This question already has answers here:
Loop through an array in JavaScript
(46 answers)
What is the difference between ( for... in ) and ( for... of ) statements?
(18 answers)
Closed last year.
I am writing a simple javascript code to parse, and verify a chess position written in Forsyth–Edwards Notation (FEN).
The default chess position in this notation is given by,
const defaultFEN = "rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR w KQkq - 0 1";
There are 6 components, I split the components by whitespace using String.split(" "), I now want to further split the first element of the resulting array by "/", which will give the state of each rank.
Running this code gives me an unintuitive result...
const defaultFEN = "rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR w KQkq - 0 1";
const locations = defaultFEN.split(" ")[0];
for (let rank in locations.split("/")) {
console.log(rank);
}
I expected the output to be 8 Strings, delimited by "/", in the first portion of the defaultFEN string. Instead I get the numbers 0-7 printing out.
Interestingly, If I manually access this array, console.log(locations.split("/")[i]), for any number i in the interval [0-7], I see the result I intended.
Why are the numbers 0-7 printing out when using the iterative loop, but it works exactly as intended if I use a normal index based for loop?
There's nothing wrong with your split, but you should use for..of (MDN) instead:
for (let rank of locations.split("/")) {
console.log(rank);
}
... as for..in loop iterates over indexes (and those are 0..7 for 8-element array).
As a sidenote, it's (usually) a good idea to use const (and not let) in this iteration, as variable is assigned a value once each loop (and usually shouldn't be reassigned):
for (const rank of locations.split("/")) {
console.log(rank);
}
Related
I have two arrays one of it is having user id and another one is having user ids. Those arrays are as follows.
1)The array which is having user id.
data[key].effective_employees Which is eaqual to [2].
Now I have another array which is having numbers of employee ids which is as follows.
data2[0].id Which is eaqual to [2,22,21].
And now I am trying to see whether the array two has number in array 1 I am using the following logic to see whether it is working or not.
if ((/^\d+$/.test(_.intersection([data2[0].id.toString()], data[key].effective_employees)))) {
let isElem = _.contains(returnStackFilterd, value);
if (isElem == false) {
returnStackFilterd.push(value);
}
} else {
returnStackFilterd = _.without(returnStackFilterd, value);
}
But this is showing true for the number 2 if the array two is having 22. Psudo code of what is happening with it is as follows.
if([2]is in[22,21]){ it is printing true} I want false here as the number two is not in the second array. The second array contains 22 and 21 which is not eaqual to 2
How do i solve this problem? The above psudo code should print false.
Let's break down your test expression and see why it doesn't work.
First off, we know that data[key].effective_employees is [2]. data2[0].id might be [2, 22, 21] or [22, 21]. If I'm understanding your question correctly, you want the whole test expression to return true in the first case and false in the second case.
Rebuilding your test expression from the bottom up, the innermost expression we find is this:
data2[0].id.toString()
This is a string with the value '2,22,21' or '22,21', depending on which case we are talking about. Next, you wrap this string in an array:
[data2[0].id.toString()]
So now we have ['2,22,21'] or ['22,21']. Note the quotes; in either case, it is an array with a single element that is a string.
Next, you take the intersection of this array with data[key].effective_employees, which we know is [2]:
_.intersection([data2[0].id.toString()], data[key].effective_employees)
So this expression is effectively
_.intersection(['2,22,21'], [2])
or
_.intersection(['22,21'], [2])
You are always taking the intersection of two arrays, where the first contains a single string and the second contains a number. They can't have any elements in common, so that's always going to produce an empty array ([]).
Finally, you test whether that empty array matches a regular expression:
/^\d+$/.test(_.intersection([data2[0].id.toString()], data[key].effective_employees))
// is effectively the same as
/^\d+$/.test([])
Regular expressions are supposed to be matched against a string, not an array. JavaScript is very lenient in situations like these and will coerce the value you're passing to a string. That means that the value [] is first converted to the empty string '' before being matched to the regular expression /^\d+$/. The empty string does not contain any digits, so this test always returns false.
This is why your test doesn't work as intended. However, let's take a few steps back, because you seem to be doing many things you don't need to do. Why convert arrays to strings (and then back to array)? Why match against a regular expression, if you just want to know whether two arrays have elements in common?
The following, simpler expression will give you the elements that data[key].effective_employees and data2[0].id have in common:
_.intersection(data[key].effective_employees, data2[0].id)
This will evaluate to either [2] or [], depending on whether data[key].effective_employees contains the number 2 or not.
I suggest saving the result of this expression to a variable, because it makes your code easier to read. Let's call it commonIds:
const commonIds = _.intersection(data[key].effective_employees, data2[0].id)
Now you can formulate different conditions, based on what exactly you want this intersection to be like. My impression is that you just want it to be nonempty (i.e., at least one element in common). In that case, you can compare its length to zero:
if (commonIds.length > 0) {
// code for when there is an overlap
} else {
// code for when there is no overlap
}
As a final note, I recommend assigning your base expressions data[key].effective_employees and data2[0].id to variables as well. Again, this makes your code more readable, and it also ensures that you need to change only one line of code if those base expressions change. Putting it all together:
const key = 'x';
const data = { [key]: {
effective_employees: [2],
}};
const data2 = [{
id: [2, 22, 21],
}];
const userId = data[key].effective_employees;
const employeeIds = data2[0].id;
const commonIds = _.intersection(userId, employeeIds);
if (commonIds.length > 0) {
console.log('userId appears in employeeIds');
} else {
console.log('userId does not appear in employeeIds');
}
<script src="https://underscorejs.org/underscore-umd-min.js"></script>
If data[key].effective_employees is the number 2, and data2[0].id is the array [2, 22, 21], the expression to test whether data2[0].id contains data[key].effective_employees is:
data2[0].id.includes(data[key].effective_employees)
From your original question, data2[0].id.toString() coerces the array to a string 2,22,21, which is no use to you. You also do not need to use Underscore for this.
This question already has answers here:
Convert [key1,val1,key2,val2] to a dict?
(12 answers)
Make dictionary from list with python [duplicate]
(5 answers)
Convert list into a dictionary [duplicate]
(4 answers)
Closed 4 years ago.
trying to figure out how to do this and have yet to find a good solution. I pulled this data out of an XML response. It was in a var tag. Now what I would like to do is create a dictionary out of it. The domain.com should be paired with the number right listed behind it.
This is the data:
[
'cb131.domain1.com', '147827',
'cb143.domain2.com', '147825',
'cb175.domain1.com', '147454',
'cb190.domain.com', '146210',
'cb201.domain.com', '146208',
'cb219.domain.com', '146042',
'cb225.domain.com', '146282',
'cb900.domain.com', '148461',
'cb901.domain.com', '148493',
'cb902.domain.com', '148495',
'cb903.domain.com', '148497',
'cb904.domain.com','148499',
'cb905.domain.com', '148501',
'cb906.domain.com', '148503',
'cb907.domain.com', '148505',
'cb908.domain.com', '148507',
'cb909.domain.com', '148509'
]
So for example cb131.domain1.com should be paired with 147827, cb143.domain2.com paired with 147825 and so on.
Drawing a blank on a good quick solution on how to do this. Hopefully someone can help.
Thanks!
Edited with answer I choose below:
I choose this answer and also to help anyone else I add a nice way to print out the results (data is the string I obtained):
import ast
i = iter(ast.literal_eval(data))
dic = dict(zip(i, i))
for key , value in dic.items():
print(key, " :: ", value)
This should do it. Assuming the list is saved to a variable l:
keys = l[::2]
vals = l[1::2]
dic = dict(zip(keys, vals))
You can create an iterator from the list after using ast.literal_eval to parse it from the input text, zip the iterator with itself, and pass the generated sequence of tuples to the dict constructor:
import ast
i = iter(ast.literal_eval(data))
dict(zip(i, i))
Assuming you have the above in a python array called data, you can do:
new_data = []
for i in range(0, len(data), 2):
new_data.append((data[i], data[i+1]))
Now new_data would be a list of tuples. You could certainly create a better data structure to hold these pairs if you want.
I do not yet know Python that I can write a snippet, but:
initialize an empty dictionary in Python
create a for loop counting index from 0 to length of your array in steps of two.
inside add a dictionary entry with key of value at index and value at index + 1
perhaps check for duplicates
Does this answer help you?
This is Python - quickly google'd:
dictionary = { }
for idx in range(0, len(data), 2)
dictionary[data[idx]] = data[idx + 1]
Does anybody know what is the meaning of this declaration in JS:
var m = [[0]];
I mean what is declarated as type, and why the zero is in the brackets?
[0] is an array with the first index is equal to 0
[[0]] Is an array with its first index equal to an array (whos first index is 0)
It would probably be easier to imagine if you had more elements and space it a bit better:
var m = [[0,1,2],[2,4,5],[1,3]]
so
m[0] = [0,1,2];
m[1] = [2,4,5];
m[2] = [1,3];
This can then be expanded for however many dimensions you need leading to collections of collections of collections.
you can then access each index and use it as the array your referencing for instance:
m[0].push(4);
m[2].join(',');
etc.
(as mentioned above) You can also then access it shorthand like:
m[0][0]
m[x][y]
m[n-1][m[0][1]]
Making it as complicated or simple as you need.
m has been declared as an array, with a nested array inside it, and the nested array contains a single element - 0
console.log(m[0][0]) will output 0
my array:
tempListArray = "[{"id":"12","value":false},{"id":"10","value":false},{"id":"9","value":false},{"id":"8","value":false}]";
To check if an element exists I would do this:
for (var i in tempListArray) {
//check flag
if (tempListArray[i].id == Id) {
flagExistsLoop = 1;
break;
}
}
Is there anyway, I can check if an Id exists without looping through the whole array. Basically I am worried about performance if say I have a 100 elements.
Thanks
No, without using custom dictionary objects (which you seriously don't want to for this) there's no faster way than doing a 'full scan' of all contained objects.
As a general rule of thumb, don't worry about performance in any language or any situation until the total number of iterations hits 5 digits, most often 6 or 7. Scanning a table of 100 elements should be a few milliseconds at worst. Worrying about performance impact before you have noticed performance impact is one of the worst kinds of premature optimization.
No, you can't know that without iterating the array.
However, note for...in loops are a bad way of iterating arrays:
There is no warranty that it will iterate the array with order
It will also iterate (enumerable) non-numeric own properties
It will also iterate (enumerable) properties that come from the prototype, i.e., defined in Array.prototype and Object.protoype.
I would use one of these:
for loop with a numeric index:
for (var i=0; i<tempListArray.length; ++i) {
if (tempListArray[i].id == Id) {
flagExistsLoop = 1;
break;
}
}
Array.prototype.some (EcmaScript 5):
var flagExistsLoop = tempListArray.some(function(item) {
return item.id == Id;
});
Note it may be slower than the other ones because it calls a function at each step.
for...of loop (EcmaScript 6):
for (var item of tempListArray) {
if (item.id == Id) {
flagExistsLoop = 1;
break;
}
}
Depending on your scenario, you may be able to use Array.indexOf() which will return -1 if the item is not present.
