Analysis of Deviance Table (Type III Wald chisquare tests) Response: Filled_pause_frequency Chisq Df Pr(>Chisq) (Intercept) 49.7207 1 1.773e-12 *** Chinese_level 1.2587 2 0.53293 Request_scenario 3.4722 1 0.06241 . Strategy_type 177.5540 2 < 2.2e-16 *** Chinese_level:Request_scenario 4.1748 2 0.12401 Chinese_level:Strategy_type 95.3308 4 < 2.2e-16 *** Request_scenario:Strategy_type 4.6560 2 0.09749 . Chinese_level:Request_scenario:Strategy_type 23.9163 4 8.302e-05 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

使用r语言中GLMsData的数据集delection，用plot函数绘制女性比例关于party的图，再用plot函数绘制女性比例关于region的图，Fit the Binomial glm with a logit link function, using the systematic component cbind(Females, Males) ~ Party * Region. Use the analysis of deviance table to select a suitable model.

好的，我会根据您的要求回答您的问题。首先，我们需要加载 R 语言中的 GLMsData 包和 MASS 包，使用以下命令： R library(GLMsData) library(MASS) 接着，我们可以使用 data() 函数来查看该包中...

优化代码KNeighborsClassifier(algorithm='auto', leaf_size=30, metric='minkowski', metric_params=None, n_jobs=1, n_neighbors=5, p=2, weights='uniform') GradientBoostingClassifier(criterion='friedman_mse', init=None, learning_rate=0.1, loss='deviance', max_depth=3, max_features=None, 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=100, presort='auto', random_state=None,subsample=1.0, verbose=0, warm_start=False)，出错TypeError: init() got an unexpected keyword argument 'min_impurity_split'

这个错误是因为在 GradientBoostingClassifier 中的参数 min_impurity_split 已经被弃用，应该使用参数 min_impurity_decrease 来代替。你只需要将参数名 min_impurity_split 改为 min_impurity_decrease 即可解决该...

用R语言先载入GLMsData包，再载入数据集belection，用数据集belection完成 1.Fit the Binomial glm with a logit link function, using the systematic component cbind(Females, Males) ~ Party * Region. Use the analysis of deviance table to select a suitable model. 2.Fit the Binomial glm with a logit link function, using the selected model in the above question. Conduct a diagnostic analysis, i.e. plot: (1) deviance residuals against transformed fitted value; (2) Cook’s distance; (3) Q-Q plot, and then comment. 3. Is overdispersion evident? 4.Find out the estimate coefficient for the PartyLabour and interpret it. 5. Determine if the saddlepoint approximation is likely to be suitable for these data. 步骤要详细

5.进行诊断分析，首先可以绘制 deviance residuals vs transformed fitted value 的散点图： R plot(model2, which = 1) 如果散点图中没有出现任何明显的模式，说明模型拟合良好。否则，可能需要进一步检查...

请读取以下链接中的数据：https://kdocs.cn/l/ccdM4odUPRAi?f=201 [文件]A4_AccidentCount_test.csv ，并帮我解决以下问题：基于A4_ accident . csv 数据，根据下式进行 logistic 回归： is _ acc ~ ST _ MP + Length + NLane + LaneWidth + LShoulderWidth + RShoulderWidth + AADT ( is _ acc ：是否发生事故，0为未发生，1为发生） 1．给出 logistic 回归的结果（附代码）2.根据各个自变量的显著性，对自变量按照非常显著、显著、不显著进行排序 3.调整 logistic 回归中的自变量，再次进行回归，并对比先后结果，选取任一指标，说明模型是否在该指标上有所提升。

Residual deviance: 81.17 on 149 degrees of freedom AIC: 95.17 Number of Fisher Scoring iterations: 7 3. 根据输出结果中的p值，可以按照非常显著（***）、显著（**）、不显著（*）对自变量进行排序，...

