));
console.log(getGrade(25));
I like the ceil solution proposed earlier, but here is another general solution in case it's helpful:
function grade(score) {
if (score < 0 || score > 30) throw RangeError(`Score ${score} out of range`);
for (let ii = 5; ii >= 0; ii--) {
if (score > 5*ii) return String.fromCharCode(70 - ii);
}
return 'F';
}
console.log(0, '=>', grade(0)) // F
console.log(4, '=>', grade(4)) // F
console.log(6, '=>', grade(6)) // E
console.log(10, '=>', grade(10)) // E
console.log(27, '=>', grade(27)) // A
console.log(30, '=>', grade(30)) // A
Related
If my Array1 is
Array1 = [1,2,3,4,5,6,7,8,9,10]
the Result should be the same as Combined_Array= [1,2,3,4,5,6,7,8,9,10]
if i got
Array2=[11,12,13,14,15,16,17,18,19,20]
the Resut should be Combined_Array =[1,2,3,4,5,11,12,13,14,15]
and if again i got
Array3=[21,22,23,24,25,26,27,28,19,30]
The Combined_array = [1,2,3,11,12,13,21,22,23,24]
and so on , Doesnt matter how much Array's i want that it should give me a Combined_Array from all the different Array with Fixed Length
Need a Function to make this work .
You could take a closure over the collected arrays and retuen an array of the parts which are difined by the count of arrays.
const
getCombined = (a) => {
const allArrays = [];
return (b => {
allArrays.push(b);
let i = 0,
p = Math.floor(10 / allArrays.length),
result = [];
while (i < allArrays.length) result.push(...allArrays[i++].slice(0, p));
while (result.length < 10) result.push(allArrays[allArrays.length - 1][p++]);
return result;
});
};
var c = [],
add = getCombined(c);
c = add([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]);
console.log(...c); // [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
c = add([11, 12, 13, 14, 15, 16, 17, 18, 19, 20]);
console.log(...c); // [1, 2, 3, 4, 5, 11, 12, 13, 14, 15]
c = add([21, 22, 23, 24, 25, 26, 27, 28, 29, 30]);
console.log(...c); // [1, 2, 3, 11, 12, 13, 21, 22, 23, 24]
You need to consider many corner cases (if result array length exceeds given arrays count, if given arrays length differs and so on).
This will work for the simple scenario:
const arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
const arr2 = [11, 12, 13, 14, 15, 16, 17, 18, 19, 20];
const arr3 = [21, 22, 23, 24, 25, 26, 27, 28, 19, 30];
const combineArrays = (arr, length) => {
let elementsCount = Math.floor(length / arr.length);
const result = arr.reduce((acc, el) =>
acc.concat(el.slice(0, elementsCount)), []);
while (result.length < length)
result.push(...arr.pop().slice(elementsCount, ++elementsCount));
return result;
};
const result1 = combineArrays([arr1], 10); // [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
const result2 = combineArrays([arr1, arr2], 10); // [1, 2, 3, 4, 5, 11, 12, 13, 14, 15]
const result3 = combineArrays([arr1, arr2, arr3], 10); // [1, 2, 3, 11, 12, 13, 21, 22, 23, 24]
Hi i have a problem when code, I want to check an even number in multidimensional array and then change to 'X' if there are even numbers in row or column more than equals to 3 times appear but I change it all to 'X' eventhough it's less than 3. my code like this :
function evenNumbers(numbers){
for(var i = 0; i < numbers.length; i++){
for(var j = 0; j < numbers[i].length; j++){
if(numbers[i][j] % 2 == 0){
numbers[i][j] = 'X'
}
}
}
return numbers
}
console.log(evenNumbers([
[1, 2, 4, 6, 5],
[6, 17, 8, 11, 10],
[8, 11, 10, 18, 16],