Granted it is probably iterating behind the scenes, but the code is much cleaner. Also note how object comparisons are done in javascript, where two objects are not equal even though their values may be equal. See below:
var tempListArray = [{"id":"12","value":false},{"id":"10","value":false},{"id":"9","value":false},{"id":"8","value":false}];
var check1 = tempListArray[2];
var check2 = {"id":"9","value":false};
doCheck(tempListArray, check1);
doCheck(tempListArray, check2);
function doCheck(array, item) {
var index = array.indexOf(item);
if (index === -1)
document.write("not in array<br/>");
else
document.write("exists at index " + index + "<br/>");
}
try to use php.js it may help while you can use same php function names and it has some useful functionalities
There is no way without iterating through the elements (that would be magic).
But, you could consider using an object instead of an array. The object would use the (presumably unique) id value as the key, and the value could have the same structure you have now (or without the redundant id property). This way, you can efficiently determine if the id already exists.
There is a possible cheat for limited cases :) and it is magic...cough cough (math)
imagine you have 3 elements:
1
2
3
and you want to know if one of these is in an array without iterating it...
we could make a number that contains a numerical flavor of the array. we do this by assigning prime numbers to the elements:
1 - 2
2 - 3
3 - 5
the array so when we add item 2 we check that the array doesn't already contain the prime associated to that item by checking (if Flavor!=0 && (Flavor%3)!=0) then adding the prime Flavor*=3;
now we can tell that the second element is in the array by looking at the number.
if Flavor!=0 && (Flavor%3)==0 // its There!
Of course this is limited to the numerical representation that can be handled by the computer. and for small array sizes (1-3 elements) it might still be faster to scan. but it's just one idea.
but the basis is pretty sound. However, this method becomes unusable if you cannot correlate elements one to one with a set of primes. You'll want to have the primes calculated in advance. and verify that the product of those is less numerical max numerical representation. (also be careful with floating-point. because they might not be able to represent the number at the higher values due to the gaps between representable values.) You probably have the best luck with an unsigned integer type.
This method will probably be too limiting. And there is something else you can do to possibly speed up your system if you don't want to iterate the entire array.
Use different structures:
dictionaries/maps/trees etc.
if your attached to the array another method can be a bloom filter. This will let you know if an element is not in your set, which can be just as useful.
I have an assignment for a Javascript course where I have to count how many of each specific types of elements occur in an array. The array is 105 elements long, and just occurrences of the numbers 1, 2, 3, 4, and 5. I have to count how many 1's, 2's, 3's, etc.
Of course there's a simple way to do this using a loop however my teacher has added the following at the end of the assignment:
Use only the length property, toString(), sort() and indexOf() methods. Please no loops or conditional statements.
I have no idea how to do this assignment without using a loop. Any help you can give me is greatly appreciated. Thanks!
this will be the answere:
var c=[1,2,3,4,5,2,3,4,5]
c.sort()
cout_of_1 = c.indexOf(2)-c.indexOf(1);
cout_of_2 = c.indexOf(3)-c.indexOf(2);
cout_of_3 = c.indexOf(4)-c.indexOf(3);
cout_of_4 = c.indexOf(5)-c.indexOf(4);
cout_of_5 = c.length-c.indexOf(5)+1;
Since this is an assignment and you haven't really shown an attempt, I'll just give you a couple of hints.
Once you sort the array, all of the 1s will be grouped together at the start of the array, then all the 2s, etc.
The indexOf function returns the first index at which a given element can be found in the array, or -1 if it is not present.
Given the index of the first occurrence of each digit in a sorted array, you should be able to calculate how many of each digit there are. I'm not sure you'll need toString() at all. Your instructor might have thrown that in to be sneaky.
First, sort the array, then grab the first occurrence of the next value. That's the amount of numbers you have, minus the ones you already have.
var array = [1,2,5,3,2,4,5,2,1,3,3,4,5,6,2];
array = array.sort();
var ones = array.indexOf(2);
var twos = array.indexOf(3) - ones;
var threes = array.indexOf(4) - ones - twos;
...
However for the last value (5) there is no next value to check. You can work around this by checking the length and subtracting one from it.
var fives = array.length - ones - twos - threes - fours;
Keep in mind that this is actually bad code, and you should use loops.