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我有一个包含 95 行和 9 列的数据集，并且想要进行 5 折交叉验证。在训练中，前 8 列（特征）用于预测第 9 列。我的测试集是正确的，但是当它应该只有 8 列时，我的 x 训练集的大小为 (4,19,9)，而当它应该有 19 行时，我的 y 训练集的大小为 (4,9)。我是否错误地索引了子数组？

kdata = data[0:95,:] # Need total rows to be divisible by 5, so ignore last 2 rows 
np.random.shuffle(kdata) # Shuffle all rows
folds = np.array_split(kdata, k) # each fold is 19 rows x 9 columns

for i in range (k-1):
    xtest = folds[i][:,0:7] # Set ith fold to be test
    ytest = folds[i][:,8]
    new_folds = np.delete(folds,i,0)
    xtrain = new_folds[:][:][0:7] # training set is all folds, all rows x 8 cols
    ytrain = new_folds[:][:][8]   # training y is all folds, all rows x 1 col
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1 回答 1

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欢迎来到堆栈溢出。

创建新折叠后，您需要使用np.row_stack().

另外，我认为您在 Python 或 Numpy 中对数组进行了错误的切片，[inclusive:exclusive]因此切片行为是，当您指定切片时，[0:7]您只取 7 列，而不是您想要的 8 个特征列。

同样，如果您在 for 循环中指定 5 折，它应该是range(k)which 给你[0,1,2,3,4]而不是range(k-1)which 只给你[0,1,2,3].

修改后的代码如下：

folds = np.array_split(kdata, k) # each fold is 19 rows x 9 columns
np.random.shuffle(kdata) # Shuffle all rows
folds = np.array_split(kdata, k)

for i in range (k):
    xtest = folds[i][:,:8] # Set ith fold to be test
    ytest = folds[i][:,8]
    new_folds = np.row_stack(np.delete(folds,i,0))
    xtrain = new_folds[:, :8]
    ytrain = new_folds[:,8]

    # some print functions to help you debug
    print(f'Fold {i}')
    print(f'xtest shape  : {xtest.shape}')
    print(f'ytest shape  : {ytest.shape}')
    print(f'xtrain shape : {xtrain.shape}')
    print(f'ytrain shape : {ytrain.shape}\n')

这将为您打印出折叠和所需形状的训练和测试集：

Fold 0
xtest shape  : (19, 8)
ytest shape  : (19,)
xtrain shape : (76, 8)
ytrain shape : (76,)

Fold 1
xtest shape  : (19, 8)
ytest shape  : (19,)
xtrain shape : (76, 8)
ytrain shape : (76,)

Fold 2
xtest shape  : (19, 8)
ytest shape  : (19,)
xtrain shape : (76, 8)
ytrain shape : (76,)

Fold 3
xtest shape  : (19, 8)
ytest shape  : (19,)
xtrain shape : (76, 8)
ytrain shape : (76,)

Fold 4
xtest shape  : (19, 8)
ytest shape  : (19,)
xtrain shape : (76, 8)
ytrain shape : (76,)
于 2020-03-09T03:44:20.043 回答