Python 过采样时如何保持/扩展索引

我有这样一个数据帧，我想对列“role”进行过采样（在实际情况中，行/列的数量要比这个最小的示例大得多）

这就是我正在做的：

X,y = smote.fit_sample(df,df[['role']])
X
       role value
0   1   1
1   1   1
2   1   2
3   1   1
4   1   1
5   1   2
6   1   1
7   2   1
8   2   1
[.........]

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这是可行的，但问题是我需要保留索引（pop_13vdpn1_site_1等），这可能吗？

首先，您需要处理df并将功能和目标标签拆分为

X_train

和

y_train

现在您可以进行过采样：

X_train_over, y_train_over = smote.fit_sample(X_train, y_train)

最后从上面的输出创建一个数据帧。比如说,

X = pd.DataFrame(X_train_over, columns=X_train.columns)
y = pd.DataFrame(y_train_over, columns=y_train.columns)

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最后，我找到了一个解决方法（可能不是最优的）

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下面的步骤应该可以做到这一点

import io
import pandas as pd
import numpy as np
from imblearn.over_sampling import SMOTE

示例数据

df = pd.read_csv(io.StringIO("""
role  value
    pop_13vdpn1_site_1 1 1
    pop_13vdpn1_site_1 1 1
    pop_13vdpn1_site_1 1 2
    pop_13vdpn1_site_1 1 1
    pop_13vdpn1_site_1 1 1
    pop_13vdpn1_site_1 1 2
    pop_13vdpn1_site_1 1 1
    pop_13vdpn1_site_1 2 1
    pop_13vdpn1_site_1 2 1
    pop_13vdpn1_site_1 2 1
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_2 2 2
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_3 2 1
    pop_13vdpn1_site_1 1 1
    pop_13vdpn1_site_1 1 1
    pop_13vdpn1_site_1 1 2
    pop_13vdpn1_site_1 1 1
    pop_13vdpn1_site_1 1 1
    pop_13vdpn1_site_1 1 2
    pop_13vdpn1_site_1 1 1
    pop_13vdpn1_site_1 2 1
    pop_13vdpn1_site_1 2 1
    pop_13vdpn1_site_1 2 1
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_2 2 2
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_2 2 1
    pop_13vdpn1_site_3 2 1
"""), sep="\s+", engine="python")

df = df.reset_index()

形状应为（40,3）：

Smote接受数组，因此我们需要定义x和y值

X_train = np.array(df['role']).reshape(40,1)
y_train = np.array(df['value']).reshape(40,)

打击行动：

from imblearn.over_sampling import SMOTE
sm = SMOTE(random_state=42)
X,y = sm.fit_resample(X_train,y_train)

将给定的

X

和

y

放入数据帧中：

ndf = pd.DataFrame({'role':X.reshape(68,), 'value':y})

重新制作原始名称

ndf['name'] = ndf['role'].apply(lambda x: 'pop_13vdpn1_site_'+str(x))

看看数据是否更平衡

from collections import Counter
Counter(df['role'])
Counter(ndf['role'])

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嗨，Giorgos，但是，如果我这样做，我会得到X和X的NaN值y@psagrera如果不提供

索引

参数，输出是什么？角色值0 1 1有趣！我遇到了一个非常相似的情况，你的建议就是我现在要尝试的。谢谢分享！

ndf = pd.DataFrame({'role':X.reshape(68,), 'value':y})

ndf['name'] = ndf['role'].apply(lambda x: 'pop_13vdpn1_site_'+str(x))

from collections import Counter
Counter(df['role'])
Counter(ndf['role'])