Function removeAllWhere

removeAllWhere<T>(array, fn, start?, stop?): number
Remove all occurrences of values which match a predicate.
Type Parameters
- T
Parameters
- array: T[]
  The array of interest.
- fn: ((value, index) => boolean)
  The predicate function to apply to the values.
  - - (value, index): boolean
    - Parameters
      value: T
      index: number
      Returns boolean
- start: number = 0
  The index of the first element in the range to be searched, inclusive. The default value is 0. Negative values are taken as an offset from the end of the array.
- stop: number = -1
  The index of the last element in the range to be searched, inclusive. The default value is -1. Negative values are taken as an offset from the end of the array.
Returns number
The number of elements removed from the array.

Notes
If stop < start the search will conceptually wrap at the end of the array, however the array will be traversed front-to-back.

Complexity
Linear.

Example
```
import { ArrayExt } from '@lumino/algorithm';

function isEven(value: number): boolean {
  return value % 2 === 0;
}

function isNegative(value: number): boolean {
  return value < 0;
}

let data = [0, 12, -13, -9, 23, 39, 14, -15, 12, 75];
ArrayExt.removeAllWhere(data, isEven);            // 4
ArrayExt.removeAllWhere(data, isNegative, 0, 3);  // 2
```
- Defined in array.ts:1455