我想在具有以下分布的給定數(shù)據(jù)集中運(yùn)行 10 次機(jī)器學(xué)習(xí)算法np.unique(x[:,24], return_counts=True)(array([1., 2.]), array([700, 300]))這意味著我 70% 的數(shù)據(jù)來自第 1 類，30% 來自第 2 類。下面是我的數(shù)據(jù)的快照。最后一列通知類標(biāo)簽（1 或 2）：1,6,4,12,5,5,3,4,1,67,3,2,1,2,1,0,0,1,0,0,1,0,0,1,12,48,2,60,1,3,2,2,1,22,3,1,1,1,1,0,0,1,0,0,1,0,0,1,24,12,4,21,1,4,3,3,1,49,3,1,2,1,1,0,0,1,0,0,1,0,1,0,11,42,2,79,1,4,3,4,2,45,3,1,2,1,1,0,0,0,0,0,0,0,0,1,11,24,3,49,1,3,3,4,4,53,3,2,2,1,1,1,0,1,0,0,0,0,0,1,24,36,2,91,5,3,3,4,4,35,3,1,2,2,1,0,0,1,0,0,0,0,1,0,14,24,2,28,3,5,3,4,2,53,3,1,1,1,1,0,0,1,0,0,1,0,0,1,12,36,2,69,1,3,3,2,3,35,3,1,1,2,1,0,1,1,0,1,0,0,0,0,14,12,2,31,4,4,1,4,1,61,3,1,1,1,1,0,0,1,0,0,1,0,1,0,12,30,4,52,1,1,4,2,3,28,3,2,1,1,1,1,0,1,0,0,1,0,0,0,22,12,2,13,1,2,2,1,3,25,3,1,1,1,1,1,0,1,0,1,0,0,0,1,21,48,2,43,1,2,2,4,2,24,3,1,1,1,1,0,0,1,0,1,0,0,0,1,22,12,2,16,1,3,2,1,3,22,3,1,1,2,1,0,0,1,0,0,1,0,0,1,11,24,4,12,1,5,3,4,3,60,3,2,1,1,1,1,0,1,0,0,1,0,1,0,21,15,2,14,1,3,2,4,3,28,3,1,1,1,1,1,0,1,0,1,0,0,0,1,11,24,2,13,2,3,2,2,3,32,3,1,1,1,1,0,0,1,0,0,1,0,1,0,24,24,4,24,5,5,3,4,2,53,3,2,1,1,1,0,0,1,0,0,1,0,0,1,11,30,0,81,5,2,3,3,3,25,1,3,1,1,1,0,0,1,0,0,1,0,0,1,12,24,2,126,1,5,2,2,4,44,3,1,1,2,1,0,1,1,0,0,0,0,0,0,24,24,2,34,3,5,3,2,3,31,3,1,2,2,1,0,0,1,0,0,1,0,0,1,14,9,4,21,1,3,3,4,3,48,3,3,1,2,1,1,0,1,0,0,1,0,0,1,11,6,2,26,3,3,3,3,1,44,3,1,2,1,1,0,0,1,0,1,0,0,0,1,11,10,4,22,1,2,3,3,1,48,3,2,2,1,2,1,0,1,0,1,0,0,1,0,12,12,4,18,2,2,3,4,2,44,3,1,1,1,1,0,1,1,0,0,1,0,0,1,14,10,4,21,5,3,4,1,3,26,3,2,1,1,2,0,0,1,0,0,1,0,0,1,11,6,2,14,1,3,3,2,1,36,1,1,1,2,1,0,0,1,0,0,1,0,1,0,14,6,0,4,1,5,4,4,3,39,3,1,1,1,1,0,0,1,0,0,1,0,1,0,13,12,1,4,4,3,2,3,1,42,3,2,1,1,1,0,0,1,0,1,0,0,0,1,12,7,2,24,1,3,3,2,1,34,3,1,1,1,1,0,0,0,0,0,1,0,0,1,11,60,3,68,1,5,3,4,4,63,3,2,1,2,1,0,0,1,0,0,1,0,0,1,22,18,2,19,4,2,4,3,1,36,1,1,1,2,1,0,0,1,0,0,1,0,0,1,11,24,2,40,1,3,3,2,3,27,2,1,1,1,1,0,0,1,0,0,1,0,0,1,1完整的數(shù)據(jù)集可以在這里找到我想將數(shù)據(jù)分成 90% 用于訓(xùn)練和 10% 用于測(cè)試。但是，對(duì)于每個(gè)拆分，我必須保持?jǐn)?shù)據(jù)的比例（例如，在訓(xùn)練和驗(yàn)證拆分中，70% 的數(shù)據(jù)必須屬于 1 類，30% 屬于 2 類）我知道如何簡單地將數(shù)據(jù)劃分為訓(xùn)練和測(cè)試，但我不知道如何使這種劃分服從我上面引用的類分布。如何在 Python 中做到這一點(diǎn)？
2 回答
慕碼人8056858
TA貢獻(xiàn)1803條經(jīng)驗(yàn) 獲得超6個(gè)贊
您可以使用RepeatedStratifiedKFold，顧名思義，重復(fù) K 折交叉驗(yàn)證器n時(shí)間。要重復(fù)處理10時(shí)間，設(shè)置，并在/大小中具有大約 n_repeats的比例，我們可以設(shè)置：9:1traintestn_splits=10
from sklearn.model_selection import RepeatedStratifiedKFold
X = a[:,:-1]
y = a[:,-1]
rskf = RepeatedStratifiedKFold(n_splits=10, n_repeats=10, random_state=2)
for train_index, test_index in rskf.split(X, y):
X_train, X_test = X[train_index], X[test_index]
y_train, y_test = y[train_index], y[test_index]
print(f'\nClass 1: {((y_train==1).sum()/len(y_train))*100:.0f}%')
print(f'\nShape of train: {X_train.shape[0]}')
print(f'Shape of test: {X_test.shape[0]}')
Class 1: 73%
Shape of train: 33
Shape of test: 4
Class 1: 73%
Shape of train: 33
Shape of test: 4
Class 1: 73%
Shape of train: 33
Shape of test: 4
Class 1: 73%
Shape of train: 33
Shape of test: 4
...
精慕HU
TA貢獻(xiàn)1845條經(jīng)驗(yàn) 獲得超8個(gè)贊
將數(shù)據(jù)拆分為訓(xùn)練和測(cè)試的一種眾所周知的方法是 scikit-learn train_test_split。
model_selection.train_test_split的 API 文檔。
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.10, random_state=42)
您可以使用random_state變量（種子），直到您的類之間的比例正確。雖然train_test_split不會(huì)強(qiáng)制執(zhí)行比例，但它通常遵循人口比例。
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