Question

我正在完成一项任务，以复制underscore.js库的部分内容。我坚持将“each”函数实现为“唯一”函数。

我的每个功能如下：

_.each = function(collection, iterator) {
//checks to see if collection is an array or object
if (Array.isArray(collection)) {
  //if array, iterates an action function on each index value
  for (var i = 0; i < collection.length; i++) {
    iterator(collection[i], i, collection);
  }
}
//else statement takes care of an object passed as argument
  else {
    //for...in loop will iterate through object keys
    for (var key in collection) {
      iterator(collection[key], key, collection)
    }
  }       
}

我能够在过滤函数中使用上面的每个函数，如下所示：

_.filter = function(collection, test) {       
     var newArray = [];
    //calling _.each function to iterate through array
    _.each(collection, function(value, index, list) {
    //applying truth test
      if (test(value, index, list)) {
      newArray.push(value);
      };
    });
  return newArray;
};

我试图在我的“独特”功能中使用每个功能，如下所示：

_.uniq = function(array) {
    //two temporary arrays, one to hold unique values and one for duplicates
  var unique = [];
  var notUnique = [];
  _.each(array, function(item, index) {
    //sort array
    var sortedArray = array.sort(function(a, b){return a-b});
    //if value is not equal to the next value after sort, it goes to the unique array
    if (sortedArray[index] !== sortedArray[index+1]) {
      unique.push(sortedArray[index]);
    }
    //if the value matches the next value, it goes to the not unique array
    else {
      notUnique.push(sortedArray[index]);
    }
  return unique;
});
}

然而，当我在测试数组上运行它时，我收到一个未定义的输出。使用像这样的嵌套函数时，数组排序方法是否不起作用？

Answer 1

您要在unique内返回_.each数组，所以基本上您的_.uniq函数默认重新调整undefined。

_.uniq = function(array) {
    //two temporary arrays, one to hold unique values and one for duplicates
    var unique = [];
    var notUnique = [];
    _.each(array, function(item, index) {
        //sort array
        var sortedArray = array.sort(function(a, b) {
            return a - b
        });
        //if value is not equal to the next value after sort, it goes to the unique array
        if (sortedArray[index] !== sortedArray[index + 1]) {
            unique.push(sortedArray[index]);
        }
        //if the value matches the next value, it goes to the not unique array
        else {
            notUnique.push(sortedArray[index]);
        }
    });
    return unique;  // return here
}

Javascript：使用“each”函数作为“唯一”函数的回调

1 个答案: