Python 我可以使用numpy数组生成交叉验证的折叠吗?
我想使用numpy数组为k-folds交叉验证任务构建折叠。取出测试片很容易,但我不知道如何返回数组的其余部分,而忽略了测试片。有没有一种有效的方法可以做到这一点Python 我可以使用numpy数组生成交叉验证的折叠吗?,python,numpy,slice,Python,Numpy,Slice,我想使用numpy数组为k-folds交叉验证任务构建折叠。取出测试片很容易,但我不知道如何返回数组的其余部分,而忽略了测试片。有没有一种有效的方法可以做到这一点 examples = range(50) classes = range(50) data = np.array(zip(classes,examples)) test_slice = data[5:10] train_on_remainder = ?? 您可以这样设置: test_slice, remainder = np.spl
examples = range(50)
classes = range(50)
data = np.array(zip(classes,examples))
test_slice = data[5:10]
train_on_remainder = ??
您可以这样设置:
test_slice, remainder = np.split(data.copy(), [test_size], axis=0)
# run test
remainder[:test_size], test_slice = test_slice, remainder[:test_size].copy()
# run test
remainder[test_size:2*test_size], test_slice = test_slice, remainder[test_size:2*test_size].copy()
# etc.
我不认为你可以少复制它
工作原理:
. full set: | 0 | 1 | 2 | 3 | 4 | 5 |
split (full copy) / \
tst / rem | 0 | | 1 | 2 | 3 | 4 | 5 |
run trial
| 1 | 2 | 3 | 4 | 5 |
swap tst and ^ |
first segment: | v
(partial copy) | 0 |
tst / rem | 1 | | 0 | 2 | 3 | 4 | 5 |
run trial
| 0 | 2 | 3 | 4 | 5 |
swap tst and ^ |
second segment: | v
(partial copy) | 1 |
tst / rem | 2 | | 0 | 1 | 3 | 4 | 5 |
run trial
| 0 | 1 | 3 | 4 | 5 |
swap tst and ^ |
third segment: | v
(partial copy) | 2 |
如你所见,它几乎是在移动褶皱。保存许多完整副本。您可以这样设置:
test_slice, remainder = np.split(data.copy(), [test_size], axis=0)
# run test
remainder[:test_size], test_slice = test_slice, remainder[:test_size].copy()
# run test
remainder[test_size:2*test_size], test_slice = test_slice, remainder[test_size:2*test_size].copy()
# etc.
我不认为你可以少复制它
工作原理:
. full set: | 0 | 1 | 2 | 3 | 4 | 5 |
split (full copy) / \
tst / rem | 0 | | 1 | 2 | 3 | 4 | 5 |
run trial
| 1 | 2 | 3 | 4 | 5 |
swap tst and ^ |
first segment: | v
(partial copy) | 0 |
tst / rem | 1 | | 0 | 2 | 3 | 4 | 5 |
run trial
| 0 | 2 | 3 | 4 | 5 |
swap tst and ^ |
second segment: | v
(partial copy) | 1 |
tst / rem | 2 | | 0 | 1 | 3 | 4 | 5 |
run trial
| 0 | 1 | 3 | 4 | 5 |
swap tst and ^ |
third segment: | v
(partial copy) | 2 |
如你所见,它几乎是在移动褶皱。保存许多完整副本。这是一个奇怪的问题,因为人们通常会使用sklearn的
train\u test\u split()
编辑:另一种方法可能是
r = np.arange(len(data))
trainX = data[r < 5 | r > 10]
当然,如果有sklearn的train\u test\u split()
,你也可以问类似的问题。如果有sklearn的train\u test\u split()
split = np.vsplit(data, np.array([5,10]))
'''This will give you a list with 3 elements'''
test_slice = split[1]
train_slice = np.vstack((split[0],split[2]))
编辑:另一种方法可能是
r = np.arange(len(data))
trainX = data[r < 5 | r > 10]
当然,如果有sklearn的train\u test\u split()
split = np.vsplit(data, np.array([5,10]))
'''This will give you a list with 3 elements'''
test_slice = split[1]
train_slice = np.vstack((split[0],split[2]))
[5 5][6][7][8][9]
[0 0][11][2 2][3][4][10][11][12
12] [13 13] [14 14] [15 15] [16 16] [17 17] [18 18]
…[47][48][49]]
[5 5][6][7][8][9]
[0 0][11][2 2][3][4][10][11][12
12] [13 13] [14 14] [15 15] [16 16] [17 17] [18 18]
…[47][48][49]]
在1d阵列上演示了两种方法:
In [64]: data = np.arange(20)
In [65]: test = data[5:10]
In [66]: rest = np.concatenate((data[:5],data[10:]),axis=0)
In [67]: rest
Out[67]: array([ 0, 1, 2, 3, 4, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19])
In [68]:
In [68]: mask = np.zeros(data.shape[0], dtype=bool)
In [69]: mask[5:10] = True
In [70]: test = data[mask]
In [71]: rest = data[~mask]
In [72]: rest
Out[72]: array([ 0, 1, 2, 3, 4, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19])
有一个np.delete
函数
In [75]: np.delete(data, np.arange(5,10))
Out[75]: array([ 0, 1, 2, 3, 4, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19])
在内部,它使用我演示的两种方法之一。在1d阵列上演示的两种方法:
In [64]: data = np.arange(20)
In [65]: test = data[5:10]
In [66]: rest = np.concatenate((data[:5],data[10:]),axis=0)
In [67]: rest
Out[67]: array([ 0, 1, 2, 3, 4, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19])
In [68]:
In [68]: mask = np.zeros(data.shape[0], dtype=bool)
In [69]: mask[5:10] = True
In [70]: test = data[mask]
In [71]: rest = data[~mask]
In [72]: rest
Out[72]: array([ 0, 1, 2, 3, 4, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19])
有一个np.delete
函数
In [75]: np.delete(data, np.arange(5,10))
Out[75]: array([ 0, 1, 2, 3, 4, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19])
在内部,它使用了我演示的两种方法之一。如果您必须手动实现任意数量拆分的k倍方法:我使用了以下解决方案(为交叉验证设置训练和验证):
如果您必须为任意数量的拆分手动实施k-fold方法:我使用了以下解决方案(为交叉验证设置训练和验证):
连接两边的切片;结果将是一个副本;结果将是一份副本。