Baseline Widely available
lastIndexOf()
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const animals = ["Dodo", "Tiger", "Penguin", "Dodo"];
console.log(animals.lastIndexOf("Dodo"));
// Expected output: 3
console.log(animals.lastIndexOf("Tiger"));
// Expected output: 1
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arr.lastIndexOf(searchElement) arr.lastIndexOf(searchElement, fromIndex)忏
searchElement
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æè¿°lastIndexOf
compares searchElement
to elements of the Array using strict equality (the same method used by the ===, or triple-equals, operator).
lastIndexOf
The following example uses lastIndexOf
to locate values in an array.
var numbers = [2, 5, 9, 2];
numbers.lastIndexOf(2); // 3
numbers.lastIndexOf(7); // -1
numbers.lastIndexOf(2, 3); // 3
numbers.lastIndexOf(2, 2); // 0
numbers.lastIndexOf(2, -2); // 0
numbers.lastIndexOf(2, -1); // 3
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The following example uses lastIndexOf
to find all the indices of an element in a given array, using push
to add them to another array as they are found.
var indices = [];
var array = ["a", "b", "a", "c", "a", "d"];
var element = "a";
var idx = array.lastIndexOf(element);
while (idx != -1) {
indices.push(idx);
idx = idx > 0 ? array.lastIndexOf(element, idx - 1) : -1;
}
console.log(indices);
// [4, 2, 0]
Note that we have to handle the case idx == 0
separately here because the element will always be found regardless of the fromIndex
parameter if it is the first element of the array. This is different from the indexOf
method.
lastIndexOf
was added to the ECMA-262 standard in the 5th edition; as such it may not be present in other implementations of the standard. You can work around this by inserting the following code at the beginning of your scripts, allowing use of lastIndexOf
in implementations which do not natively support it. This algorithm is exactly the one specified in ECMA-262, 5th edition, assuming Object
, TypeError
, Number
, Math.floor
, Math.abs
, and Math.min
have their original values.
// Production steps of ECMA-262, Edition 5, 15.4.4.15
// Reference: http://es5.github.io/#x15.4.4.15
if (!Array.prototype.lastIndexOf) {
Array.prototype.lastIndexOf = function (searchElement /*, fromIndex*/) {
"use strict";
if (this === void 0 || this === null) {
throw new TypeError();
}
var n,
k,
t = Object(this),
len = t.length >>> 0;
if (len === 0) {
return -1;
}
n = len - 1;
if (arguments.length > 1) {
n = Number(arguments[1]);
if (n != n) {
n = 0;
} else if (n != 0 && n != 1 / 0 && n != -(1 / 0)) {
n = (n > 0 || -1) * Math.floor(Math.abs(n));
}
}
for (k = n >= 0 ? Math.min(n, len - 1) : len - Math.abs(n); k >= 0; k--) {
if (k in t && t[k] === searchElement) {
return k;
}
}
return -1;
};
}
Again, note that this implementation aims for absolute compatibility with lastIndexOf
in Firefox and the SpiderMonkey JavaScript engine, including in several cases which are arguably edge cases. If you intend to use this in real-world applications, you may be able to calculate from
with less complicated code if you ignore those cases.
-0
. For example, [0].lastIndexOf(0, -0)
will now always return +0
(Firefox bug 1242043).RetroSearch is an open source project built by @garambo | Open a GitHub Issue
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