[18, 12, 19, 27, 21],
[22, 10, 12, 22, 12]
]));
expected output :
[1, 'X', 'X', 'X', 5],
['X', 17, 'X', 11, 10],
['X', 11, 'X', 'X', 'X'],
['X', 12, 19, 27 , 21],
[''X, 'X', 'X', 'X', 'X']
what I got :
[ [ 1, 'X', 'X', 'X', 5 ],
[ 'X', 17, 'X', 11, 'X' ],
[ 'X', 11, 'X', 'X', 'X' ],
[ 'X', 'X', 19, 27, 21 ],
[ 'X', 'X', 'X', 'X', 'X' ] ]
Thanks, in advance
Help me with no ES6 built-in just regular JS, please. Thanks again
You could use some to make the check, then map
const arr = [
[1, 2, 4, 6, 5],
[6, 17, 8, 11, 10],
[8, 11, 10, 18, 16],
[18, 12, 19, 27, 21],
[22, 10, 12, 22, 12]
];
function xEvens(arr) {
return arr.map(row => {
let i = 0;
let shouldChange = row.some(n => n % 2 === 0 && ++i >= 3);
return shouldChange ? row.map(n => n % 2 === 0 ? 'X' : n) : row;
});
}
console.log(xEvens(arr));
Pre es6 version
var arr = [
[1, 2, 4, 6, 5],
[6, 17, 8, 11, 10],
[8, 11, 10, 18, 16],
[18, 12, 19, 27, 21],
[22, 10, 12, 22, 12]
];
function xEvens(arr) {
return arr.map(function(row) {
var i = 0;
var shouldChange = row.some(function(n) {
return n % 2 === 0 && ++i >= 3;
});
return shouldChange ? row.map(function(n) {
return n % 2 === 0 ? 'X' : n
}) : row;
});
}
console.log(xEvens(arr));
You can try following approach:
Idea:
Loop on array and return an array where X is replaced for even numbers.
Keep a track of count of replacements.
If count matches the threshold, return this new array.
If not, return the original array.
This way you will not need an extra loop just to check the validity for replacement.
function replaceEvensWithCondition(arr, minCount, replaceChar) {
return arr.map(function(row) {
var count = 0;
var updatedArray = row.map(function(n) {
var isEven = n % 2 === 0;
count += +isEven;
return isEven ? replaceChar : n;
});
return count >= minCount ? updatedArray : row
});
}
var arr = [ [1, 2, 4, 6, 5], [6, 17, 8, 11, 10], [8, 11, 10, 18, 16], [18, 12, 19, 27, 21], [22, 10, 12, 22, 12] ];
console.log(replaceEvensWithCondition(arr, 3, 'X'));
I have an array [1, 5, 20, 17, 6, 12, 13, 20, 1, 14, 20].
Whenever the element of the array is equal to 20 I would like to wrap it with an asterisk on either side like this [1, 5, *20*, 17, 6, 12, 13, *20*, 1 , 14, *20*].
How can I achieve this?
You can use map
let arr = [1, 5, 20, 17, 6, 12, 13, 20, 1, 14, 20]
let result = arr.map(o => o === 20 ? '*20*' : o);
console.log(result);
Doc: map()
You can use Arrays forEach to modify the elements of the array. elem is each element and i is the respective index. We are using forEach to modify the existing array. Since this is what you desired..
let arr = [1, 5, 20, 17, 6, 12, 13, 20, 1, 14, 20]
arr.forEach((elem, i) => {
if (elem === 20) {
arr[i] = "*20*"
}
})
console.log(arr)
function rollDice(max, times, bonus) {
var rolls = [1, 5, 20, 17, 6, 12, 13, 20, 1, 14, 20];
rolls.forEach((elem, i) => { if (elem === 20) { rolls[i] = "twenty" } });
for (var i = 0; times > i; i++)
{
max = Math.floor(max);
rolls.push(Math.floor(Math.random() * max) + 1 | + parseInt(bonus));
}
console.log(rolls);
}
rollDice(20, 5);
The problem you are experiencing is that you need to convert the integer number into strings. JavaScript has several ways to cleverly do this behind-the-scenes, but if you are still gaining an understanding of that, it's better to be explicit about what data types you start with (integers), and what data types you expect to end with (strings).
