List of utility methods to do Array Min Value
double | min(double[] array) min if (array.length == 0) throw new IllegalArgumentException("length cannot be 0."); double maxElement = Double.POSITIVE_INFINITY; for (double d : array) { if (d < maxElement) maxElement = d; return maxElement; ... |
double | min(double[] array) Returns the minimum value in an array of double s.
double min = Double.POSITIVE_INFINITY; for (int i = 0; i < array.length; i++) { if (array[i] < min) { min = array[i]; return min; |
double | min(double[] array) Returns the minimum value in an array. if (array == null) { throw new IllegalArgumentException("The Array must not be null"); } else if (array.length == 0) { throw new IllegalArgumentException("Array cannot be empty."); double min = array[0]; for (int i = 1; i < array.length; i++) { min = min(array[i], min); ... |
double | min(double[] array) min if (array.length == 0) return 0; double res = array[0]; for (int i = 1; i < array.length; i++) { res = Math.min(res, array[i]); return res; |
double | min(double[] array) Double.NaN is ignored. double min = Double.NaN; int n = -1; for (int i = 0; i < array.length; i++) { if (!Double.isNaN(array[i])) { min = array[i]; n = i; break; if (n == -1) { return Double.NaN; for (int i = n + 1; i < array.length; i++) { if (min > array[i]) { min = array[i]; return min; |
double | min(double[] da) Returns the minimum value in vector of doubles. if (da.length == 0) { return Double.NaN; double out = Double.POSITIVE_INFINITY; for (int i = 0; i < da.length; i++) { out = Math.min(out, da[i]); return out; ... |
double | min(double[] data) Find the minimum of all elements in the array, ignoring elements that are NaN. double min = Double.NaN; for (int i = 0; i < data.length; i++) { if (Double.isNaN(data[i])) continue; if (Double.isNaN(min) || data[i] < min) min = data[i]; return min; ... |
double | min(double[] matrix) Compute the minimum value of a 1d array. double min = matrix[0]; for (int i = 0; i < matrix.length; i++) { if (min > matrix[i]) { min = matrix[i]; return min; |
int | min(double[] numbers) min if (numbers.length == 0) return -1; double min = numbers[0]; int id = 0; for (int i = 1; i < numbers.length; i++) { double n = numbers[i]; if (n < min) { min = n; ... |
double | min(double[] series) Finds the minimal value in timeseries. if (countNaN(series) == series.length) { return Double.NaN; double min = Double.MAX_VALUE; for (int i = 0; i < series.length; i++) { if (min > series[i]) { min = series[i]; return min; |