基于A4_ accident . csv 数据，根据下式进行 logistic 回归： is _ acc ~ ST _ MP + Length + NLane + LaneWidth + LShoulderWidth + RShoulderWidth + AADT ( is _ acc ：是否发生事故，0为未发生，1为发生） 1．给出 logistic 回归的结果（附代码）2.根据各个自变量的显著性，对自变量按照非常显著、显著、不显著进行排序 3.调整 logistic 回归中的自变量，再次进行回归，并对比先后结果，选取任一指标，说明模型是否在该指标上有所提升。

Residual deviance: 3768.8 on 3052 degrees of freedom AIC: 3784.8 Number of Fisher Scoring iterations: 5 其次，根据各个自变量的显著性，对自变量按照非常显著、显著、不显著进行排序：按照显著性由高...

Parametric coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 0.52314 0.01395 37.5 <2e-16 * --- Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 Approximate significance of smooth terms: edf Ref.df F p-value s(X1) 1.976 1.999 11.034 0.000196 * s(X2) 1.000 1.000 22.669 3.73e-05 * s(X3) 1.434 1.670 2.187 0.097375 . s(X4) 1.000 1.000 17.832 0.000178 * s(X5) 1.875 1.974 6.487 0.007730 ** --- Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 R-sq.(adj) = 0.943 Deviance explained = 95.4% GCV = 0.01 Scale est. = 0.0079798 n = 41解释下

The "Deviance explained" is the percentage of deviance in the response variable that is explained by the model. The "GCV" is the generalized cross-validation score, which is a measure of predictive ...

Family: gaussian Link function: identity Formula: Y ~ s(X1) + s(X2) + s(X3) + s(X4) Parametric coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 0.580617 0.001752 331.4 <2e-16 * --- Signif. codes: 0 ‘*’ 0.001 ‘**’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 Approximate significance of smooth terms: edf Ref.df F p-value s(X1) 7.931 8.730 36.52 < 2e-16 s(X2) 1.000 1.000 18.02 0.000959 * s(X3) 8.952 8.995 81.75 < 2e-16 * s(X4) 8.985 8.999 34.34 < 2e-16 * --- Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 R-sq.(adj) = 0.999 Deviance explained = 100% GCV = 0.00039298 Scale est. = 0.00012586 n = 41解释

这是一个广义线性模型，使用高斯分布作为响应变量的分布，使用恒等函数作为联系函数。该模型的公式是Y ~ s(X1) + s(X2) + s(X3) + s(X4)，其中s()代表平滑函数。模型的参数估计值和标准误差都被列出来了，以及t值和p...

Parametric coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 0.52314 0.01395 37.5 <2e-16 * --- Signif. codes: 0 ‘*’ 0.001 ‘**’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 Approximate significance of smooth terms: edf Ref.df F p-value s(X1) 1.976 1.999 11.034 0.000196 s(X2) 1.000 1.000 22.669 3.73e-05 * s(X3) 1.434 1.670 2.187 0.097375 . s(X4) 1.000 1.000 17.832 0.000178 * s(X5) 1.875 1.974 6.487 0.007730 --- Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 R-sq.(adj) = 0.943 Deviance explained = 95.4% GCV = 0.01 Scale est. = 0.0079798 n = 41运行输的结果图进行解释

Approximate significance of smooth terms部分给出了每个平滑项的平滑度（edf）、参考自由度（Ref.df）、F值和p值。在这个例子中，X1、X2、X4和X5的平滑项是显著的（即p值小于0.05），而X3的平滑项不显著。 R-sq....

Parametric coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 0.53529 0.03186 16.8 <2e-16 * --- Signif. codes: 0 ‘*’ 0.001 ‘**’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 Approximate significance of smooth terms: edf Ref.df F p-value s(X1) 1.818 2.243 0.687 0.5401 s(X2) 1.729 2.080 2.880 0.0675 . s(X3) 1.740 2.110 0.942 0.4003 s(X4) 1.516 1.828 0.973 0.2865 s(X5) 1.727 2.098 0.011 0.9968 s(X6) 2.063 2.478 0.390 0.7171 s(X7) 1.419 1.682 2.056 0.2184 --- Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 R-sq.(adj) = 0.213 Deviance explained = 42.3% GCV = 0.065102 Scale est. = 0.046688 n = 46这个拟合结果，怎么解释