You transform the array, "mapping" over each item, transforming it to a string, and then if the string matches "20", you add the asterisks.
const start_array = [1, 5, 20, 17, 6, 12, 13, 20, 1, 14, 20];
const new_array = start_array.map((integer) => {
let number_string = integer.toString();
if (number_string === "20") {
number_string = "*" + number_string + "*";
}
return number_string;
})
console.log(new_array);
somehow arr[i].sort() not sorting last nested array and i getting bad result.
tryed FOR and WHILE, different operators, nothing helped. what im doing wrong?
it have return biggest numbers in array.
function largestOfFour(arr) {
var i = 0;
while (i != arr.length) {
arr[i] = arr[i].sort().pop();
i++;
}
return arr;
}
largestOfFour([
[4, 5, 8, 1, 3],
[13, 27, 18, 26],
[32, 35, 37, 39],
[1000, 1001, 817, 1]
]);
By default sort() orders the elements as strings, not numbers, and "1001" < "817". If you want a numeric sort you need to provide a comparison function that performs numeric ordering.
function largestOfFour(arr) {
var i = 0;
while (i != arr.length) {
arr[i] = arr[i].sort(function(a, b) {
return a - b;
}).pop();
i++;
}
return arr;
}
console.log(largestOfFour([
[4, 5, 8, 1, 3],
[13, 27, 18, 26],
[32, 35, 37, 39],
[1000, 1001, 817, 1]
]));
Not the cause of the problem you are seeing (which #Barmar answered), but you can use Array.map with Math.max.apply to get the max number of each list:
function largestOfFour(arr) {
return arr.map(function(list) {
return Math.max.apply(null, list);
});
}
console.log(largestOfFour([
[4, 5, 8, 1, 3],
[13, 27, 18, 26],
[32, 35, 37, 39],
[1000, 1001, 817, 1]
]));
Or a bit more concise with ES6:
const largestOfFour = arr => arr.map(list => Math.max(...list))
The problem is in sort defaults - it sorts the array as strings - so "8..." > "1..."
Try following:
function cmp(a,b){
return a-b;
}
function largestOfFour(arr) {
var i = 0;
while (i != arr.length) {
arr[i] = arr[i].sort(cmp).pop();
i++;
}
return arr;
}
largestOfFour([
[4, 5, 8, 1, 3],
[13, 27, 18, 26],
[32, 35, 37, 39],
[1000, 1001, 817, 1]
]);
Output: [ 8, 27, 39, 1001 ]
I currently have the following array set up:
var TicketInfo =
{
t1: {
1: [7, 12, 35,39,41, 43],
2: [7, 15, 20,34,45, 48],
3: [3, 7, 10, 13, 22, 43],
4: [2, 4, 5,23,27, 33]
},
t2: {
1: [10, 12, 17,44,48, 49],
2: [13, 15, 17, 18, 32, 39],
3: [16, 17, 20, 45, 48, 49],
4: [6, 16, 18, 21, 32, 40]
}
}
What I want to do is iterate through these to bring back the arrays under.
As a test I've tried something like this:
for(t in TicketInfo["t1"])
{
i++;
Write(t.i);
}
But it's obviously not working how I want it to.
Any ideas?
I want to be able to output the arrays like [7, 12, 35,39,41, 43]
Thanks
var TicketInfo =
{
t1: {
1: [7, 12, 35,39,41, 43],
2: [7, 15, 20,34,45, 48],
3: [3, 7, 10, 13, 22, 43],
4: [2, 4, 5,23,27, 33]
},
t2: {
1: [10, 12, 17,44,48, 49],
2: [13, 15, 17, 18, 32, 39],
3: [16, 17, 20, 45, 48, 49],
4: [6, 16, 18, 21, 32, 40]
}
}
for(var j in TicketInfo )
{
for(var p in TicketInfo[j] )
{
for(var i = 0; i < TicketInfo[j][p].length; i++ )
{
console.log(TicketInfo[j][p][i]);
}
}
}
http://jsfiddle.net/J6rTj/
If you're here from google trying to find a way to do a quick print for debugging, here's a one liner for you:
console.log(myArray.join("\n"))
Example:
var myArray = [[1,2,3],[4,5,6],[7,8,9]];
console.log(myArray.join("\n"));
Output:
1,2,3
4,5,6
7,8,9
Example with proper brackets:
var myArray = [[1,2,3],[4,5,6],[7,8,9]];
console.log("[[" + myArray.join("],\n[") + "]]");
Output:
[[1,2,3],
[4,5,6],
[7,8,9]]
Answer to OP's question:
obj = {
1: [7, 12, 35,39,41, 43],
2: [7, 15, 20,34,45, 48],
3: [3, 7, 10, 13, 22, 43],
4: [2, 4, 5,23,27, 33],
}
var keys = Object.keys(obj);
keys.sort();
console.log(keys);
var listFromObj = []
for (var i = 0; i < keys.length; i++) {
if (obj.hasOwnProperty(keys[i])) listFromObj.push(obj[keys[i]]);
}
console.log("[" + listFromObj.join("]\n[") + "]");
Output:
[7,12,35,39,41,43]
[7,15,20,34,45,48]
[3,7,10,13,22,43]
[2,4,5,23,27,33]
The syntax is TicketInfo["t1"]["1"][0].
That example will give you 7.
TicketInfo["t1"]["1"] will give you the array you're after at the base of your question.
In your code t just represents the key.
Try following code:
var TicketInfo =
{
t1: {
1: [7, 12, 35,39,41, 43],
2: [7, 15, 20,34,45, 48],
3: [3, 7, 10, 13, 22, 43],
4: [2, 4, 5,23,27, 33]
},
t2: {
1: [10, 12, 17,44,48, 49],
2: [13, 15, 17, 18, 32, 39],
3: [16, 17, 20, 45, 48, 49],
4: [6, 16, 18, 21, 32, 40]
}
}
for(t in TicketInfo["t1"])
{
i++;
console.log(TicketInfo["t1"][t]);
}
Do I understand that you want to output entire table in order? Since you use objects on t1/t2 level, you'll have to do extra steps for that.
First, see if you can simply replace objects with real arrays:
var TicketInfoArrays = {
t1: [
[7, 12, 35,39,41, 43],
[7, 15, 20,34,45, 48],
[3, 7, 10, 13, 22, 43],
[2, 4, 5,23,27, 33]
]
}
var t1 = TicketInfoArrays.t1
for(var idx = 0, len = t1.length; idx<len; idx++){
var line = idx+": ["
var nested = t1[idx]
for(var idx2 = 0, len2 = nested.length; idx2<len2; idx2++){
line += ((idx2 > 0 ? ', ':'') + nested[idx2])
}
console.log(line + ']')
}
If that's somehow impossible, but you sure that keys in those objects always start at some specific number and go ascending without gaps, you can simply itreate over properties until you hit empty element:
var TicketInfo = {
t1: {
1: [7, 12, 35,39,41, 43],
2: [7, 15, 20,34,45, 48],
3: [3, 7, 10, 13, 22, 43],
4: [2, 4, 5,23,27, 33]
}
}
var t1 = TicketInfo.t1
var idx = 1
var nested
while(nested = t1[idx]){
var line = idx+": ["
var nested = t1[idx]
for(var idx2 = 0, len2 = nested.length; idx2<len2; idx2++){
line += ((idx2 > 0 ? ', ':'') + nested[idx2])
}
console.log(line + ']')
idx++
}
Finally, if you can't guarantee even that, you will have to manually collect all keys, sort them and then iterate over this sorted list.
var TicketInfoUnordered = {
t1: {
8: [7, 12, 35,39,41, 43],
20: [7, 15, 20,34,45, 48],
45: [3, 7, 10, 13, 22, 43],
3: [2, 4, 5,23,27, 33]
}
}
var t1 = TicketInfoUnordered.t1
var keys = []
for(key in t1){
if(t1.hasOwnProperty(key)){ keys.push(key) }
}
keys.sort(function(a, b){ return a - b })
for(var idx = 0, len = keys.length; idx<len; idx++){
var line = keys[idx]+": ["
var nested = t1[keys[idx]]
for(var idx2 = 0, len2 = nested.length; idx2<len2; idx2++){
line += ((idx2 > 0 ? ', ':'') + nested[idx2])
}
console.log(line + ']')
}