现在我有一个二次多项式线性混合效应模型，使用lme4包，有四个固定因子：时间（10个时间点：1-10，多边因子），Age_Group1（成人和儿童），类型（两种语义类型单词：重复和占有粒子）和Word（四个单词，四个音调：T1，T2，T3，T4）和随机因素：参与者。
我想要儿童和成人的音高曲线＆＃39;跨类型和单词的制作，特别是它们的斜率（线性趋势）和锐度（二次趋势）。
该模型如下：
model <- lmer(Pitch ~ poly(Time,2)*Type*Word*Age_Group+(1|Participant),data = sub_t0_f0_4yrs, REML = F)
我得到的结果是＆＃34;摘要＆＃34;功能：
Fixed effects:
Estimate Std. Error df t value Pr(>|t|)
(Intercept) 24.7762 0.5283 96.0000 46.897 < 2e-16 ***
poly(Time, 2)1 -390.5239 13.4959 6141.0000 -28.937 < 2e-16 ***
poly(Time, 2)2 -104.0100 13.4959 6141.0000 -7.707 1.49e-14 ***
TypeReduplicative 0.8123 0.2409 6142.0000 3.371 0.000753 ***
WordT2 4.9607 0.2469 6142.0000 20.088 < 2e-16 ***
WordT3 3.2516 0.2425 6142.0000 13.411 < 2e-16 ***
WordT4 -2.3748 0.2436 6141.0000 -9.750 < 2e-16 ***
Age_Group1Adults -7.3116 0.8168 96.0000 -8.952 2.73e-14 ***
poly(Time, 2)1:TypeReduplicative 115.9732 18.9821 6141.0000 6.110 1.06e-09 ***
poly(Time, 2)2:TypeReduplicative 21.5677 18.9821 6141.0000 1.136 0.255912
poly(Time, 2)1:WordT2 106.4190 19.4239 6141.0000 5.479 4.45e-08 ***
poly(Time, 2)2:WordT2 -13.1036 19.4239 6141.0000 -0.675 0.499949
poly(Time, 2)1:WordT3 362.7334 19.0861 6141.0000 19.005 < 2e-16 ***
poly(Time, 2)2:WordT3 42.8257 19.0861 6141.0000 2.244 0.024880 *
poly(Time, 2)1:WordT4 80.1560 19.1942 6141.0000 4.176 3.01e-05 ***
poly(Time, 2)2:WordT4 -1.0699 19.1942 6141.0000 -0.056 0.955552
TypeReduplicative:WordT2 -0.7602 0.3439 6142.0000 -2.211 0.027102 *
TypeReduplicative:WordT3 -0.3809 0.3407 6142.0000 -1.118 0.263664
TypeReduplicative:WordT4 -1.2234 0.3437 6142.0000 -3.559 0.000375 ***
poly(Time, 2)1:Age_Group1Adults 39.1559 20.7488 6141.0000 1.887 0.059188 .
poly(Time, 2)2:Age_Group1Adults 30.5261 20.7488 6141.0000 1.471 0.141281
TypeReduplicative:Age_Group1Adults -0.7886 0.3714 6141.0000 -2.123 0.033758 *
WordT2:Age_Group1Adults -0.7191 0.3753 6142.0000 -1.916 0.055389 .
WordT3:Age_Group1Adults -1.5218 0.3724 6141.0000 -4.087 4.42e-05 ***
WordT4:Age_Group1Adults -1.0807 0.3731 6141.0000 -2.897 0.003783 **
poly(Time, 2)1:TypeReduplicative:WordT2 0.8514 27.0859 6141.0000 0.031 0.974924
poly(Time, 2)2:TypeReduplicative:WordT2 -25.8572 27.0859 6141.0000 -0.955 0.339800
poly(Time, 2)1:TypeReduplicative:WordT3 -107.0904 26.8447 6141.0000 -3.989 6.71e-05 ***
poly(Time, 2)2:TypeReduplicative:WordT3 5.8849 26.8447 6141.0000 0.219 0.826487
poly(Time, 2)1:TypeReduplicative:WordT4 -111.4500 27.0717 6141.0000 -4.117 3.89e-05 ***
poly(Time, 2)2:TypeReduplicative:WordT4 -14.2141 27.0717 6141.0000 -0.525 0.599565
poly(Time, 2)1:TypeReduplicative:Age_Group1Adults -116.7790 29.2756 6141.0000 -3.989 6.71e-05 ***
poly(Time, 2)2:TypeReduplicative:Age_Group1Adults -26.3431 29.2756 6141.0000 -0.900 0.368246
poly(Time, 2)1:WordT2:Age_Group1Adults 32.8847 29.5640 6141.0000 1.112 0.266044
poly(Time, 2)2:WordT2:Age_Group1Adults 21.4939 29.5640 6141.0000 0.727 0.467237
poly(Time, 2)1:WordT3:Age_Group1Adults -17.5598 29.3432 6141.0000 -0.598 0.549576
