How to Get First N Number of Elements from an Array

How to get first N number of elements from an array

I believe what you're looking for is:

// ...inside the render() function

var size = 3;
var items = list.slice(0, size).map(i => {
    return <myview item={i} key={i.id} />
});                       
return (
  <div>
    {items}
  </div>   
)

Ruby on Rails - Get first n number of elements from array with condition

When you only want to show the first 3 comments that match a certain condition then I would do this:

<% matching_comments = @comments.select { |comment| comment["postId"] == post["id"] } %>
<% matching_comments.first(3).each do |comment| %>
  <td><%= comment["body"] %></td>
<% end %>
<td><%= matching_comments.size %></td>

You might load the @comments with the condition in the controller to make the view easier to read.

Javascript Array - get first 10 items

This will do the job. Replace [startIndex] and [endIndex] with 0 and 10 to get the first 10.

data.slice([startIndex], [endIndex]).map((item, i) => {
  placeIDs.push(item.place_id);
});

Get the first N elements of an array?

Use array_slice()

This is an example from the PHP manual: array_slice

$input = array("a", "b", "c", "d", "e");
$output = array_slice($input, 0, 3);   // returns "a", "b", and "c"

There is only a small issue

If the array indices are meaningful to you, remember that array_slice will reset and reorder the numeric array indices. You need the preserve_keys flag set to trueto avoid this. (4th parameter, available since 5.0.2).

Example:

$output = array_slice($input, 2, 3, true);

Output:

array([3]=>'c', [4]=>'d', [5]=>'e');

jq: get first n elements of an array field

If you are looking to print only those specific fields on the first 3 results, use the slice operator. See jqplay for demo

.identifier[:3] | map({system, value})

return the first n elements of a k sized array in O(1) time

If you must work with raw arrays and not ArrayList then Arrays has what you need. If you look at the source code, these are the absolutely best ways to get a copy of an array. They do have a good bit of defensive programming because the System.arraycopy() method throws lots of unchecked exceptions if you feed it illogical parameters.

You can use either Arrays.copyOf() which will copy from the first to Nth element to the new shorter array.

public static <T> T[] copyOf(T[] original, int newLength)

Copies the specified array, truncating or padding with nulls (if
necessary) so the copy has the specified length. For all indices that
are valid in both the original array and the copy, the two arrays will
contain identical values. For any indices that are valid in the copy
but not the original, the copy will contain null. Such indices will
exist if and only if the specified length is greater than that of the
original array. The resulting array is of exactly the same class as
the original array.

2770
2771    public static <T,U> T[] More ...copyOf(U[] original, int newLength, Class<? extends T[]> newType) {
2772        T[] copy = ((Object)newType == (Object)Object[].class)
2773            ? (T[]) new Object[newLength]
2774            : (T[]) Array.newInstance(newType.getComponentType(), newLength);
2775        System.arraycopy(original, 0, copy, 0,
2776                         Math.min(original.length, newLength));
2777        return copy;
2778    }

or Arrays.copyOfRange() will also do the trick:

public static <T> T[] copyOfRange(T[] original, int from, int to)

Copies the specified range of the specified array into a new array.
The initial index of the range (from) must lie between zero and
original.length, inclusive. The value at original[from] is placed into
the initial element of the copy (unless from == original.length or
from == to). Values from subsequent elements in the original array are
placed into subsequent elements in the copy. The final index of the
range (to), which must be greater than or equal to from, may be
greater than original.length, in which case null is placed in all
elements of the copy whose index is greater than or equal to
original.length - from. The length of the returned array will be to -
from. The resulting array is of exactly the same class as the original
array.

3035    public static <T,U> T[] More ...copyOfRange(U[] original, int from, int to, Class<? extends T[]> newType) {
3036        int newLength = to - from;
3037        if (newLength < 0)
3038            throw new IllegalArgumentException(from + " > " + to);
3039        T[] copy = ((Object)newType == (Object)Object[].class)
3040            ? (T[]) new Object[newLength]
3041            : (T[]) Array.newInstance(newType.getComponentType(), newLength);
3042        System.arraycopy(original, from, copy, 0,
3043                         Math.min(original.length - from, newLength));
3044        return copy;
3045    }

As you can see, both of these are just wrapper functions over System.arraycopy with defensive logic that what you are trying to do is valid.

System.arraycopy is the absolute fastest way to copy arrays.

How to find the number of the elements of an array that are in a specific range of values

can you try this one.
in this way it do not sort, but it will display values, that are between the numbers the user enters

#include <stdio.h>
#define SIZE 5
int main(void){
    float grades[SIZE];
    float g;
    int i;
    for (i=0;i<SIZE;i++){
        printf("Enter a grade: ");
        scanf("%f", &g);
        grades[i]= g;
    }
    float a,b;
    printf("Enter a grade range: \n");
    scanf("%f %f", &a,&b);
    
    
    for (i=0;i<SIZE;i++){
        if(a>b){
            if(grades[i]>=b && grades[i]<=a){
                printf("%f ", grades[i]);
            }
        }else{
            if(grades[i]>=a && grades[i]<=b){
                printf("%f ", grades[i]);
            }
        }
    }
    
    return 0;
}

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