[英]Get number of non-empty elements from nested arrays
我有十个空值的数组
onTable[0 to 10];
看
["Jean5", "Jean3", "Paul2", "Jean6", "", "Paul4", "Jean", "peirre4", ""]
["Paul5", "peirre6", "peirre3", "", "Jean4", "Paul", "peirre5", "Jean2", ""]
...
我希望获得每个数组的长度没有空值,并且没有创建十个变量来检查。
我已经测试了这个解决方案计算数组中的空值,但我不想让make 10变量。 即: count1, count2,...
我也检查过根据长度比较两个数组:跳过空值但不是我想要的。
如果可能的话,我希望看起来像
onTable[0].length(exclude(""))
制作它的好方法是什么?
使用filter
与Boolean
过滤从子阵列非空元素,并使用length
就可以了。
onTable[0].filter(Boolean).length
由于空字符串在JavaScript中是假的,它将从过滤后的数组中删除。
演示:
var arr = [ ["Jean5", "Jean3", "Paul2", "Jean6", "", "Paul4", "Jean", "peirre4", ""], ["Paul5", "peirre6", "peirre3", "", "Jean4", "Paul", "peirre5", "Jean2", ""] ]; var len = arr[1].filter(Boolean).length; document.write(len);
使用prototype
:
Array.prototype.lengthWihtoutEmptyValues = function () { var initialLength = this.length; var finalLength = initialLength; for (var i = 0; i < initialLength; i++) { if (this[i] == "") { finalLength--; } } return finalLength; } var arrays = [ ["Jean5", "Jean3", "Paul2", "Jean6", "", "Paul4", "Jean", "peirre4", ""], ["Paul5", "peirre6", "peirre3", "", "Jean4", "Paul", "peirre5", "Jean2", ""] ]; var arrayLength = arrays[0].lengthWihtoutEmptyValues(); $("#arrayLength").html(arrayLength);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div id="arrayLength"></div>
您可以根据需要使用过滤函数:将值检查为undefined
或null
等。
var arr = [ ["Jean5", "Jean3", "Paul2", "Jean6", "", "Paul4", "Jean", "peirre4", ""], ["Paul5", "peirre6", "peirre3", "", "Jean4", "Paul", "peirre5", "Jean2", ""] ]; var len = arr[1].filter(function(x){ return x != ""}).length; document.write(len);
你应该避免“浪费”记忆并诱导过多的GC。 您可以reduce()
每个子数组reduce()
到它的非空值的计数:
sub.reduce(function(prev, cur) {
return prev + (!!cur);
}, 0);
var arr = [ ["Jean5", "Jean3", "Paul2", "Jean6", "", "Paul4", "Jean", "peirre4", ""], ["Paul5", "peirre6", "peirre3", "", "Jean4", "Paul", "peirre5", "Jean2", ""], ["Just1", "", ""] ]; var lengths = arr.map(function(sub) { return sub.reduce(function(prev, cur) { return prev + (!!cur); }, 0); }); document.write('[' + lengths.join('], [') + ']');
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