R 如何在泊松回归中组合组来估计对比度？
我不确定这更像是一个编程或统计（即我缺乏理解）问题 我有一个泊松混合模型，我想用它来比较不同时间段各组的平均计数R 如何在泊松回归中组合组来估计对比度？,r,statistics,emmeans,R,Statistics,Emmeans,我不确定这更像是一个编程或统计（即我缺乏理解）问题 我有一个泊松混合模型，我想用它来比较不同时间段各组的平均计数 mod <- glmer(Y ~ TX_GROUP * time + (1|ID), data = dat, family = poisson) mod_em <- emmeans(mod, c("TX_GROUP","time"), type = "response") TX_GROUP time rate SE df asymp.LCL a
mod <- glmer(Y ~ TX_GROUP * time + (1|ID), data = dat, family = poisson)
mod_em <- emmeans(mod, c("TX_GROUP","time"), type = "response")
TX_GROUP time rate SE df asymp.LCL asymp.UCL
0 1 5.743158 0.4566671 Inf 4.914366 6.711723
1 1 5.529303 0.4639790 Inf 4.690766 6.517741
0 2 2.444541 0.2981097 Inf 1.924837 3.104564
1 2 1.467247 0.2307103 Inf 1.078103 1.996855
0 3 4.570218 0.4121428 Inf 3.829795 5.453790
1 3 1.676827 0.2472920 Inf 1.255904 2.238826
如果我尝试使该值与组合组的简单平均值不匹配。要使用包中的示例数据，似乎没有问题，但我会在公式中使用分组
> warp.lm <- lm(breaks ~ wool*tension, data = warpbreaks)
> warp.emm <- emmeans(warp.lm, c("tension", "wool"))
> warp.emm
tension wool emmean SE df lower.CL upper.CL
L A 44.55556 3.646761 48 37.22325 51.88786
M A 24.00000 3.646761 48 16.66769 31.33231
H A 24.55556 3.646761 48 17.22325 31.88786
L B 28.22222 3.646761 48 20.88992 35.55453
M B 28.77778 3.646761 48 21.44547 36.11008
H B 18.77778 3.646761 48 11.44547 26.11008
Confidence level used: 0.95
>warp.lm warp.emm warp.emm
张力羊毛emmean SE df lower.CL upper.CL
LA 44.55556 3.646761 48 37.22325 51.88786
M A 24.000003.646761 48 16.66769 31.33231
H A 24.55556 3.646761 48 17.22325 31.88786
L B 28.22223.646761 48 20.88992 35.55453
M B 28.77778 3.646761 48 21.44547 36.11008
H B 18.77778 3.646761 48 11.44547 26.11008
使用的置信水平：0.95
> contrast(warp.emm, list(A.LM = c(1, 1, 0, 0, 0, 0),
+ B.LM = c(0, 0, 0, 1, 1, 0)))
contrast estimate SE df t.ratio p.value
A.LM 68.55556 5.157299 48 13.293 <.0001
B.LM 57.00000 5.157299 48 11.052 <.0001
A.LM 68.55556 5.157299 48 13.293要使用包中的示例数据，似乎还可以，不过我会在公式中使用分组
> warp.lm <- lm(breaks ~ wool*tension, data = warpbreaks)
> warp.emm <- emmeans(warp.lm, c("tension", "wool"))
> warp.emm
tension wool emmean SE df lower.CL upper.CL
L A 44.55556 3.646761 48 37.22325 51.88786
M A 24.00000 3.646761 48 16.66769 31.33231
H A 24.55556 3.646761 48 17.22325 31.88786
L B 28.22222 3.646761 48 20.88992 35.55453
M B 28.77778 3.646761 48 21.44547 36.11008
H B 18.77778 3.646761 48 11.44547 26.11008
Confidence level used: 0.95
>warp.lm warp.emm warp.emm
张力羊毛emmean SE df lower.CL upper.CL
LA 44.55556 3.646761 48 37.22325 51.88786
M A 24.000003.646761 48 16.66769 31.33231
H A 24.55556 3.646761 48 17.22325 31.88786
L B 28.22223.646761 48 20.88992 35.55453
M B 28.77778 3.646761 48 21.44547 36.11008
H B 18.77778 3.646761 48 11.44547 26.11008
使用的置信水平：0.95
> contrast(warp.emm, list(A.LM = c(1, 1, 0, 0, 0, 0),
+ B.LM = c(0, 0, 0, 1, 1, 0)))
contrast estimate SE df t.ratio p.value
A.LM 68.55556 5.157299 48 13.293 <.0001
B.LM 57.00000 5.157299 48 11.052 <.0001
A.LM 68.55556 5.157299 48 13.293首先，我建议您将两种对比放在一个列表中，例如
contr = list(`2+2|0` = c(0, 0, 1, 0, 1, 0),
`2+3|1` = c(0, 0, 0, 1, 0, 1))
mod_con = update(contrast(mod_emm, contr), tran = "log")
summary(mod_con, type = "response")
updatecontrastmod_emmr = regrid(mod_emm)
contrast(mod_emmr, contr)
pigs> pigs.lm <- lm(log(conc) ~ source + factor(percent), data = pigs)
group> pigs.emm = add_grouping(ref_grid(pigs.lm), "group", "percent", c("1&2","1&2","3&4","3&4"))
> str(pigs.emm)
'emmGrid' object with variables:
source = fish, soy, skim
percent = 9, 12, 15, 18
group = 1&2, 3&4
Nesting structure: percent %in% group
Transformation: “log”
组> emmeans(pigs.emm, "group")
group emmean SE df lower.CL upper.CL
1&2 3.535084 0.02803816 23 3.477083 3.593085
3&4 3.703931 0.03414907 23 3.633288 3.774574
Results are averaged over the levels of: source, percent
Results are given on the log (not the response) scale.