Error）、t值（t value）和p值（Pr(>|t|)）等信息。其中，t值和p值用于检验每个参数的显著性，p值小于0.05（或0.01、0.001）表示该参数的效应在统计上是显著的。拟合结果中还包含了模型的拟合优度指标，比如调整...

type.measure='deviance

type.measure='deviance'是一种用来衡量模型拟合优度的指标。在统计学中，模型的拟合优度是指模型对观测数据的拟合程度，也就是模型在拟合数据时产生的误差的大小。对于这种类型的模型，衡量其拟合优度的指标是...

raise ValueError( ValueError: For evaluating multiple scores, use sklearn.model_selection.cross_validate instead. ['accuracy', 'adjusted_mutual_info_score', 'adjusted_rand_score', 'average_precision', 'balanced_accuracy', 'completeness_score', 'explained_variance', 'f1', 'f1_macro', 'f1_micro', 'f1_samples', 'f1_weighted', 'fowlkes_mallows_score', 'homogeneity_score', 'jaccard', 'jaccard_macro', 'jaccard_micro', 'jaccard_samples', 'jaccard_weighted', 'matthews_corrcoef', 'max_error', 'mutual_info_score', 'neg_brier_score', 'neg_log_loss', 'neg_mean_absolute_error', 'neg_mean_absolute_percentage_error', 'neg_mean_gamma_deviance', 'neg_mean_poisson_deviance', 'neg_mean_squared_error', 'neg_mean_squared_log_error', 'neg_median_absolute_error', 'neg_negative_likelihood_ratio', 'neg_root_mean_squared_error', 'normalized_mutual_info_score', 'positive_likelihood_ratio', 'precision', 'precision_macro', 'precision_micro', 'precision_samples', 'precision_weighted', 'r2', 'rand_score', 'recall', 'recall_macro', 'recall_micro', 'recall_samples', 'recall_weighted', 'roc_auc', 'roc_auc_ovo', 'roc_auc_ovo_weighted', 'roc_auc_ovr', 'roc_auc_ovr_weighted', 'top_k_accuracy', 'v_measure_score'] was passed.

这个错误是因为你在调用某个函数时传入了多个评估指标，而该函数不支持同时对多个指标进行评估。建议使用 sklearn.model_selection.cross_validate 函数来对多个指标进行评估。你可以将评估指标作为参数传递给该函数...

Only deviance, C available as type.measure for Cox models; deviance used instead

Only deviance, C available as type.measure for Cox models; deviance used instead 这是因为在 Cox 回归中，只有对数偏差和倾向指数可用作性能度量标准，但是 summary() 函数默认使用了平均残差平方和...

['accuracy', 'adjusted_mutual_info_score', 'adjusted_rand_score', 'average_precision', 'balanced_accuracy', 'completeness_score', 'explained_variance', 'f1', 'f1_macro', 'f1_micro', 'f1_samples', 'f1_weighted', 'fowlkes_mallows_score', 'homogeneity_score', 'jaccard', 'jaccard_macro', 'jaccard_micro', 'jaccard_samples', 'jaccard_weighted', 'matthews_corrcoef', 'max_error', 'mutual_info_score', 'neg_brier_score', 'neg_log_loss', 'neg_mean_absolute_error', 'neg_mean_absolute_percentage_error', 'neg_mean_gamma_deviance', 'neg_mean_poisson_deviance', 'neg_mean_squared_error', 'neg_mean_squared_log_error', 'neg_median_absolute_error', 'neg_negative_likelihood_ratio', 'neg_root_mean_squared_error', 'normalized_mutual_info_score', 'positive_likelihood_ratio', 'precision', 'precision_macro', 'precision_micro', 'precision_samples', 'precision_weighted', 'r2', 'rand_score', 'recall', 'recall_macro', 'recall_micro', 'recall_samples', 'recall_weighted', 'roc_auc', 'roc_auc_ovo', 'roc_auc_ovo_weighted', 'roc_auc_ovr', 'roc_auc_ovr_weighted', 'top_k_accuracy', 'v_measure_score']哪些适合xgboost回归

对于XGBoost回归问题，以下指标适合用于评估模型性能： - explained_variance：解释方差 - neg_mean_absolute_error：负均绝对误差 - neg_mean_squared_error：负均方误差 - neg_mean_squared_log_error：负对数...

frailty <- rgamma(2139,1,1) data$new_col <- frailty cox_model <- coxph(Surv(days, cens) ~ arms + cd40 + age + wtkg + homo + race + gender+frailty, data = data) HR <- exp(coef(cox_model)["arms"]) cox_snell_residuals <- residuals(cox_model, type = "deviance")阐述这段代码

这段代码的作用是构建一个基于Cox比例风险模型的生存分析，并计算生存分析中的相应统计量。首先，使用R中的rgamma函数生成一个2139个元素的Gamma分布的随机向量，其中参数1...这里使用了“deviance”类型的残差计算。

from sklearn.ensemble import GradientBoostingClassifier gbc = GradientBoostingClassifier(loss='deviance', learning_rate=0.1, n_estimators=5, subsample=1, min_samples_split=2, min_samples_leaf=1, max_depth=3) gbc.fit(X_train, y_train)什么意思

- loss：损失函数，可以选择 'deviance' 表示对数损失函数（Logistic Regression），或者 'exponential' 表示指数损失函数； - learning_rate：学习率，控制每个基分类器的权重缩减程度，一般取值较小，如 0.1、...

library(coxme) cox_model <- coxme(Surv(days, cens) ~ arms + cd40 + age + wtkg + homo + race + gender + (1|frailty), data = data, frailty = TRUE, dist = "gamma") HR <- exp(coef(cox_model)["arms"]) cox_snell_residuals <- residuals(cox_model, type = "deviance")这个代码有问题吗

这段代码的语法是正确的，但它的正确性取决于你的数据和研究问题。这段代码使用了 R 语言中的 coxme 包，该包提供了一个用于拟合 Cox 比例风险模型的函数 coxme()。模型中包括了一个 Frailty 随机效应，用于考虑...

using dataset:grazing of R 4. Show that overdispersion exists. 5. Fit a Quasi-Poisson model and fit a negative binomial GLM. 6. Compare the estimated coefficients, standard errors, significant tests of all three fitted models.

dispersion (resid(model_pois, type = "pearson")^2) / df.residual(model_pois) dispersion If the dispersion parameter is greater than 1, then overdispersion exists. 5. To fit a Quasi-Poisson model...

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type.measure='deviance

Only deviance, C available as type.measure for Cox models; deviance used instead

frailty <- rgamma(2139,1,1) data$new_col <- frailty cox_model <- coxph(Surv(days, cens) ~ arms + cd40 + age + wtkg + homo + race + gender+frailty, data = data) HR <- exp(coef(cox_model)["arms"]) cox_snell_residuals <- residuals(cox_model, type = "deviance")阐述这段代码

from sklearn.ensemble import GradientBoostingClassifier gbc = GradientBoostingClassifier(loss='deviance', learning_rate=0.1, n_estimators=5, subsample=1, min_samples_split=2, min_samples_leaf=1, max_depth=3) gbc.fit(X_train, y_train)什么意思

library(coxme) cox_model <- coxme(Surv(days, cens) ~ arms + cd40 + age + wtkg + homo + race + gender + (1|frailty), data = data, frailty = TRUE, dist = "gamma") HR <- exp(coef(cox_model)["arms"]) cox_snell_residuals <- residuals(cox_model, type = "deviance")这个代码有问题吗

using dataset:grazing of R 4. Show that overdispersion exists. 5. Fit a Quasi-Poisson model and fit a negative binomial GLM. 6. Compare the estimated coefficients, standard errors, significant tests of all three fitted models.

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