poly(Time, 2)2:WordT3:Age_Group1Adults 15.0965 29.3432 6141.0000 0.514 0.606934
poly(Time, 2)1:WordT4:Age_Group1Adults -8.2570 29.4136 6141.0000 -0.281 0.778934
poly(Time, 2)2:WordT4:Age_Group1Adults 31.1921 29.4136 6141.0000 1.060 0.288975
TypeReduplicative:WordT2:Age_Group1Adults 1.2192 0.5273 6141.0000 2.312 0.020792 *
TypeReduplicative:WordT3:Age_Group1Adults 2.2395 0.5252 6141.0000 4.264 2.03e-05 ***
TypeReduplicative:WordT4:Age_Group1Adults 3.7471 0.5271 6141.0000 7.108 1.31e-12 ***
poly(Time, 2)1:TypeReduplicative:WordT2:Age_Group1Adults 14.9473 41.5588 6141.0000 0.360 0.719109
poly(Time, 2)2:TypeReduplicative:WordT2:Age_Group1Adults -26.2315 41.5588 6141.0000 -0.631 0.527940
poly(Time, 2)1:TypeReduplicative:WordT3:Age_Group1Adults 127.8737 41.4020 6141.0000 3.089 0.002020 **
poly(Time, 2)2:TypeReduplicative:WordT3:Age_Group1Adults 15.8382 41.4020 6141.0000 0.383 0.702069
poly(Time, 2)1:TypeReduplicative:WordT4:Age_Group1Adults 137.2644 41.5495 6141.0000 3.304 0.000960 ***
poly(Time, 2)2:TypeReduplicative:WordT4:Age_Group1Adults -23.2233 41.5495 6141.0000 -0.559 0.576229
很明显，Word Type和Age_Group之间存在重要的三方互动。我通常会进行事后测试，以便在不同条件下比较成人和儿童，例如：
lsmeans (model, pairwise ~ Age_Group1|Words*Type*Time)
我得到了：
$contrasts
Type = Possessive, Word = T1, Time = 5.5:
contrast estimate SE df t.ratio p.value
4yrs - Adults 7.751045 0.8696730 122.81 8.913 <.0001
Type = Reduplicative, Word = T1, Time = 5.5:
contrast estimate SE df t.ratio p.value
4yrs - Adults 8.160387 0.8687632 122.30 9.393 <.0001
Type = Possessive, Word = T2, Time = 5.5:
contrast estimate SE df t.ratio p.value
4yrs - Adults 8.779598 0.8726068 124.46 10.061 <.0001
Type = Reduplicative, Word = T2, Time = 5.5:
contrast estimate SE df t.ratio p.value
4yrs - Adults 7.592119 0.8687632 122.30 8.739 <.0001
Type = Possessive, Word = T3, Time = 5.5:
contrast estimate SE df t.ratio p.value
4yrs - Adults 9.490212 0.8696530 122.80 10.913 <.0001
Type = Reduplicative, Word = T3, Time = 5.5:
contrast estimate SE df t.ratio p.value
4yrs - Adults 7.888047 0.8687632 122.30 9.080 <.0001
Type = Possessive, Word = T4, Time = 5.5:
contrast estimate SE df t.ratio p.value
4yrs - Adults 9.280836 0.8706023 123.33 10.660 <.0001
Type = Reduplicative, Word = T4, Time = 5.5:
contrast estimate SE df t.ratio p.value
4yrs - Adults 5.608742 0.8705994 123.33 6.442 <.0001
只比较成人和儿童的截距，而不是像汇总函数那样的线性和二次参数。 所以我想知道如何比较线性趋势：poly（时间，2）1和四元趋势：成人和孩子之间的poly（时间，2）2，是否有任何解决方案？
答案 0 :(得分：0)
您可以使用 lsmeans 包中提供的多项式对比来完成此操作;然后产生交互对比（对比的对比）。在第一步中，您必须使用at来指定Time的不同级别，因为它是一个协变量，默认情况下它会降低到它的平均值（在您的情况下为5.5）。
model.lsm <- lsmeans(model, ~ Time * Age_Group1 | Words * Type,
at = list(Time = .5 + 4:6))
model.lsm # show the estimates)
contrast(model.lsm, interaction = c("poly", "pairwise"))
请注意，"poly"规范会在Time的指定级别之间创建正交多项式对比。它们的缩放比模型中的正交缩放poly()项不同。在创建model.lsm时，您可以使用at = list(Time = 1:10)（或等效cov.reduce = FALSE）来包含Time的所有级别。然而，它会产生对比度达到6度（由于你的模型，估计为零的第三个）。如果仅显示三个值，您将获得相同的P值，但不会出现混乱。
有关详细信息，请参阅contrast的帮助页面。