我需要查找数组b
中包含数组arr
的字符串数。我得到了输出,但我按此顺序需要它。[[3,6,0],[1,3,1]]
我的代码在这里。
var arr = [["00","00","00","01","01","01","01","01","01"],["000","100","01","01","01"]];
var b = ["00","01",10];
var cc = [];
for (var i=0;i<b.length;i++) {
var k = [];
for (var y=0;y<arr.length;y++) {
var a = 0;
for (var x=0;x<arr[y].length;x++) {
if ((arr[y][x].substring(0,2)).indexOf(b[i]) != -1) {
a++;
}
}
k.push(a)
}
cc.push(k);
}
console.log(JSON.stringify(cc));// output :[[3,1],[6,3],[0,1]]
实际输出:[[3,1],[6,3],[0,1]]
预期产出:[[3,6,0],[1,3,1]]
我想在javascript或jquery中得到结果。
答案 0 :(得分:2)
正如b
号码10
中所示,您需要将其转换为String
然后搜索数组,因为arr
仅包含字符串
var arr = [
["00","00","00","01","01","01","01","01","01"],
["000","100","01","01","01"]
];
var b = ["00", "01", 10];
var len, i, j, key, result = [], counts = [], count = 0;
for (i = 0, len = arr.length; i < len; i++) {
for (j = 0; j < b.length; j++) {
count = 0;
key = String(b[j]);
count = arr[i].filter(function (el) {
return el.slice(0, 2) === key;
}).length;
counts.push(count);
}
result.push(counts);
counts = [];
}
console.log(JSON.stringify(result));
&#13;
IE < 9
的版本,其中没有.filter
方法
var arr = [
["00","00","00","01","01","01","01","01","01"],
["000","100","01","01","01"]
];
var b = ["00", "01", 10];
var len,
key,
result = [],
counts = [],
i, j, k, count;
for (i = 0, len = arr.length; i < len; i++) {
for (j = 0; j < b.length; j++) {
count = 0;
key = String(b[j]);
for (k = 0; k < arr[i].length; k++) {
if (arr[i][k].slice(0, 2) === key) {
count++;
}
}
counts.push(count);
}
result.push(counts);
counts = [];
}
console.log(JSON.stringify(result));
&#13;
答案 1 :(得分:1)
好像你的样本输入中有一些拼写错误。以下代码可能有所帮助。
var arr = [["00","00","00","01","01","01","01","01","01"],["00","10","01","01","01"]];
var b = ["00","01","10"];
var cc = [];
arr.forEach(function(ar,i){
cc[i] = [];
b.forEach(function(a,j){
cc[i][j] = ar.filter(function(d){ return d==a }).length;
});
});
alert(JSON.stringify(cc));
&#13;
或者
var arr = [
["00", "00", "00", "01", "01", "01", "01", "01", "01"],
["00", "10", "01", "01", "01"]
];
var b = ["00", "01", "10"];
var cc = arr.map(function(ar) {
return b.map(function(a) {
return ar.filter(function(d) {
return d == a
}).length;
})
});
alert(JSON.stringify(cc));
&#13;