我有一个熊猫数据框架,我想把它分为3个单独的集。我知道使用sklearn中的train_test_split。交叉验证,可以将数据分为两组(训练和测试)。然而,我无法找到将数据分成三组的任何解决方案。最好是有原始数据的下标。
我知道一个解决办法是使用train_test_split两次,并以某种方式调整索引。但是是否有一种更标准/内置的方法将数据分成3组而不是2组?
我有一个熊猫数据框架,我想把它分为3个单独的集。我知道使用sklearn中的train_test_split。交叉验证,可以将数据分为两组(训练和测试)。然而,我无法找到将数据分成三组的任何解决方案。最好是有原始数据的下标。
我知道一个解决办法是使用train_test_split两次,并以某种方式调整索引。但是是否有一种更标准/内置的方法将数据分成3组而不是2组?
当前回答
我能想到的最简单的方法是将分割分数映射到数组下标,如下所示:
train_set = data[:int((len(data)+1)*train_fraction)]
test_set = data[int((len(data)+1)*train_fraction):int((len(data)+1)*(train_fraction+test_fraction))]
val_set = data[int((len(data)+1)*(train_fraction+test_fraction)):]
其中data = random.shuffle(data)
其他回答
将数据集分割为训练集和测试集,如在其他答案中一样,使用
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42)
然后,如果您适合您的模型,您可以添加validation_split作为参数。这样就不需要提前创建验证集。例如:
from tensorflow.keras import Model
model = Model(input_layer, out)
[...]
history = model.fit(x=X_train, y=y_train, [...], validation_split = 0.3)
验证集旨在作为训练集训练期间的代表运行测试集,完全来自训练集,无论是通过k-fold交叉验证(推荐)还是通过validation_split;然后,您不需要单独创建一个验证集,仍然可以将数据集分为您所要求的三个集。
考虑到df id你的原始数据帧:
1 -首先你在训练和测试之间分割数据(10%):
my_test_size = 0.10
X_train_, X_test, y_train_, y_test = train_test_split(
df.index.values,
df.label.values,
test_size=my_test_size,
random_state=42,
stratify=df.label.values,
)
2 -然后你在训练和验证之间分割训练集(20%):
my_val_size = 0.20
X_train, X_val, y_train, y_val = train_test_split(
df.loc[X_train_].index.values,
df.loc[X_train_].label.values,
test_size=my_val_size,
random_state=42,
stratify=df.loc[X_train_].label.values,
)
3 -然后,根据上述步骤中生成的索引对原始数据帧进行切片:
# data_type is not necessary.
df['data_type'] = ['not_set']*df.shape[0]
df.loc[X_train, 'data_type'] = 'train'
df.loc[X_val, 'data_type'] = 'val'
df.loc[X_test, 'data_type'] = 'test'
结果是这样的:
注意:此解决方案使用问题中提到的解决方案。
我能想到的最简单的方法是将分割分数映射到数组下标,如下所示:
train_set = data[:int((len(data)+1)*train_fraction)]
test_set = data[int((len(data)+1)*train_fraction):int((len(data)+1)*(train_fraction+test_fraction))]
val_set = data[int((len(data)+1)*(train_fraction+test_fraction)):]
其中data = random.shuffle(data)
回答任意数量的子集:
def _separate_dataset(patches, label_patches, percentage, shuffle: bool = True):
"""
:param patches: data patches
:param label_patches: label patches
:param percentage: list of percentages for each value, example [0.9, 0.02, 0.08] to get 90% train, 2% val and 8% test.
:param shuffle: Shuffle dataset before split.
:return: tuple of two lists of size = len(percentage), one with data x and other with labels y.
"""
x_test = patches
y_test = label_patches
percentage = list(percentage) # need it to be mutable
assert sum(percentage) == 1., f"percentage must add to 1, but it adds to sum{percentage} = {sum(percentage)}"
x = []
y = []
for i, per in enumerate(percentage[:-1]):
x_train, x_test, y_train, y_test = train_test_split(x_test, y_test, test_size=1-per, shuffle=shuffle)
percentage[i+1:] = [value / (1-percentage[i]) for value in percentage[i+1:]]
x.append(x_train)
y.append(y_train)
x.append(x_test)
y.append(y_test)
return x, y
这适用于任何比例。在本例中,您应该执行percentage = [train_percentage, val_percentage, test_percentage]。
然而,将数据集分为train、test、cv(0.6、0.2、0.2)的一种方法是使用train_test_split方法两次。
from sklearn.model_selection import train_test_split
x, x_test, y, y_test = train_test_split(xtrain,labels,test_size=0.2,train_size=0.8)
x_train, x_cv, y_train, y_cv = train_test_split(x,y,test_size = 0.25,train_size =0.75)