我正在使用sklearn
模块来查找最佳拟合模型和模型参数。但是,我在下面有一个意外的索引错误:
> IndexError Traceback (most recent call
> last) <ipython-input-38-ea3f99e30226> in <module>
> 22 s = mean_squared_error(y[ts], best_m.predict(X[ts]))
> 23 cv[i].append(s)
> ---> 24 print(np.mean(cv, 1))
> IndexError: tuple index out of range
我想做的是找到最合适的回归变量及其参数,但是我遇到了以上错误。我调查了SO
并尝试了this solution,但仍然出现了同样的错误。有解决此错误的想法吗?谁能指出我为什么会发生此错误?有什么想法吗?
我的代码:
from sklearn.model_selection import KFold
from sklearn.metrics import mean_squared_error
from sklearn.linear_model import LinearRegression, Ridge, Lasso
from xgboost.sklearn import XGBRegressor
from sklearn.datasets import make_regression
models = [SVR(), RandomForestRegressor(), LinearRegression(), Ridge(), Lasso(), XGBRegressor()]
params = [{'C': [0.01, 1]}, {'n_estimators': [10, 20]}]
X, y = make_regression(n_samples=10000, n_features=20)
with warnings.catch_warnings():
warnings.filterwarnings("ignore")
cv = [[] for _ in range(len(models))]
fold = KFold(5,shuffle=False)
for tr, ts in fold.split(X):
for i, (model, param) in enumerate(zip(models, params)):
best_m = GridSearchCV(model, param)
best_m.fit(X[tr], y[tr])
s = mean_squared_error(y[ts], best_m.predict(X[ts]))
cv[i].append(s)
print(np.mean(cv, 1))
所需的输出:
如果有一种方法可以解决上述错误,那么我希望选择具有参数的最佳拟合模型,然后将其用于估算。有什么想法可以改善上述尝试吗?谢谢
答案 0 :(得分:1)
定义时
cv = [[] for _ in range(len(models))]
每个型号都有一个空列表。
但是,在循环中,您经过enumerate(zip(models, params))
仅具有两个元素,因为您的params
列表具有两个元素(因为list(zip(x,y))
has length等于min(len(x),len(y)
)。
因此,您得到一个IndexError
,因为在用cv
计算平均值时,np.mean
中的某些列表为空(除了前两个以外)。
解决方案:
如果您不需要在其余模型上使用GridSearchCV
,则可以使用空字典扩展params
列表:
params = [{'C': [0.01, 1]}, {'n_estimators': [10, 20]}, {}, {}, {}, {}]
答案 1 :(得分:1)
问题的根本原因是,当您要求评估GridSearchCV
中的6个模型时,您仅提供前两个模型的参数:
models = [SVR(), RandomForestRegressor(), LinearRegression(), Ridge(), Lasso(), XGBRegressor()]
params = [{'C': [0.01, 1]}, {'n_estimators': [10, 20]}]
在此设置中enumerate(zip(models, params))
的结果,即:
for i, (model, param) in enumerate(zip(models, params)):
print((model, param))
是
(SVR(C=1.0, cache_size=200, coef0=0.0, degree=3, epsilon=0.1, gamma='auto',
kernel='rbf', max_iter=-1, shrinking=True, tol=0.001, verbose=False), {'C': [0.01, 1]})
(RandomForestRegressor(bootstrap=True, criterion='mse', max_depth=None,
max_features='auto', max_leaf_nodes=None,
min_impurity_decrease=0.0, min_impurity_split=None,
min_samples_leaf=1, min_samples_split=2,
min_weight_fraction_leaf=0.0, n_estimators=10, n_jobs=1,
oob_score=False, random_state=None, verbose=0, warm_start=False), {'n_estimators': [10, 20]})
即,仅忽略了最后4个模型,因此您在cv
中会得到空的条目:
print(cv)
# result:
[[5950.6018771284835, 5987.293514740653, 6055.368320208183, 6099.316091619069, 6146.478702335218], [3625.3243553665975, 3301.3552182952058, 3404.3321983193728, 3521.5160621260898, 3561.254684271113], [], [], [], []]
会在尝试获取np.mean(cv, 1)
时导致下游错误。
Psi在其答案中已经正确指出,解决方案是在您实际上不执行任何CV搜索的模型中使用空字典。省略XGBRegressor
(尚未安装),结果如下:
models = [SVR(), RandomForestRegressor(), LinearRegression(), Ridge(), Lasso()]
params2 = [{'C': [0.01, 1]}, {'n_estimators': [10, 20]}, {}, {}, {}]
cv = [[] for _ in range(len(models))]
fold = KFold(5,shuffle=False)
for tr, ts in fold.split(X):
for i, (model, param) in enumerate(zip(models, params2)):
best_m = GridSearchCV(model, param)
best_m.fit(X[tr], y[tr])
s = mean_squared_error(y[ts], best_m.predict(X[ts]))
cv[i].append(s)
其中print(cv)
给出:
[[4048.660483326826, 3973.984055352062, 3847.7215568088545, 3907.0566348092684, 3820.0517432992765], [1037.9378737329769, 1025.237441119364, 1016.549294695313, 993.7083268195154, 963.8115632611381], [2.2948917095935095e-26, 1.971022007799432e-26, 4.1583774042712844e-26, 2.0229469068846665e-25, 1.9295075684919642e-26], [0.0003350178681602639, 0.0003297411022124562, 0.00030834076832371557, 0.0003355298330301431, 0.00032049282437794516], [10.372789356303688, 10.137748082073076, 10.136028304131141, 10.499159069700834, 9.80779910439471]]
和print(np.mean(cv, 1))
可以正常工作,给出:
[3.91949489e+03 1.00744890e+03 6.11665355e-26 3.25824479e-04
1.01907048e+01]
因此,就您而言,您确实应该将params
更改为:
params = [{'C': [0.01, 1]}, {'n_estimators': [10, 20]}, {}, {}, {}, {}]
如Psi所建议。