Question

这是我数据的结构

 > dput(test)
structure(list(MAT = c(4.9, 4.9, 15.5, 14.1, 14.1, 14.1, 11.5, 
11.5, 11.5, 17, 6.1, 2.7, 2.2, 2.2, 14.1, 14.1, 14.1, 9.5, 9.5, 
9.5, 9.5, 9.3, 8.3, 8.266666651, 8.266666651, 4.3, 4.3, 22.3, 
14.1, 14.1, 14.1, 8.5, 8.5, 8.5, 8.5, 21.5, 21.5, 3.8, 3.8, 6, 
6, 6, 6, 6), es = c(0.29603085763985, 0.421393627439682, 0.189653473156549, 
0.226685054608428, 0.291373762079697, 0.166533544378467, 0.250586529054368, 
0.146320008054403, 0.199565119644333, -0.0819047677231083, 0.15963948187092, 
-0.154628141843561, 0.201121044198443, 0.0867981239977565, 0.543870310978598, 
0.34547921143505, 0.37557241352574, -0.287318919407836, 0.207937483228907, 0.190143660810163, 0.276182673435993, 0.128596803172119, 0.454753165843559, 
0.399237234440439, 0.32075358541748, 0.362664873575803, -0.0865925288159671, 
0.51290512543514, 0.186308318839249, 0.147936083867325, 0.243792477087184, 
0.625169403695832, 0.110317782120045, 0.217836235313289, 0.171468156841181, 
0.50548821117127, 0.164418265301427, -0.00246305543239786, 0.325552346507191, 
0.381240606108843, 0.19337350462531, 0.0408803528990759, 0.321815078821239, 
0.307642815014319), var = c(0.00496277337027962, 0.0130962311273343, 
0.0180149624217804, 0.0134568083459063, 0.00139708925143695, 
0.000725862546533828, 0.00670831011660164, 0.0190783110089115, 
0.0641568910090007, 0.0121596544795352, 0.0653909966557582, 0.0514610437228611, 
0.0231592619167496, 0.0108989891148006, 0.0588577146414195, 0.0695760532112402, 
0.0744256820906048, 0.00997789089155498, 0.00928124381998638, 
0.0145009450673482, 0.00652956018299188, 0.0111886178917916, 
0.0265943757419349, 0.142676904340634, 0.110705177803624, 0.0576538348777718, 
0.0625171635976251, 0.0131652117394448, 0.00947904166717649, 
0.00813569411386797, 0.00444289889858652, 0.0673007030900184,0.00545169559098343, 0.240046081413733, 0.00561125010476281, 
    0.0185516235174018, 0.0179989506841957, 0.0496806959944248, 0.022478393723115, 
    0.0521209786580004, 0.282298667080106, 0.0151428845076692, 0.00992945920656693, 0.0145544965304081), MAP = c(810, 810, 1140, 1750, 1750, 1750, 
    1034, 1034, 1034, 720, 645, 645, 645, 645, 1000, 1000, 1000, 
    691, 691, 691, 691, 1134, 1750, 1326, 1326, 1140, 1140, 1310, 
    1750, 1750, 1750, 1003, 1003, 1003, 1003, 1750, 1750, 1750, 1750, 
    1750, 1750, 1750, 1750, 1750), CO2dif = c(162L, 162L, 190L, 165L, 
    165L, 165L, 200L, 200L, 200L, 150L, 335L, 335L, 335L, 335L, 348L, 
    348L, 348L, 200L, 200L, 200L, 200L, 220L, 350L, 350L, 350L, 350L, 
    350L, 350L, 180L, 180L, 180L, 130L, 130L, 130L, 130L, 320L, 320L, 
    360L, 360L, 345L, 345L, 350L, 348L, 348L)), row.names = c(NA, 
    -44L), class = "data.frame", .Names = c("MAT", "es", "var", "MAP", 
    "CO2dif"))

我使用元分析进行模型选择，预测效果大小的最佳模型是：

library(metafor)

summary(rma(es, var, data=test ,control=list(stepadj=.5), mods= ~ 1 + log(MAT) + MAP + CO2dif + log(MAT):CO2dif, knha=TRUE))

Model Results:

                 estimate      se     tval    pval    ci.lb    ci.ub     
intrcpt            1.2556  0.3719   3.3758  0.0017   0.5033   2.0080   **
log(MAT)          -0.5740  0.1694  -3.3882  0.0016  -0.9167  -0.2313   **
MAP                0.0001  0.0001   2.5181  0.0160   0.0000   0.0003    *
CO2dif            -0.0042  0.0013  -3.2932  0.0021  -0.0067  -0.0016   **
log(MAT):CO2dif    0.0020  0.0005   3.7500  0.0006   0.0009   0.0031  ***

现在，我想以es = 1200

来假设MAT = 1200 mm

来绘制MAP vs CO2dif，并举例说明此模型

MAPi <- 1200
CO2i <- 350
make_pct <- function(x) (exp(x) - 1) * 100

ggplot(test, aes(x = log(MAT), y = make_pct(es))) +
  geom_abline(aes(intercept = make_pct(1.2556 + 0.0001 * MAPi - 0.0042 * CO2i),
                  slope = make_pct(log(0.0020 * CO2i)) - make_pct(log(0.5740))) , 
                  color = "red", size=0.8) +
  geom_point() +
  theme_classic()

效果大小（es）采用log格式，我想要百分比，因此我使用函数make_pct对其进行转换。另一方面，MAT必须在模型输出中指示的图中进行对数变换。上面的ggplot的斜率是否与对数和百分比变换一致？在我看来，坡度相当低。我对这种类型的图和变换不是很熟悉，所以欢迎任何提示。谢谢

Answer 1

exp(es)-1与解释变量log(MAT)之间的关系不是线性的。
对于MAP和CO2dif的给定值集，此关系的格式为：
y = exp(es)-1 = k1*exp(k2*log(MAT))。
此功能可以如下绘制：

library(metafor)
library(ggplot2)

modfit <- rma(es, var, data=test ,control=list(stepadj=.5), 
   mods= ~ 1 + MAP + log(MAT)*CO2dif, knha=TRUE)
pars <- coef(modfit)

MAPi <- 1200
CO2i <- 350

make_pct <- function(x) (exp(x) - 1) * 100

mod_fun <- function(MAP, MAT, CO2dif, pars) {
    y <- pars[1]+pars[2]*MAP+pars[3]*log(MAT)+
         pars[4]*CO2dif+pars[5]*log(MAT)*CO2dif
    make_pct(y)
}

test$ESpct <- mod_fun(MAPi, test$MAT, CO2i, coef(modfit))

ggplot(test, aes(x = log(MAT), y = make_pct(es))) +
  geom_line(aes(y=ESpct), color = "red", size=0.8) +
  geom_point() + theme_classic()

绘制具有对数和百分比变换的模型的斜率和截距

1 个答案: