如何以不同的方式合并 javascript 中的数组?
How to merge arrays in javascript in a different way?
我想以稍微不同的方式合并数组。
我有 2 个或更多数组,例如:
var array1 = ["apple", "banana"];
var array2 = ["apple", "apple", "orange"];
我想要输出:
var array3 = ["apple", "apple", "banana", "orange"];
因此,如果任何给定数组中不止一次包含一个变量,合并算法应该将所有变量都从该数组中保留下来。
我看到了一些防止重复的代码,但它给出的输出如下:
var array3 = ["apple", "banana", "orange"];
更多示例:
var arr1 = [1,2,3,4];
var arr2 = [1,1,2,4,5,5,5];
var arr3 = [1,3,3,5,5];
我想要输出:
var array4 = [1,1,2,3,3,4,5,5,5];
我该怎么做?
未经测试且不是 JS,但我认为这就是您要找的。
我刚刚对其进行了手动测试,它适用于您的测试用例。
while items in lista or items in listb
compare a.head, b.head
if a.head is smaller or b.is_empty then
append a.head to output
a.drophead
else if b.head is smaller or a.is_empty then
append b.head to output
b.drophead
else
append b.head to output
b.drophead
a.drophead
这是一种计算每个数组中每个项目出现的次数的方法:
var arr1 = [1,2,3,4];
var arr2 = [1,1,2,4,5,5,5];
var arr3 = [1,3,3,5,5];
function joinCommon(/* list of arrays */) {
var arr, arrayCounts, masterList = {}, item, output;
// for each array passed in
for (var i = 0; i < arguments.length; i++) {
arr = arguments[i];
arrayCounts = {};
// iterate each array
for (var j = 0; j < arr.length; j++) {
item = arr[j];
if (!arrayCounts[item]) {
arrayCounts[item] = 1;
} else {
++arrayCounts[item];
}
// now keep master list and master counts
if (!masterList[item]) {
masterList[item] = {cnt: 1, val: item};
} else {
masterList[item].cnt = Math.max(masterList[item].cnt, arrayCounts[item]);
}
}
}
// now output result
output = [];
for (var i in masterList) {
for (var j = 0; j < masterList[i].cnt; j++) {
output.push(masterList[i].val);
}
}
return output;
}
var results = joinCommon(arr1, arr2, arr3);
我喜欢用 ramda (http://ramdajs.com/docs/index.html) 来处理这些东西
var arr1 = [1,2,3,4];
var arr2 = [1,1,2,4,5,5,5];
var arr3 = [1,3,3,5,5];
var allArrays = [arr1, arr2, arr3];
var allValues = R.compose(R.uniq, R.flatten)(allArrays);
var getItemCounts = R.countBy(function(item) {
return item;
});
var itemCounts = R.map(function(arr) {
return getItemCounts(arr);
})(allArrays);
var combined = [];
R.forEach(function(item) {
var countsForItem = R.pluck(item, itemCounts);
var maxCount = R.max(countsForItem);
combined.push.apply(combined, R.repeatN(item, maxCount));
})(allValues);
console.log(combined.sort());
JSFiddle:http://jsfiddle.net/pcr0q1xa/3/
这是一个使用 ECMA5 的解决方案。
Javascript
function indexOf(items, value) {
return items.map(function (subitem) {
return subitem.value;
}).indexOf(value);
}
function countItems(previous, item) {
var atIndex = indexOf(previous, item);
if (atIndex !== -1) {
previous[atIndex].count += 1;
} else {
previous.push({
value: item,
count: 1
});
}
return previous;
}
function mergeCounts(item) {
var atIndex = indexOf(this, item.value);
if (atIndex === -1) {
this.push(item);
} else if (this[atIndex].count < item.count) {
this[atIndex] = item;
}
}
function expandCounts(previous, item) {
var iter;
for (iter = 0; iter < item.count; iter += 1) {
previous.push(item.value);
}
return previous;
}
function mergeArg(items, arg) {
arg.reduce(countItems, []).forEach(mergeCounts, items);
return items;
}
function mergeMaxItems() {
return [].reduce.call(arguments, mergeArg, []).reduce(expandCounts, []);
}
var arr1 = [1, 2, 3, 4],
arr2 = [1, 1, 2, 4, 5, 5, 5],
arr3 = [1, 3, 3, 5, 5];
document.body.appendChild(document.createTextNode(mergeMaxItems(arr1, arr2, arr3)));
Ramda是你的朋友。
function merge () {
return R.chain(R.apply(R.repeat), R.toPairs(R.reduce(
R.mergeWith(R.max),
{},
R.map(R.countBy(R.identity), arguments)
)))
}
var array1 = ["apple", "banana"];
var array2 = ["apple", "apple", "orange"];
console.log(JSON.stringify(merge(array1, array2)))
var arr1 = [1,2,3,4];
var arr2 = [1,1,2,4,5,5,5];
var arr3 = [1,3,3,5,5];
console.log(JSON.stringify(merge(arr1, arr2, arr3)))
<script src="http://cdnjs.cloudflare.com/ajax/libs/ramda/0.22.1/ramda.min.js"></script>
我想以稍微不同的方式合并数组。 我有 2 个或更多数组,例如:
var array1 = ["apple", "banana"];
var array2 = ["apple", "apple", "orange"];
我想要输出:
var array3 = ["apple", "apple", "banana", "orange"];
因此,如果任何给定数组中不止一次包含一个变量,合并算法应该将所有变量都从该数组中保留下来。
我看到了一些防止重复的代码,但它给出的输出如下:
var array3 = ["apple", "banana", "orange"];
更多示例:
var arr1 = [1,2,3,4];
var arr2 = [1,1,2,4,5,5,5];
var arr3 = [1,3,3,5,5];
我想要输出:
var array4 = [1,1,2,3,3,4,5,5,5];
我该怎么做?
未经测试且不是 JS,但我认为这就是您要找的。 我刚刚对其进行了手动测试,它适用于您的测试用例。
while items in lista or items in listb
compare a.head, b.head
if a.head is smaller or b.is_empty then
append a.head to output
a.drophead
else if b.head is smaller or a.is_empty then
append b.head to output
b.drophead
else
append b.head to output
b.drophead
a.drophead
这是一种计算每个数组中每个项目出现的次数的方法:
var arr1 = [1,2,3,4];
var arr2 = [1,1,2,4,5,5,5];
var arr3 = [1,3,3,5,5];
function joinCommon(/* list of arrays */) {
var arr, arrayCounts, masterList = {}, item, output;
// for each array passed in
for (var i = 0; i < arguments.length; i++) {
arr = arguments[i];
arrayCounts = {};
// iterate each array
for (var j = 0; j < arr.length; j++) {
item = arr[j];
if (!arrayCounts[item]) {
arrayCounts[item] = 1;
} else {
++arrayCounts[item];
}
// now keep master list and master counts
if (!masterList[item]) {
masterList[item] = {cnt: 1, val: item};
} else {
masterList[item].cnt = Math.max(masterList[item].cnt, arrayCounts[item]);
}
}
}
// now output result
output = [];
for (var i in masterList) {
for (var j = 0; j < masterList[i].cnt; j++) {
output.push(masterList[i].val);
}
}
return output;
}
var results = joinCommon(arr1, arr2, arr3);
我喜欢用 ramda (http://ramdajs.com/docs/index.html) 来处理这些东西
var arr1 = [1,2,3,4];
var arr2 = [1,1,2,4,5,5,5];
var arr3 = [1,3,3,5,5];
var allArrays = [arr1, arr2, arr3];
var allValues = R.compose(R.uniq, R.flatten)(allArrays);
var getItemCounts = R.countBy(function(item) {
return item;
});
var itemCounts = R.map(function(arr) {
return getItemCounts(arr);
})(allArrays);
var combined = [];
R.forEach(function(item) {
var countsForItem = R.pluck(item, itemCounts);
var maxCount = R.max(countsForItem);
combined.push.apply(combined, R.repeatN(item, maxCount));
})(allValues);
console.log(combined.sort());
JSFiddle:http://jsfiddle.net/pcr0q1xa/3/
这是一个使用 ECMA5 的解决方案。
Javascript
function indexOf(items, value) {
return items.map(function (subitem) {
return subitem.value;
}).indexOf(value);
}
function countItems(previous, item) {
var atIndex = indexOf(previous, item);
if (atIndex !== -1) {
previous[atIndex].count += 1;
} else {
previous.push({
value: item,
count: 1
});
}
return previous;
}
function mergeCounts(item) {
var atIndex = indexOf(this, item.value);
if (atIndex === -1) {
this.push(item);
} else if (this[atIndex].count < item.count) {
this[atIndex] = item;
}
}
function expandCounts(previous, item) {
var iter;
for (iter = 0; iter < item.count; iter += 1) {
previous.push(item.value);
}
return previous;
}
function mergeArg(items, arg) {
arg.reduce(countItems, []).forEach(mergeCounts, items);
return items;
}
function mergeMaxItems() {
return [].reduce.call(arguments, mergeArg, []).reduce(expandCounts, []);
}
var arr1 = [1, 2, 3, 4],
arr2 = [1, 1, 2, 4, 5, 5, 5],
arr3 = [1, 3, 3, 5, 5];
document.body.appendChild(document.createTextNode(mergeMaxItems(arr1, arr2, arr3)));
Ramda是你的朋友。
function merge () {
return R.chain(R.apply(R.repeat), R.toPairs(R.reduce(
R.mergeWith(R.max),
{},
R.map(R.countBy(R.identity), arguments)
)))
}
var array1 = ["apple", "banana"];
var array2 = ["apple", "apple", "orange"];
console.log(JSON.stringify(merge(array1, array2)))
var arr1 = [1,2,3,4];
var arr2 = [1,1,2,4,5,5,5];
var arr3 = [1,3,3,5,5];
console.log(JSON.stringify(merge(arr1, arr2, arr3)))
<script src="http://cdnjs.cloudflare.com/ajax/libs/ramda/0.22.1/ramda.min.js"></script>