我不确定你到底想要什么。可能是这个???
lm(mpg ~ disp * gear, data = mtcars) %>%
lsmeans("disp", at = list(disp = c(90, 110, 130)))
(即,要求"disp"
,而不是"gear"
)这将返回平均gear
3.6875的估计值
disp lsmean SE df lower.CL upper.CL
90 25.24867 0.9998285 28 23.20061 27.29673
110 24.40467 0.9219954 28 22.51605 26.29330
130 23.56068 0.8515002 28 21.81646 25.30490
或者您是否也需要对每个齿轮进行估算,在这种情况下,您可能应该将其建模为一个因素:
lm(mpg ~ disp * factor(gear), data = mtcars) %>% lsmeans("gear")
这给了你
gear lsmean SE df lower.CL upper.CL
3 18.56976 0.9620189 26 16.592300 20.54722
4 11.37039 2.2772013 26 6.689537 16.05125
5 19.94602 1.1954875 26 17.488656 22.40338
还是您真的想要两者的组合?
lm(mpg ~ disp * factor(gear), data = mtcars) %>%
lsmeans(~ gear * disp, at = list(disp = c(90, 110, 130)))
gear disp lsmean SE df lower.CL upper.CL
3 90 22.19623 1.8414008 26 18.41117 25.98128
4 90 28.56839 0.9946413 26 26.52387 30.61290
5 90 27.09119 1.7048444 26 23.58683 30.59554
3 110 21.68082 1.7067310 26 18.17258 25.18905
4 110 26.12414 0.7884968 26 24.50336 27.74491
5 110 26.07568 1.5477849 26 22.89417 29.25720
3 130 21.16541 1.5739549 26 17.93010 24.40072
4 130 23.67988 0.7573308 26 22.12317 25.23660
5 130 25.06018 1.4087160 26 22.16452 27.95584
还是齿轮和排量的各自组合?
.Last.value[c(1,5,9)]
gear disp lsmean SE df lower.CL upper.CL
3 90 22.19623 1.8414008 26 17.50882 26.88364
4 110 26.12414 0.7884968 26 24.11696 28.13131
5 130 25.06018 1.4087160 26 21.47420 28.64616
但是你说了一些关于齿轮差异的事情;这就是你想要的吗?
lm(mpg ~ disp * factor(gear), data = mtcars) %>%
lsmeans(~ gear | disp, at = list(disp = c(90, 110, 130))) %>%
pairs
disp = 90:
contrast estimate SE df t.ratio p.value
3 - 4 -6.37216328 2.092861 26 -3.045 0.0141
3 - 5 -4.89496043 2.509432 26 -1.951 0.1449
4 - 5 1.47720285 1.973780 26 0.748 0.7373
disp = 110:
contrast estimate SE df t.ratio p.value
3 - 4 -4.44331881 1.880069 26 -2.363 0.0646
3 - 5 -4.39486710 2.304033 26 -1.907 0.1567
4 - 5 0.04845172 1.737057 26 0.028 0.9996
disp = 130:
contrast estimate SE df t.ratio p.value
3 - 4 -2.51447435 1.746678 26 -1.440 0.3360
3 - 5 -3.89477376 2.112301 26 -1.844 0.1755
4 - 5 -1.38029941 1.599384 26 -0.863 0.6679
P value adjustment: tukey method for comparing a family of 3 estimates