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Checking if an array is element of a list of arrays


Checking if an array is element of a list of arrays

By : Peter G.
Date : November 17 2020, 11:55 AM
like below fixes the issue When you do obj in list python compares obj for equality with each of the members of list. The problem is that the equality operator in numpy doesn't behave in a consistent way. For example:
code :
>>> obj_3 == obj_6
False
>>> obj_7==obj_6
array([False, False, False, False], dtype=bool)
for i in [obj_1, obj_2, obj_3, obj_4, obj_5, obj_6, obj_7, obj_8, obj_9, obj_10, obj_11, obj_12, obj_13]:
    print any(np.array_equal(i, j) for j in [obj_1, obj_2, obj_3, obj_4, obj_5, obj_6, obj_7, obj_8, obj_9])
True
True
True
True
True
True
True
True
True
False
False
False
False


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Checking if each element of array is in multiple arrays

Checking if each element of array is in multiple arrays


By : Redline
Date : March 29 2020, 07:55 AM
Hope that helps This looks like a job for a recursion (might not be the most efficient but definitely the easiest solution and if your data is this small it shouldn't matter):
code :
var check = function(input, colors) {
    if (!input.length) {
        return true;
    }
    var input_color = input.pop();
    var ok = false;
    for (var i = 0; i < colors.length; i++) {
        var color = colors[i];
        if (!color) {
            break;
        }
        if (color.indexOf(input_color) !== -1) {
            colors.splice(i, 1);
            ok = check(input, colors);
            if (!ok) {
                colors.splice(i, 0, color);
            } else {
                break;
            }
        }
    }
    if (!ok) {
        input.push(input_color);
    }
    return ok;
};
var colors = [
  ['white', 'blue'],
  ['green', 'yellow'],
  ['black'],
  ['yellow', 'blue', 'pink'],
  ['orange', 'red'],
  ['brown', 'white']
];
check(['white', 'blue', 'pink', 'black', 'orange', 'yellow'], colors);
sorting an array list of int arrays in terms of the first and the second element of the int arrays

sorting an array list of int arrays in terms of the first and the second element of the int arrays


By : Cristian Trujillo
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further , try this
code :
if(a[0] - b[0]==0) //Like the first row of the matrix if so, we proceed to the next to know which is lower
Collections.sort(segment, new Comparator<int[]>() {
         public int compare(int[] a, int[] b) {
              if(a[0] - b[0]==0) //if equals
              {
                  return a[1]-b[1];//recompare 
              }
              else
                  return a[0]-b[0];
         }
    });
Insert a list, element by element, into the elements of an array of arrays

Insert a list, element by element, into the elements of an array of arrays


By : Tanuj Kumar
Date : March 29 2020, 07:55 AM
Hope that helps I have an array of arrays: , You have a list of arrays, plus another list or array:
code :
In [656]: parameters = [np.array([1,2,3]) for _ in range(5)]
In [657]: temps=np.arange(5)
In [659]: [np.concatenate((arr,[t])) for arr, t in zip(parameters, temps)]
Out[659]: 
[array([1, 2, 3, 0]),
 array([1, 2, 3, 1]),
 array([1, 2, 3, 2]),
 array([1, 2, 3, 3]),
 array([1, 2, 3, 4])]
 [np.append(arr,t) for arr, t in zip(parameters,temps)]
for i,(arr,t) in enumerate(zip(parameters,temps)):
    parameters[i] = np.append(arr,t)
In [663]: np.hstack((np.vstack(parameters),temps[:,None]))
Out[663]: 
array([[1, 2, 3, 0],
       [1, 2, 3, 1],
       [1, 2, 3, 2],
       [1, 2, 3, 3],
       [1, 2, 3, 4]])
checking adjacent elements of an element in an array of arrays in java

checking adjacent elements of an element in an array of arrays in java


By : cozzbp
Date : March 29 2020, 07:55 AM
help you fix your problem So, i have a char table that looks like , The most elegant way i can think of is this:
code :
public static void main(String[] args) {

    char[] a = {'.', '.', '*', '.', '*'};
    char[] b = {'.', '*', '*', '.', '*'};
    char[] c = {'.', '.', '*', '.', '*'};
    char[] d = {'.', '*', '*', '.', '*'};
    char[] e = {'*', '.', '*', '.', '*'};
    char[][] ae = {a, b, c, d, e};

    char[][] numberArray = new char[5][5];


    for (int i = 0; i < ae.length; i++) {
        for (int j = 0; j < ae[i].length;  j++) {
            numberArray[i][j] = checkAdjacentField(i, j, ae);
        }
    }
    StringBuilder matrix = new StringBuilder();

    for (char[] aNumberArray : numberArray) {
        StringBuilder bld = new StringBuilder("{");
        for (char character : aNumberArray) {
            bld.append(character).append(",");
        }
        bld.deleteCharAt(bld.length() - 1);
        bld.append("}");
        matrix.append(bld.toString()).append("\n");
    }
    System.out.println(matrix.toString());
}

private static char checkAdjacentField(int i, int j, char[][] ae) {
    int count = 0;
    if (j <= ae[i].length - 2) { // to the right
        count += ae[i][j + 1] == '*' ? 1 : 0;
    }
    if (j <= ae[i].length - 2 && i <= ae.length -2) { // move to top right
        count += ae[i + 1][j + 1] == '*' ? 1 : 0;
    }
    if (j <= ae[i].length - 2 && i > 0) { // move to bottom right
        count += ae[i - 1][j + 1] == '*' ? 1 : 0;
    }
    if (j > 0) { // to the left
        count += ae[i][j - 1] == '*' ? 1 : 0;
    }
    if (j > 0 && i <= ae.length -2) { // to top left
        count += ae[i + 1][j - 1] == '*' ? 1 : 0;
    }
    if (j > 0 && i > 0) { // to bottom left
        count += ae[i - 1][j - 1] == '*' ? 1 : 0;
    }
    if (i <= ae.length -2) { // move to top
        count += ae[i +1][j] == '*' ? 1 : 0;
    }
    if (i > 0) { // move top bottom
        count += ae[i - 1][j] == '*' ? 1 : 0;
    }
    System.out.printf("field %s, %s has %s Adjacent fields with a * \n", i, j , count);
    String stringValue = String.valueOf(count);
    return stringValue.charAt(0);
}
Looking up an element or checking if it exists in an array/list?

Looking up an element or checking if it exists in an array/list?


By : Samar Hosseinzadegan
Date : March 29 2020, 07:55 AM
will be helpful for those in need If you want to check for a specific room and throw an Exception if it doesn't exist then you need to provide the Exception supplier to orElseThrow() method, e.g.:
code :
List<Room> rooms = new ArrayList<>();
Room room = rooms.stream()
        .filter(r -> r.getRoomNumber() == 101)
        .findFirst()
        .orElseThrow(() -> new NoSuchElementException());
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