小编mai*_*con的帖子

请问,我怎样才能获得所有这些二进制排列,但在Python中没有重复？

 a = list(itertools.permutations([1, 1, 0, 0]))
 for i in range(len(a)):
     print a[i]

    (1, 1, 0, 0)
    (1, 1, 0, 0)
    (1, 0, 1, 0)
    ...

如果它大致有效将会很棒,因为我必须使用这样的30个元素列表.

请,我只需要为(X_test,y_test)数据的每次拆分明确地获得交叉验证统计信息.

所以,为了尝试这样做我做了:

kf = KFold(n_splits=n_splits)

X_train_tmp = []
y_train_tmp = []
X_test_tmp = []
y_test_tmp = []
mae_train_cv_list = []
mae_test_cv_list = []

for train_index, test_index in kf.split(X_train):

    for i in range(len(train_index)):
        X_train_tmp.append(X_train[train_index[i]])
        y_train_tmp.append(y_train[train_index[i]])

    for i in range(len(test_index)):
        X_test_tmp.append(X_train[test_index[i]])
        y_test_tmp.append(y_train[test_index[i]])

    model.fit(X_train_tmp, y_train_tmp) # FIT the model = SVR, NN, etc.

    mae_train_cv_list.append( mean_absolute_error(y_train_tmp, model.predict(X_train_tmp)) # MAE of the train part of the KFold.

    mae_test_cv_list.append( mean_absolute_error(y_test_tmp, model.predict(X_test_tmp)) ) # MAE of the test part of the KFold.

    X_train_tmp = []
    y_train_tmp = …