复杂数据集拆分 - StratifiedGroupShuffleSplit

ami*_*jad 13 python machine-learning dataset scikit-learn

我有一个大约 2m 观测值的数据集,我需要以 60:20:20 的比例将其拆分为训练、验证和测试集。我的数据集的简化摘录如下所示:

+---------+------------+-----------+-----------+
| note_id | subject_id | category  |   note    |
+---------+------------+-----------+-----------+
|       1 |          1 | ECG       | blah ...  |
|       2 |          1 | Discharge | blah ...  |
|       3 |          1 | Nursing   | blah ...  |
|       4 |          2 | Nursing   | blah ...  |
|       5 |          2 | Nursing   | blah ...  |
|       6 |          3 | ECG       | blah ...  |
+---------+------------+-----------+-----------+
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有多个类别——它们并不均衡——所以我需要确保训练、验证和测试集都具有与原始数据集中相同的类别比例。这部分很好,我可以StratifiedShuffleSplitsklearn库中使用。

但是,我还需要确保每个主题的观察结果不会分散在训练、验证和测试数据集上。来自给定主题的所有观察结果都需要在同一个桶中,以确保我的训练模型在验证/测试之前从未见过该主题。例如,subject_id 1 的每个观察都应该在训练集中。

我想不出一种方法来确保按类别分层拆分,防止跨数据集的subject_id污染(因为想要更好的词),确保 60:20:20 拆分并确保以某种方式对数据集进行混洗。任何帮助,将不胜感激!

谢谢!


编辑:

我现在了解到,也可以sklearn通过该GroupShuffleSplit函数完成按类别分组和跨数据集拆分将组保持在一起。所以基本上,我需要的是一个组合的分层和分组洗牌拆分,即StratifiedGroupShuffleSplit不存在。Github 问题:https : //github.com/scikit-learn/scikit-learn/issues/12076

Jua*_*tiz 6

这个问题在 scikit-learn 1.0 中通过StratifiedGroupKFold解决了

在此示例中,您在洗牌后生成 3 个折叠,将组保持在一起并进行分层(尽可能多)

import numpy as np
from sklearn.model_selection import StratifiedGroupKFold

X = np.ones((30, 2))
y = np.array([0, 0, 1, 1, 1, 1, 1, 1, 0, 0,
              0, 0, 0, 0, 0, 0, 0, 1, 1, 1,
              1, 1, 1, 0, 0, 0, 0, 1, 1, 1,])
groups = np.array([1, 1, 2, 2, 3, 3, 3, 4, 5, 5,
                   5, 5, 6, 6, 7, 8, 8, 9, 9, 9,
                   10, 11, 11, 12, 12, 12, 13, 13,
                   13, 13])
print("ORIGINAL POSITIVE RATIO:", y.mean())
cv = StratifiedGroupKFold(n_splits=3, shuffle=True)
for fold, (train_idxs, test_idxs) in enumerate(cv.split(X, y, groups)):
    print("Fold :", fold)
    print("TRAIN POSITIVE RATIO:", y[train_idxs].mean())
    print("TEST POSITIVE RATIO :", y[test_idxs].mean())
    print("TRAIN GROUPS        :", set(groups[train_idxs]))
    print("TEST GROUPS         :", set(groups[test_idxs]))
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在输出中,您可以看到折叠中正例的比率保持接近原始正例的比率,并且同一组永远不会出现在两个集合中。当然,你拥有的群体越少/越大(即你的班级越不平衡),保持接近原始班级分布就越困难。

输出:

ORIGINAL POSITIVE RATIO: 0.5
Fold : 0
TRAIN POSITIVE RATIO: 0.4375
TEST POSITIVE RATIO : 0.5714285714285714
TRAIN GROUPS        : {1, 3, 4, 5, 6, 7, 10, 11}
TEST GROUPS         : {2, 8, 9, 12, 13}
Fold : 1
TRAIN POSITIVE RATIO: 0.5
TEST POSITIVE RATIO : 0.5
TRAIN GROUPS        : {2, 4, 5, 7, 8, 9, 11, 12, 13}
TEST GROUPS         : {1, 10, 3, 6}
Fold : 2
TRAIN POSITIVE RATIO: 0.5454545454545454
TEST POSITIVE RATIO : 0.375
TRAIN GROUPS        : {1, 2, 3, 6, 8, 9, 10, 12, 13}
TEST GROUPS         : {11, 4, 5, 7}
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  • 这非常接近我所需要的,但仅用于交叉验证(即每个分割大小相同,而不是可配置为 60:20:20) (4认同)
  • 是的,这是为了交叉验证。不过,您仍然可以通过稍微滥用它来获得具有所需比率的单个分割:设置“n_splits=int(1/desired_test_ratio)”。之后,您可以随机选择任何创建的折叠。如果您还想要训练-验证分割,您可以使用相应的“desired_val_ratio”在训练分割中重复该过程。 (4认同)

小智 5

这已经一年多了,但我发现自己处于类似的情况,我有标签和组,并且由于组的性质,一组数据点可以仅在测试中或仅在训练中,我'我使用 pandas 和 sklearn 编写了一个小算法,希望这会有所帮助

from sklearn.model_selection import GroupShuffleSplit
groups = df.groupby('label')
all_train = []
all_test = []
for group_id, group in groups:
    # if a group is already taken in test or train it must stay there
    group = group[~group['groups'].isin(all_train+all_test)]
    # if group is empty 
    if group.shape[0] == 0:
        continue
    train_inds, test_inds = next(GroupShuffleSplit(
        test_size=valid_size, n_splits=2, random_state=7).split(group, groups=group['groups']))

    all_train += group.iloc[train_inds]['groups'].tolist()
    all_test += group.iloc[test_inds]['groups'].tolist()



train= df[df['groups'].isin(all_train)]
test= df[df['groups'].isin(all_test)]

form_train = set(train['groups'].tolist())
form_test = set(test['groups'].tolist())
inter = form_train.intersection(form_test)

print(df.groupby('label').count())
print(train.groupby('label').count())
print(test.groupby('label').count())
print(inter) # this should be empty
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ami*_*jad 4

