Question

我正在编写一个带有对象和数组的函数。我遍历数组，并使用该值作为键来查找对象。

我希望一旦找到1个值就停止循环并返回该值。运行该功能时，我总是不知所措。

const searchInObject = function(obj, keys) {
  //if array => loop and return the first value that is found.
  //if not array and type is string find in object
  // if no array or undefined return default value
  if (Array.isArray(keys)) {
    keys.map(key => {
      if (obj[key]) return obj[key];
    })
  }
};


const obj = {
  a: '1',
  b: '2'
};
console.log(searchInObject(obj, ['a', 'b']));

Answer 1

只需遍历键并检查键是否在obj中存在。如果是，则返回值

const searchInObject = function(obj, keys) {
  if (!Array.isArray(keys))
    return;

  for (const key of keys) {
    if (key in obj)
      return obj[key]
  }
};

const obj = { a: '1', b: '2' };

console.log(searchInObject(obj, ['a', 'b']));

Answer 2

尝试一下：

const searchInObject = function(obj, keys) {
  if (Array.isArray(keys)) { //if array => loop and return the first value that is found.
    for(let i = 0; i < keys.length; i++){
      if (obj[keys[i]]) return obj[keys[i]];
    }
  } else if (typeof keys === 'string') { //if not array and type is string find in object
    return obj[keys];
  } else { // if no array or undefined return default value
    return "default value"; // change this to the default value you want
  }
};


const obj = {
  a: '1',
  b: '2'
};
console.log(searchInObject(obj, ['a', 'b']));

Answer 3

您可以改为使用Array.find()函数。这将返回第一个找到的值；如果找不到任何内容，则返回undefined。

const searchInObject = (obj, keys) => {
    if (Array.isArray(keys))
        return keys.find(key => obj.hasOwnProperty(key))
    else return undefined;
}

console.log(searchInObject({a: '1', b: '2'}, ['a','b']))

Answer 4

我想您想使用importances = tree.feature_importances_ #std = np.std([tree.feature_importances_ for tree in forest.estimators_], # axis=0) indices = np.argsort(importances)[::-1] # Print the feature ranking print("Feature ranking:") for f in range(X.shape[1]): print("%d. feature %d (%f)" % (f + 1, indices[f], importances[indices[f]])) # Plot the feature importances of the forest plt.figure() plt.title("Feature importances") plt.bar(range(X.shape[1]), importances[indices], color="r", yerr=std[indices], align="center") plt.xticks(range(X.shape[1]), [feature_cols[i] for i in indices]) plt.xlim([-1, X.shape[1]]) plt.show()而不是filter。

下面是代码：

map

以上代码即使在const searchInObject = function(obj, keys) { if(keys instanceof Array) { keys = keys.filter(key => { // This will return undefined if key is not present in object return obj[key]; }); // The filtered keys array will contain only existing keys // If length of keys is non zero then just return the value with first key // Else return -1 or any other value return keys.length > 0 ? obj[keys[0]] : -1; } // If keys is not an array, return -1 (or anything else if you want to) return -1; };中找到第一个keys，仍然会遍历整个key数组。要对其进行优化，不必使用obj或map，而可以迭代filter并返回第一个现有值。

代码如下：

keys

Answer 5

实际上，不可能使用 Map 函数中断其迭代。尝试使用for循环或尝试以下类似方法的更好方法。

return keys.map(key => {
      if (obj[key]) return obj[key];
    })[0]

在对象中查找数组值

5 个答案: