如何将 kfold.split() 应用于列表字典？答案

【问题标题】：How do I apply kfold.split() to a dictionary of lists?如何将 kfold.split() 应用于列表字典？
【发布时间】：2021-02-03 15:51:57
【问题描述】：

我想使用 CrossValidation 训练 Keras 模型，但我的数据是列表的字典。

我想要 10 次折叠，所以我想要每个验证步骤中 10% 的 dict 键的子集，以及接下来的 10%（带有随机播放）。

示例：对于第一个验证步骤：

pairs_train = {'0': list1,
        '1': list2,
        '2': list3,
        '3': list4,
        '4': list5,
        '5': list6,
        '6': list7,
        '7': list8,
        '8': list9,
        }

 pairs_val = {'9': list10,
        }

这是我的功能：

def crossValidation(self, k_folds=10):
    cv_accuracy_train = []
    cv_accuracy_val = []
    cv_loss_train = []
    cv_loss_val = []

    s = pd.Series(pairs)

    idx = 0
    for train_idx, val_idx in kfold.split(s):
        print("=========================================")
        print("====== K Fold Validation step => %d/%d =======" % (idx, k_folds))
        print("=========================================")

        train_gen = DataGenerator(pairs=s[train_idx], batch_size=self.param_grid['batch_size'],
                                  nr_files=len(self.Data.all_files), nr_tests=len(self.Data.all_tests),
                                  negative_ratio=self.param_grid['negative_ratio'])

        val_gen = DataGenerator(pairs=s[val_idx], batch_size=self.param_grid['batch_size'],
                                nr_files=len(self.Data.all_files), nr_tests=len(self.Data.all_tests),
                                negative_ratio=self.param_grid['negative_ratio'])

        # Train
        h = self.model.fit(train_gen,
                           validation_data=val_gen,
                           epochs=self.param_grid['nb_epochs'],
                           verbose=2)

        cv_accuracy_train.append(np.array(h.history['mae'])[-1])
        cv_accuracy_val.append(np.array(h.history['val_mae'])[-1])
        cv_loss_train.append(np.array(h.history['loss'])[-1])
        cv_loss_val.append(np.array(h.history['val_loss'])[-1])
        idx += 1

追溯：

    File "/Users/joaolousada/Documents/5ºAno/Master-Thesis/main/Prioritizer/Prioritizer.py", line 173, in crossValidation
    train_gen = DataGenerator(pairs=s[train_idx], batch_size=self.param_grid['batch_size'],
  File "/Users/joaolousada/opt/anaconda3/lib/python3.7/site-packages/pandas/core/series.py", line 908, in __getitem__
    return self._get_with(key)
  File "/Users/joaolousada/opt/anaconda3/lib/python3.7/site-packages/pandas/core/series.py", line 943, in _get_with
    return self.loc[key]
  File "/Users/joaolousada/opt/anaconda3/lib/python3.7/site-packages/pandas/core/indexing.py", line 879, in __getitem__
    return self._getitem_axis(maybe_callable, axis=axis)
  File "/Users/joaolousada/opt/anaconda3/lib/python3.7/site-packages/pandas/core/indexing.py", line 1099, in _getitem_axis
    return self._getitem_iterable(key, axis=axis)
  File "/Users/joaolousada/opt/anaconda3/lib/python3.7/site-packages/pandas/core/indexing.py", line 1037, in _getitem_iterable
    keyarr, indexer = self._get_listlike_indexer(key, axis, raise_missing=False)
  File "/Users/joaolousada/opt/anaconda3/lib/python3.7/site-packages/pandas/core/indexing.py", line 1254, in _get_listlike_indexer
    self._validate_read_indexer(keyarr, indexer, axis, raise_missing=raise_missing)
  File "/Users/joaolousada/opt/anaconda3/lib/python3.7/site-packages/pandas/core/indexing.py", line 1298, in _validate_read_indexer
    raise KeyError(f"None of [{key}] are in the [{axis_name}]")
KeyError: "None of [Int64Index([   0,    1,    2,    3,    4,    5,    6,    7,    8,    9,\n            ...\n            3257, 3258, 3261, 3262, 3263, 3265, 3266, 3267, 3268, 3269],\n           dtype='int64', length=2943)] are in the [index]"

【问题讨论】：

标签： python keras cross-validation

【解决方案1】：

如果有 dict 和 list 值。例如

pairs = {'0': [1,2,3],
        '1': [1,2,3],
        '2': [4,6,8],
        '3': [2,1,9],
        '4': [9,7,8],
        '5': [4,6,8],
        '6': [9,7,8],
        '7': [9,7,8],
        '8': [1,2,3],
        '9': [4,6,8],
        }