Confidence level used: 0.95
> summary(.Last.value, type = "response")
group response SE df lower.CL upper.CL
1&2 34.29790 0.961650 23 32.36517 36.34605
3&4 40.60662 1.386678 23 37.83703 43.57893
Results are averaged over the levels of: source, percent
Confidence level used: 0.95
Intervals are back-transformed from the log scale
这些是平均值，而不是总和，但除此之外它是有效的，并且转换不会像在
对比度（）中那样被破坏。首先，我建议您将两个对比度放在一个列表中，例如
contr = list(`2+2|0` = c(0, 0, 1, 0, 1, 0),
`2+3|1` = c(0, 0, 0, 1, 0, 1))
您必须决定何时要反变换。请参阅并注意关于“时间就是一切”的讨论。两个基本选项是：
一个选项：获取日志计数的边际平均值，然后进行反向转换：
mod_con = update(contrast(mod_emm, contr), tran = "log")
summary(mod_con, type = "response")
[需要调用update
，因为contrast
除去了特殊情况下的变换，因为它并不总是知道分配给任意线性函数的比例。例如，两个平方根的差不在平方根比例上。]
第二个选项：反向变换预测，然后求和：
mod_emmr = regrid(mod_emm)
contrast(mod_emmr, contr)
这些结果之间的区别与几何平均值（选项1）和算术平均值（选项2）之间的区别相同。我怀疑它们中的任何一个是否会产生与原始边际平均数相同的结果，因为它们是基于您的模型的预测。就我个人而言，我认为第一种选择是更好的选择，因为总和是线性运算，模型在对数尺度上是线性的
补遗
实际上还有第三个选项，即创建分组变量。我将用pigs
数据集进行说明
> pigs.lm <- lm(log(conc) ~ source + factor(percent), data = pigs)
现在，让我们创建一个分组因子group
：
> pigs.emm = add_grouping(ref_grid(pigs.lm), "group", "percent", c("1&2","1&2","3&4","3&4"))
> str(pigs.emm)
'emmGrid' object with variables:
source = fish, soy, skim
percent = 9, 12, 15, 18
group = 1&2, 3&4
Nesting structure: percent %in% group
Transformation: “log”
现在获取组
的EMM，注意它们只是各个级别的平均值：
> emmeans(pigs.emm, "group")
group emmean SE df lower.CL upper.CL
1&2 3.535084 0.02803816 23 3.477083 3.593085
3&4 3.703931 0.03414907 23 3.633288 3.774574
Results are averaged over the levels of: source, percent
Results are given on the log (not the response) scale.
Confidence level used: 0.95
以下是反应量表的总结：
> summary(.Last.value, type = "response")
group response SE df lower.CL upper.CL
1&2 34.29790 0.961650 23 32.36517 36.34605
3&4 40.60662 1.386678 23 37.83703 43.57893
Results are averaged over the levels of: source, percent
Confidence level used: 0.95
Intervals are back-transformed from the log scale
这些是平均值，而不是求和值，但在其他方面它是有效的，并且转换不会像在对比度（）中那样被破坏谢谢。第二种方法适用于我，但第一种方法不适用（这似乎更直观）-它似乎不会返回转换后的值：
(mod_em_inj <- emmeans(mod_inj, c("TX_GROUP","time"), type = "response"))
TX_GROUP time rate SE df asymp.LCL asymp.UCL
0 1 5.743158 0.4566671 Inf 4.914366 6.711723
1 1 5.529303 0.4639790 Inf 4.690766 6.517741
0 2 2.444541 0.2981097 Inf 1.924837 3.104564
1 2 1.467247 0.2307103 Inf 1.078103 1.996855
0 3 4.570218 0.4121428 Inf 3.829795 5.453790
1 3 1.676827 0.2472920 Inf 1.255904 2.238826
# Marginal means for combined period (7 - 24 months) - Method 1
(mod_em_inj2 <- emmeans(mod_inj, c("TX_GROUP","time")))
TX_GROUP time emmean SE df asymp.LCL asymp.UCL
0 1 1.7480092 0.07951497 Inf 1.59216273 1.9038557
1 1 1.7100619 0.08391274 Inf 1.54559591 1.8745278
0 2 0.8938574 0.12194916 Inf 0.65484147 1.1328734
1 2 0.3833880 0.15724024 Inf 0.07520279 0.6915732
0 3 1.5195610 0.09018011 Inf 1.34281119 1.6963107
1 3 0.5169035 0.14747615 Inf 0.22785558 0.8059515
contr = list(`2+3|0` = c(0, 0, 1, 0, 1, 0),
`2+3|1` = c(0, 0, 0, 1, 0, 1))
summary(contrast(mod_em_inj2, contr), type = "response")
contrast estimate SE df z.ratio p.value
2+3|0 2.4134184 0.1541715 Inf 15.654 <.0001
2+3|1 0.9002915 0.2198023 Inf 4.096 <.0001
# Marginal means for combined period (7 - 24 months) - Method 2
mod_emmr = regrid(mod_em_inj)
contrast(mod_emmr, contr)
contrast estimate SE df z.ratio p.value
2+3|0 7.014759 0.5169870 Inf 13.569 <.0001
2+3|1 3.144075 0.3448274 Inf 9.118 <.0001
（mod_em_inj谢谢。第二种方法对我有效，但不是第一种（看起来更直观）-它似乎不会返回转换后的值：
(mod_em_inj <- emmeans(mod_inj, c("TX_GROUP","time"), type = "response"))
TX_GROUP time rate SE df asymp.LCL asymp.UCL
0 1 5.743158 0.4566671 Inf 4.914366 6.711723
1 1 5.529303 0.4639790 Inf 4.690766 6.517741
0 2 2.444541 0.2981097 Inf 1.924837 3.104564
1 2 1.467247 0.2307103 Inf 1.078103 1.996855
0 3 4.570218 0.4121428 Inf 3.829795 5.453790
1 3 1.676827 0.2472920 Inf 1.255904 2.238826
# Marginal means for combined period (7 - 24 months) - Method 1
(mod_em_inj2 <- emmeans(mod_inj, c("TX_GROUP","time")))
TX_GROUP time emmean SE df asymp.LCL asymp.UCL
0 1 1.7480092 0.07951497 Inf 1.59216273 1.9038557
1 1 1.7100619 0.08391274 Inf 1.54559591 1.8745278
0 2 0.8938574 0.12194916 Inf 0.65484147 1.1328734
1 2 0.3833880 0.15724024 Inf 0.07520279 0.6915732
0 3 1.5195610 0.09018011 Inf 1.34281119 1.6963107
1 3 0.5169035 0.14747615 Inf 0.22785558 0.8059515
contr = list(`2+3|0` = c(0, 0, 1, 0, 1, 0),
`2+3|1` = c(0, 0, 0, 1, 0, 1))
summary(contrast(mod_em_inj2, contr), type = "response")
contrast estimate SE df z.ratio p.value
2+3|0 2.4134184 0.1541715 Inf 15.654 <.0001
2+3|1 0.9002915 0.2198023 Inf 4.096 <.0001
# Marginal means for combined period (7 - 24 months) - Method 2
mod_emmr = regrid(mod_em_inj)
contrast(mod_emmr, contr)
contrast estimate SE df z.ratio p.value
2+3|0 7.014759 0.5169870 Inf 13.569 <.0001
2+3|1 3.144075 0.3448274 Inf 9.118 <.0001
（mod_em_inj需要更多的细节。我们需要一个可复制的例子，也要看看你得到了什么。你的话中有一个不匹配的地方；你的方程有一个和，但你说你没有得到简单的平均值。这些论坛刺激了改进。例如，这个论坛建议对比（）
当?