本质上我需要的StratifiedGroupShuffleSplit是不存在的(Github问题)。这是因为这样的函数的行为尚不清楚,并且完成此操作以生成既分组又分层的数据集并不总是可能的(也在此处讨论) - 特别是对于像我这样的严重不平衡的数据集。就我而言,我希望严格进行分组,以确保组之间没有任何重叠,同时分层和数据集比例拆分为 60:20:20 ,即尽可能地进行。

正如 Ghanem 提到的,我别无选择,只能自己构建一个函数来分割数据集,如下所示:

def StratifiedGroupShuffleSplit(df_main):

    df_main = df_main.reindex(np.random.permutation(df_main.index)) # shuffle dataset

    # create empty train, val and test datasets
    df_train = pd.DataFrame()
    df_val = pd.DataFrame()
    df_test = pd.DataFrame()

    hparam_mse_wgt = 0.1 # must be between 0 and 1
    assert(0 <= hparam_mse_wgt <= 1)
    train_proportion = 0.6 # must be between 0 and 1
    assert(0 <= train_proportion <= 1)
    val_test_proportion = (1-train_proportion)/2

    subject_grouped_df_main = df_main.groupby(['subject_id'], sort=False, as_index=False)
    category_grouped_df_main = df_main.groupby('category').count()[['subject_id']]/len(df_main)*100

    def calc_mse_loss(df):
        grouped_df = df.groupby('category').count()[['subject_id']]/len(df)*100
        df_temp = category_grouped_df_main.join(grouped_df, on = 'category', how = 'left', lsuffix = '_main')
        df_temp.fillna(0, inplace=True)
        df_temp['diff'] = (df_temp['subject_id_main'] - df_temp['subject_id'])**2
        mse_loss = np.mean(df_temp['diff'])
        return mse_loss

    i = 0
    for _, group in subject_grouped_df_main:

        if (i < 3):
            if (i == 0):
                df_train = df_train.append(pd.DataFrame(group), ignore_index=True)
                i += 1
                continue
            elif (i == 1):
                df_val = df_val.append(pd.DataFrame(group), ignore_index=True)
                i += 1
                continue
            else:
                df_test = df_test.append(pd.DataFrame(group), ignore_index=True)
                i += 1
                continue

        mse_loss_diff_train = calc_mse_loss(df_train) - calc_mse_loss(df_train.append(pd.DataFrame(group), ignore_index=True))
        mse_loss_diff_val = calc_mse_loss(df_val) - calc_mse_loss(df_val.append(pd.DataFrame(group), ignore_index=True))
        mse_loss_diff_test = calc_mse_loss(df_test) - calc_mse_loss(df_test.append(pd.DataFrame(group), ignore_index=True))

        total_records = len(df_train) + len(df_val) + len(df_test)

        len_diff_train = (train_proportion - (len(df_train)/total_records))
        len_diff_val = (val_test_proportion - (len(df_val)/total_records))
        len_diff_test = (val_test_proportion - (len(df_test)/total_records)) 

        len_loss_diff_train = len_diff_train * abs(len_diff_train)
        len_loss_diff_val = len_diff_val * abs(len_diff_val)
        len_loss_diff_test = len_diff_test * abs(len_diff_test)

        loss_train = (hparam_mse_wgt * mse_loss_diff_train) + ((1-hparam_mse_wgt) * len_loss_diff_train)
        loss_val = (hparam_mse_wgt * mse_loss_diff_val) + ((1-hparam_mse_wgt) * len_loss_diff_val)
        loss_test = (hparam_mse_wgt * mse_loss_diff_test) + ((1-hparam_mse_wgt) * len_loss_diff_test)

        if (max(loss_train,loss_val,loss_test) == loss_train):
            df_train = df_train.append(pd.DataFrame(group), ignore_index=True)
        elif (max(loss_train,loss_val,loss_test) == loss_val):
            df_val = df_val.append(pd.DataFrame(group), ignore_index=True)
        else:
            df_test = df_test.append(pd.DataFrame(group), ignore_index=True)

        print ("Group " + str(i) + ". loss_train: " + str(loss_train) + " | " + "loss_val: " + str(loss_val) + " | " + "loss_test: " + str(loss_test) + " | ")
        i += 1

    return df_train, df_val, df_test

df_train, df_val, df_test = StratifiedGroupShuffleSplit(df_main)
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我基于两件事创建了一些任意损失函数:

  1. 与整个数据集相比,每个类别的百分比表示的平均平方差
  2. 数据集的比例长度与根据提供的比率 (60:20:20) 应有的长度之间的平方差

损失函数的这两个输入的加权是由静态超参数完成的hparam_mse_wgt。对于我的特定数据集,值 0.1 效果很好,但如果您使用此函数,我会鼓励您尝试使用它。将其设置为 0 将优先仅维持分流比并忽略分层。将其设置为 1 则反之亦然。

然后,使用此损失函数,我会迭代每个主题(组),并根据损失函数最高的数据集将其附加到适当的数据集(训练、验证或测试)。

它并不是特别复杂,但它适合我。它不一定适用于每个数据集,但数据集越大,机会就越大。希望其他人会发现它有用。