以下函数将返回索引以按索引拆分字典

def kfold_split(pairs:dict, perc:float, shuffle:bool) -> list:
    
    keys = list(pairs.keys())
    sets = len(keys)
    cv_perc = int(sets*perc)
    folds = int(sets/cv_perc)
    
    indices = [] 
    
    for fold in range(folds):
    
        # If you want to generate random keys
        if shuffle:

            # Choose random keys 
            random_keys = list(np.random.choice(keys, cv_perc))
            
            other_keys = list(set(keys) - set(random_keys)) 
            
            indices.append((other_keys, random_keys))
            
        else: 
            
            if fold == 0: 
                fold_keys = keys[-cv_perc*(fold+1):]
            else:
                fold_keys = keys[-cv_perc*(fold+1):-cv_perc*(fold)]
            
            other_keys = list(set(keys) - set(fold_keys)) 
            
            indices.append((other_keys, fold_keys))             
    
    return indices

您可以检索随机索引

kfold_split(pairs, perc=.2, shuffle=True)
>>>
[(['6', '2', '1', '5', '4', '7', '0', '3'], ['9', '8']),
 (['6', '1', '9', '5', '4', '7', '0', '3'], ['8', '2']),
 (['2', '1', '8', '9', '5', '4', '7', '3'], ['6', '0']),
 (['2', '8', '9', '5', '4', '7', '0', '3'], ['1', '6']),
 (['6', '2', '8', '5', '4', '7', '0', '3'], ['9', '1'])]

或订单索引

kfold_split(pairs, perc=.2, shuffle=False)
>>>
[(['6', '2', '1', '5', '4', '7', '0', '3'], ['8', '9']),
 (['2', '1', '8', '9', '5', '4', '0', '3'], ['6', '7']),
 (['6', '2', '1', '8', '9', '7', '0', '3'], ['4', '5']),
 (['6', '1', '8', '9', '5', '4', '7', '0'], ['2', '3']),
 (['6', '2', '8', '9', '5', '4', '7', '3'], ['0', '1'])]

然后您可以根据这些索引过滤您的字典，如下所示

for indices in result:
    train_indices, test_indices = indices
    
    # Filter dict by indices
    pair_test = {k:v for k,v in pairs.items() if k in test_indices}
                   
    # Train data
    pair_train = {k:v for k,v in pairs.items() if k not in train_indices}
    
    # Some other stuff here

【讨论】：

很好的答案，但不是我想要的。我想将整个 dict 拆分为 80/20%，然后排列这 5 个数据部分
@johnnylousas ，我认为当您不想加载 sklearn 库时，这可能是一个好方法。很抱歉误解了您的问题
是的，这是正确的！感谢您的贡献

【解决方案2】：

我设法为自己的问题找到了解决方案，将所有字典键作为np.array 并在kf.split() 中使用它们。然后使用获得的索引访问我想要的 dict 键。不确定是否更优化/pythonic 解决方案，但它工作正常。

def crossValidation(self, k_folds=10):
    cv_accuracy_train = []
    cv_accuracy_val = []
    cv_loss_train = []
    cv_loss_val = []

    s = np.array(list(self.Data.pairs.keys()))

    kfold = KFold(n_splits=k_folds, shuffle=True)

    idx = 0
    for train_idx, val_idx in kfold.split(s):
        print("=========================================")
        print("====== K Fold Validation step => %d/%d =======" % (idx, k_folds))
        print("=========================================")
        pairs_train = {s[key]: self.Data.pairs[s[key]] for key in train_idx}
        pairs_val = {s[key]: self.Data.pairs[s[key]] for key in val_idx}

        train_gen = DataGenerator(pairs=pairs_train, batch_size=self.param_grid['batch_size'],
                                  nr_files=len(self.Data.all_files), nr_tests=len(self.Data.all_tests),
                                  negative_ratio=self.param_grid['negative_ratio'])

        val_gen = DataGenerator(pairs=pairs_val, batch_size=self.param_grid['batch_size'],
                                nr_files=len(self.Data.all_files), nr_tests=len(self.Data.all_tests),
                                negative_ratio=self.param_grid['negative_ratio'])

        # Train
        h = self.model.fit(train_gen,
                           validation_data=val_gen,
                           epochs=self.param_grid['nb_epochs'],
                           verbose=2)

        cv_accuracy_train.append(np.array(h.history['accuracy'])[-1])
        cv_accuracy_val.append(np.array(h.history['val_accuracy'])[-1])
        cv_loss_train.append(np.array(h.history['loss'])[-1])
        cv_loss_val.append(np.array(h.history['val_loss'])[-1])
        idx += 1

【讨论】：