指示 DataFrame 的 KFold 拆分方法 return 是 iloc 还是 loc？
Indices that KFold split method return for a DataFrame is it iloc or loc?
当我们使用 _KFold.split(X) 时，其中 X 是一个 DataFrame，生成的索引将数据拆分为训练集和测试集，它是 iloc（纯粹基于整数位置的索引，供选择位置）还是 loc（按标签排列的行和列组的位置）？
您需要 DataFrame.iloc 个位置 select 行：
样本:
np.random.seed(100)
df = pd.DataFrame(np.random.random((10,5)), columns=list('ABCDE'))
#changed default index values
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
当我们使用 _KFold.split(X) 时，其中 X 是一个 DataFrame，生成的索引将数据拆分为训练集和测试集，它是 iloc（纯粹基于整数位置的索引，供选择位置）还是 loc（按标签排列的行和列组的位置）？
您需要 DataFrame.iloc 个位置 select 行：
样本:
np.random.seed(100)
df = pd.DataFrame(np.random.random((10,5)), columns=list('ABCDE'))
#changed default index values
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