Question

我已经获得了训练集和测试集的索引，代码如下。

df = pandas.read_pickle(filepath + filename)
kf = KFold(n_splits = n_splits, shuffle = shuffle, random_state = 
randomState)

result = next(kf.split(df), None)

#train can be accessed with result[0]
#test can be accessed with result[1]

我想知道是否有更快的方法将它们分别用我检索的行索引分成2个数据帧。

Answer 1

按位置选择行需要DataFrame.iloc：

<强>示例：

np.random.seed(100)
df = pd.DataFrame(np.random.random((10,5)), columns=list('ABCDE'))
df.index = df.index * 10
print (df)
           A         B         C         D         E
0   0.543405  0.278369  0.424518  0.844776  0.004719
10  0.121569  0.670749  0.825853  0.136707  0.575093
20  0.891322  0.209202  0.185328  0.108377  0.219697
30  0.978624  0.811683  0.171941  0.816225  0.274074
40  0.431704  0.940030  0.817649  0.336112  0.175410
50  0.372832  0.005689  0.252426  0.795663  0.015255
60  0.598843  0.603805  0.105148  0.381943  0.036476
70  0.890412  0.980921  0.059942  0.890546  0.576901
80  0.742480  0.630184  0.581842  0.020439  0.210027
90  0.544685  0.769115  0.250695  0.285896  0.852395

from sklearn.model_selection import KFold

#added some parameters
kf = KFold(n_splits = 5, shuffle = True, random_state = 2)
result = next(kf.split(df), None)
print (result)
(array([0, 2, 3, 5, 6, 7, 8, 9]), array([1, 4]))

train = df.iloc[result[0]]
test =  df.iloc[result[1]]

print (train)
           A         B         C         D         E
0   0.543405  0.278369  0.424518  0.844776  0.004719
20  0.891322  0.209202  0.185328  0.108377  0.219697
30  0.978624  0.811683  0.171941  0.816225  0.274074
50  0.372832  0.005689  0.252426  0.795663  0.015255
60  0.598843  0.603805  0.105148  0.381943  0.036476
70  0.890412  0.980921  0.059942  0.890546  0.576901
80  0.742480  0.630184  0.581842  0.020439  0.210027
90  0.544685  0.769115  0.250695  0.285896  0.852395

print (test)
           A         B         C         D         E
10  0.121569  0.670749  0.825853  0.136707  0.575093
40  0.431704  0.940030  0.817649  0.336112  0.175410

使用sklearn的KFold分离pandas数据帧

1 个答案: