Javascript 如何在javascript中找到嵌套数组的最大值/最小值？
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How to find the max/min of a nested array in javascript?
提问by nachocab
I want to find the maximum of a nested array, something like this:
我想找到嵌套数组的最大值，如下所示：
a = [[1,2],[20,3]]
d3.max(d3.max(a)) // 20
but my array contains a text field that I want to discard:
但我的数组包含一个我想丢弃的文本字段：
a = [["yz",1,2],["xy",20,3]]
d3.max(a) // 20
回答by mbostock
If you have a nested array of numbers (arrays = [[1, 2], [20, 3]]), nest d3.max:
如果您有一个嵌套的数字数组 ( arrays = [[1, 2], [20, 3]])，请嵌套d3.max：
var max = d3.max(arrays, function(array) {
return d3.max(array);
});
Or equivalently, use array.map:
或者等效地，使用array.map：
var max = d3.max(arrays.map(function(array) {
return d3.max(array);
}));
If you want to ignore string values, you can use array.filterto ignore strings:
如果要忽略字符串值，可以使用array.filter忽略字符串：
var max = d3.max(arrays, function(array) {
return d3.max(array.filter(function(value) {
return typeof value === "number";
}));
});
Alternatively, if you know the string is always in the first position, you could use array.slicewhich is a bit more efficient:
或者，如果您知道字符串总是在第一个位置，您可以使用更高效的array.slice：
var max = d3.max(arrays, function(array) {
return d3.max(array.slice(1));
});
Yet another option is to use an accessor function which returns NaNfor values that are not numbers. This will cause d3.max to ignore those values. Conveniently, JavaScript's built-in Numberfunction does exactly this, so you can say:
另一种选择是使用访问器函数，该函数返回NaN非数字的值。这将导致 d3.max 忽略这些值。方便的是，JavaScript 的内置Number函数正是这样做的，所以你可以说：
var max = d3.max(arrays, function(array) {
return d3.max(array, Number);
});
回答by kbec
Use this:
用这个：
function arrmax(arrs) {
var toplevel = [];
var f = function(v) {
return !isNaN(v);
};
for (var i = 0, l = arrs.length; i<l; i++) {
toplevel.push(Math.max.apply(window, arrs[i].filter(f)));
}
return Math.max.apply(window, toplevel);
}
or better:
或更好：
function arrmax(arrs) {
if (!arrs || !arrs.length) return undefined;
var max = Math.max.apply(window, arrs[0]), m,
f = function(v){ return !isNaN(v); };
for (var i = 1, l = arrs.length; i<l; i++) {
if ((m = Math.max.apply(window, arrs[i].filter(f)))>max) max=m;
}
return max;
}
See MDNfor Array.filter method details.
有关Array.filter 方法的详细信息，请参阅MDN。
回答by kennebec
You can flatten an array and apply a function to each member
您可以展平数组并将函数应用于每个成员
Array.prototype.flatten= function(fun){
if(typeof fun!= 'function') fun= '';
var A= [], L= this.length, itm;
for(var i= 0; i<L; i++){
itm= this[i];
if(itm!= undefined){
if(!itm.flatten){
if(fun) itm= fun(itm);
if(itm) A.push(itm);
}
else A= A.concat(itm.flatten(fun));
}
}
return A;
}
var a= [["yz", 1, 2], ["xy", 20, 3]], max=-Infinity;
var max=Math.max.apply(a, a.flatten(Number));
回答by Rick
If you now exactly what columns you want to test, you can use:
如果您现在确切要测试哪些列，则可以使用：
var columns = ["ColumnA", "ColumnB", "ColumnC"];
var max = selectedMax(columns,dataset);
var min = selectedMin(columns,dataset)
function selectedMax(columns, dataset) {
var max;
columns.forEach(function(element, index, array) {
var tmpmax = d3.max(dataset, function(d) {
return +d[element];
});
max = (tmpmax > max || max === undefined) ? tmpmax : max;
});
return max;
}
function selectedMin(columns, dataset) {
var min;
columns.forEach(function(element, index, array) {
var tmpmin = d3.min(dataset, function(d) {
return +d[element];
});
min = (tmpmin < min || min === undefined) ? tmpmin : min;
});
return min;
}
回答by Xavier Guihot
d3.arrayprovides d3.mergewhich flattens an array of arrays.
Coupled with d3.maxand javascript's Numberas an accessor:
加上d3.max和 javascriptNumber作为访问器：
var max = d3.max(d3.merge(arrays), Number);
For example:
例如：
var input = [["yz", 1, 2], ["xy", 20, 3]];
var max = d3.max(d3.merge(input), Number);
console.log(max);<script src="https://d3js.org/d3-array.v2.min.js"></script>回答by btown
It's a cruel hack, but looking at the source codefor d3.max, your best bet might be to define a d3.max1that discards the first element by copying that code, but replacing i=-1with i=0. The code at that link is excerpted here. Note that I'm not a regular d3.js user, but from what I know of the library, you're going to want make sure your version has an f.callcase like this function does, so that it can respond to live updates correctly.
这是一个残酷的黑客，而是看源代码的d3.max，你最好的选择可能是定义d3.max1该丢弃通过复制代码，但更换的第一个元素i=-1用i=0。该链接上的代码摘录在此处。请注意，我不是一个普通的 d3.js 用户，但根据我对这个库的了解，你会想要确保你的版本有一个f.call像这个函数那样的情况，以便它可以正确响应实时更新。
d3.max = function(array, f) {
var i = -1,
n = array.length,
a,
b;
if (arguments.length === 1) {
while (++i < n && ((a = array[i]) == null || a != a)) a = undefined;
while (++i < n) if ((b = array[i]) != null && b > a) a = b;
} else {
while (++i < n && ((a = f.call(array, array[i], i)) == null || a != a)) a = undefined;
while (++i < n) if ((b = f.call(array, array[i], i)) != null && b > a) a = b;
}
return a;
};
Then it would just be d3.max(d3.max1(a)).
那么就可以了d3.max(d3.max1(a))。
