I decided to get shuffled values from array. for that i used this function, i got it from net, it works fine. But i don't know, how it's works...
any one can help me to understand this?
my code is :
function rand(ar){
return 0.5-Math.random();
}
var ar = [5,10,15,20,25]
ar.sort(rand);
console.log(ar)
I am using this function for getting new shuffled array values from the declared one.
This code is using the supplied rand function as the comparison operator for the Array.Sort method (http://msdn.microsoft.com/en-us/library/4b4fbfhk(VS.85).aspx). Since the Math.random (http://msdn.microsoft.com/en-us/library/41336409.aspx) function returns a value from 0 (inclusive) to 1 (exclusive), the rand function will return a value from 0.5 (inclusive) to -0.5 (exclusive).
Normally the sortFunction supplied to the Sort method takes 2 arguments that are compared. The sortFunction compares them and returns a value that means:
Negative - The first item is less than the second
Zero - The items are equal
Positive - The first item is greater than the second
As the sort method runs, it uses this comparison to determine which array values should go before the others.
In the case of your code, the rand function's return value is random and has no correlation to the data. This means that, whenever the sort function tries to compare two values in the array, half of the time it will say the first item is less than the second and half the second item will be less than the first. As this is done over the entire length of the array, items are swapped randomly and the whole array becomes randomized.
array.sort() has an optional parameter that is a sorting function, you can pass a function reference to change the order of the array.
Maybe this page can be helpful http://www.javascriptkit.com/javatutors/arraysort.shtml
Related
def large_prime(n)
return [] if n==1
factor = (2..n).find {|x| n % x == 0}
[factor] + large_prime(n/factor)
end
I got this solution from somewhere else. I don't understand the 4th line of code where large_prime is called recursively and appended onto factor.
When I change the first line "return []" and leave out the '[]' after the return, I get an error message for on line 4, that says '+':no implicit conversion of nil into Array.
So why does this code work? Thanks
P.S. I'm obviously a noob and everything is very new to me.
The 3rd line finds the first divisor of n between 2 and n. This line itself does not involve recursion.
I don't really get the code you modified, but it seems to return nil in some case, while the original method always return an Array.
You must return an empty array when passed 1 to terminate the recursion. Any positive argument other than one will result in another call to large_prime, but an argument of 1 results in large_prime simply returning an empty array.
At each level of recursion, the program adds an array with the single factor it found to an array consisting of all factors found for the value n/factor. When the last factor (other than 1) is found, the final call to large_prime is made with an argument of 1, large_prime returns an empty array which is then added to the array containing the last factor, giving an array containing just the last factor. This array is then returned and you have
[next-to-last-factor] + [last-factor], giving a return array of [next-to-last-factor, last-factor] which is added to [next-to-next-to-last-factor] giving [next-to-next-to-last-factor, next-to-last-factor, last-factor]. This is then added to an array [next-to-next-to-next-to-last-factor], giving... lather, rinse, repeat until we reach the largest factor and add it in.
You must return an empty array because you can't add nil to an array in Ruby.
I have a 250*2001 matrix. I want to find the location for the maximum value for a(:,i) where i takes 5 different values: i = i + 256
a(:,256)
a(:,512)
a(:,768)
a(:,1024)
a(:,1280)
I tried using MAXLOC, but since I'm new to fortran, I couldn't get it right.
Try this
maxloc(a(:,256:1280:256))
but be warned, this call will return a value in the range 1..5 for the second dimension. The call will return the index of the maxloc in the 2001*5 array section that you pass to it. So to get the column index of the location in the original array you'll have to do some multiplication. And note that since the argument in the call to maxloc is a rank-2 array section the call will return a 2-element vector.
Your question is a little unclear: it could be either of two things you want.
One value for the maximum over the entire 250-by-5 subarray;
One value for the maximum in each of the 5 250-by-1 subarrays.
Your comments suggest you want the latter, and there is already an answer for the former.
So, in case it is the latter:
b(1:5) = MAXLOC(a(:,256:1280:256), DIM=1)
Need to know is there a way to count the frequency of items in a array without using two loops. This is without knowing the size of the array. If I know the size of the array I can use switch without looping. But I need more versatile than that. I think modifying the quicksort may give better results.
Array[n];
TwoDArray[n][2];
First loop will go on Array[], while second loop is to find the element and increase it count in two-d array.
max = 0;
for(int i=0;i<Array.length;i++){
found= false;
for(int j=0;j<TwoDArray[max].length;j++){
if(TwoDArray[j][0]==Array[i]){
TwoDArray[j][1]+=;
found = true;
break;
}
}
if(found==false){
TwoDArray[max+1][0]=Array[i];
TwoDArray[max+1][1]=1;
max+=;
}
If you can comment or provide better solution would be very helpful.
Use map or hash table to implement this. Insert key as the array item and value as the frequency.
Alternatively you can use array too if the range of array elements are not too large. Increase the count of value at indexes corresponding to the array element.
I would build a map keyed by the item in the array and with a value that is the count of that item. One pass over the array to build the map that contains the counts. For each item, look it's count up in the map, increment the count, and put the new count back into the map.
The map put and get operations can be constant time (e.g., if you use a hash map implementation with a good hash function and properly sized backing store). This means you can compute the frequencies in time proportional to the number of elements in your array.
I'm not saying this is better than using a map or hash table (especially not when there are lots of duplicates, though in that case you can get close to O(n) sorting with certain techniques, so this is not too bad either), it's just an alternative.
Sort the array
Use a (single) for-loop to iterate through the sorted array
If you find the same element as the previous one, increment the current count
If you find a different element, store the previous element and its count and set the count to 1
At the end of the loop, store the previous element and its count
I'm attempting to solve http://projecteuler.net/problem=1.
I want to create a method which takes in an integer and then creates an array of all the integers preceding it and the integer itself as values within the array.
Below is what I have so far. Code doesn't work.
def make_array(num)
numbers = Array.new num
count = 1
numbers.each do |number|
numbers << number = count
count = count + 1
end
return numbers
end
make_array(10)
(1..num).to_a is all you need to do in Ruby.
1..num will create a Range object with start at 1 and end at whatever value num is. Range objects have to_a method to blow them up into real Arrays by enumerating each element within the range.
For most purposes, you won't actually need the Array - Range will work fine. That includes iteration (which is what I assume you want, given the problem you're working on).
That said, knowing how to create such an Array "by hand" is valuable learning experience, so you might want to keep working on it a bit. Hint: you want to start with an empty array ([]) instead with Array.new num, then iterate something num.times, and add numbers into the Array. If you already start with an Array of size num, and then push num elements into it, you'll end up with twice num elements. If, as is your case, you're adding elements while you're iterating the array, the loop never exits, because for each element you process, you add another one. It's like chasing a metal ball with the repulsing side of a magnet.
To answer the Euler Question:
(1 ... 1000).to_a.select{|x| x%3==0 || x%5==0}.reduce(:+) # => 233168
Sometimes a one-liner is more readable than more detailed code i think.
Assuming you are learning Ruby by examples on ProjectEuler, i'll explain what the line does:
(1 ... 1000).to_a
will create an array with the numbers one to 999. Euler-Question wants numbers below 1000. Using three dots in a Range will create it without the boundary-value itself.
.select{|x| x%3==0 || x%5==0}
chooses only elements which are divideable by 3 or 5, and therefore multiples of 3 or 5. The other values are discarded. The result of this operation is a new Array with only multiples of 3 or 5.
.reduce(:+)
Finally this operation will sum up all the numbers in the array (or reduce it to) a single number: The sum you need for the solution.
What i want to illustrate: many methods you would write by hand everyday are already integrated in ruby, since it is a language from programmers for programmers. be pragmatic ;)
I'm a beginner at groovy and I can't seem to understand this code. Can you please tell me how does this code operate?
def list = [ [1,0], [0,1,2] ]
list = list.sort { a,b -> a[0] <=> b[0] }
assert list == [ [0,1,2], [1,0] ]
what I know is the second line should return the value of 1 because of the spaceship operator but what is the use of that? and what type of sort is this? (there are 6 sort methods in the gdk api and i'm not really sure which is one is used here)
The code is using Collection#sort(Closure). Notice that this method has two variants:
If the closure is binary (i.e. it takes two parameters), sort uses it as the typical comparator interface: it should return an negative integer, zero or a positive integer when the first parameter is less than, equal, or grater than the second parameter respectively.
This is the variant that is being used in that piece of code. It is comparing the elements of the list, which are, in turn, lists, by their first element.
If the closure is unary (i.e. it takes only one parameter) it is used to generate the values that are then going to be used for comparison (in some languages this is called a "key" function).
Therefore, the snippet of code you posted can be rewritten as:
def list = [[1,0], [0,1,2]]
list = list.sort { it[0] } // or { it.first() }
assert list == [[0,1,2], [1,0]]
Notice that using this unary-closure variant is very convenient when you want to compare the elements by some value or some "weight" that is calculated the same way for every element.
The sort in your code snippet uses the comparator argument method call - see http://groovy.codehaus.org/groovy-jdk/java/util/Collection.html#sort(java.util.Comparator)
So, you are sorting the collection using your own comparator. Now the comparator simply uses the first element of the inner collection to decide the order of the outer collection.