如何在没有 ski-kit learn 的情况下为 K-Fold 交叉验证创建训练集?
How to create Training Sets for K-Fold Cross Validation without ski-kit learn?
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,但是我的 x 训练集的大小是 (4,19,9),而它应该只有 8 列,而我的 y 训练集是 (4,9),而它应该有 19 行。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
欢迎来到 Stack Overflow。
创建新折叠后,您需要使用 np.row_stack()
将它们按行堆叠。
此外,我认为您对数组的切片不正确,在 Python 或 Numpy 中,切片行为是 [inclusive:exclusive]
因此,当您将切片指定为 [0:7]
时,您只是采用 7 列,而不是您预期的 8 列特征。
同样,如果您在 for 循环中指定 5 倍,它应该是 range(k)
,它给您 [0,1,2,3,4]
而不是 range(k-1)
,它只给您 [0,1,2,3]
.
修改后的代码:
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')
这将为您打印出训练集和测试集的折叠和所需形状:
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,)
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,但是我的 x 训练集的大小是 (4,19,9),而它应该只有 8 列,而我的 y 训练集是 (4,9),而它应该有 19 行。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
欢迎来到 Stack Overflow。
创建新折叠后,您需要使用 np.row_stack()
将它们按行堆叠。
此外,我认为您对数组的切片不正确,在 Python 或 Numpy 中,切片行为是 [inclusive:exclusive]
因此,当您将切片指定为 [0:7]
时,您只是采用 7 列,而不是您预期的 8 列特征。
同样,如果您在 for 循环中指定 5 倍,它应该是 range(k)
,它给您 [0,1,2,3,4]
而不是 range(k-1)
,它只给您 [0,1,2,3]
.
修改后的代码:
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')
这将为您打印出训练集和测试集的折叠和所需形状:
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,)