我有8个数据集进行了插值，因此x轴间距相同，但长度不同，从800-1200点不等。我想做的是计算每个y值的平均值和标准偏差。理想情况下，会有1200个点的输出，平均值和标准偏差在单独的列中，从那里我可以绘制平均值y的图表，并添加误差条。我不能给出数据本身，但这是设置。如有任何帮助或指示，我们将不胜感激！
#s1 k = 0.4 800 data points
S1dat <- data.frame(x=S[[1]][["K"]], S[[1]][["ShearStress"]])
S1datapprox <- data.frame(approx(S1dat$x, S1dat$y, n = 800))
#s2 k = 0.6 1200 data points
S2dat <- data.frame(x=S[[2]][["K"]], S[[2]][["ShearStress"]])
S2datapprox <- data.frame(approx(S2dat$x, S2dat$y, n = 1200))
#s3 k = 0.34 680 data points
S3dat <- data.frame(x=S[[3]][["K"]], S[[3]][["ShearStress"]])
S3datapprox <- data.frame(approx(S3dat$x, S3dat$y, n = 800))
#s4 k = 0.5 1000 data points
S4dat <- data.frame(x=S[[4]][["K"]], S[[4]][["ShearStress"]])
S4datapprox <- data.frame(approx(S4dat$x, S4dat$y, n = 1000))
#s5 k = 0.4 800 data points
S5dat <- data.frame(x=S[[5]][["K"]], S[[5]][["ShearStress"]])
S5datapprox <- data.frame(approx(S5dat$x, S5dat$y, n = 800))
#s6 k = 0.36 720 data points
S6dat <- data.frame(x=S[[6]][["K"]], S[[6]][["ShearStress"]])
S6datapprox <- data.frame(approx(S6dat$x, S6dat$y, n = 720))
#s7 k = 0.4 800 data points
S7dat <- data.frame(x=S[[7]][["K"]], S[[7]][["ShearStress"]])
S7datapprox <- data.frame(approx(S7dat$x, S7dat$y, n = 800))
#s8 k = 0.54 1080 data points
S8dat <- data.frame(x=S[[8]][["K"]], S[[8]][["ShearStress"]])
S8datapprox <- data.frame(approx(S8dat$x, S8dat$y, n = 1080))
str(S)
List of 8
$ :'data.frame': 805 obs. of 3 variables:
..$ K : num [1:805] 0 0.000498 0.000996 0.001494 0.001992 ...
..$ ShearStress : num [1:805] 0 178 356 578 841 ...
..$ NormalStress: num [1:805] 0 -1.77 -5.35 -7.14 -11 -15 -16.7 -20.4 -22 -23.6 ...
$ :'data.frame': 1500 obs. of 3 variables:
..$ K : num [1:1500] 0 0.0004 0.000801 0.001201 0.001602 ...
..$ ShearStress : num [1:1500] 0 23.6 38.3 43.7 68.3 ...
..$ NormalStress: num [1:1500] 0.1 -1.34 -2.49 -4.04 -5.7 -7.28 -9.08 -10.7 -12.5 -14.3 ...
$ :'data.frame': 812 obs. of 3 variables:
..$ K : num [1:812] 0 0.000419 0.000838 0.001257 0.001676 ...
..$ ShearStress : num [1:812] 0 243 547 973 1280 ...
..$ NormalStress: num [1:812] 0 -0.89 -3.63 -6.05 -8.7 -11.5 -14.1 -16.9 -19.3 -22.1 ...
$ :'data.frame': 853 obs. of 3 variables:
..$ K : num [1:853] 0 0.000587 0.001174 0.00176 0.002347 ...
..$ ShearStress : num [1:853] 0 246 480 756 1246 ...
..$ NormalStress: num [1:853] 0 1 3 3 4 ...
$ :'data.frame': 916 obs. of 3 variables:
..$ K : num [1:916] 0 0.000437 0.000874 0.001312 0.001749 ...
..$ ShearStress : num [1:916] 0 44.4 67.1 89.2 119.1 ...
..$ NormalStress: num [1:916] 0 -0.01 -2.08 -5.06 -7.06 -9.4 -11.9 -14.6 -17.2 -20.1 ...
$ :'data.frame': 329 obs. of 3 variables:
..$ K : num [1:329] 0 0.000213 0.000536 0.001105 0.001871 ...
..$ ShearStress : num [1:329] 0 52.7 174.7 415.4 740.5 ...
..$ NormalStress: num [1:329] 0 -29.1 -30.4 -31.8 -33 -34.2 -35.3 -36.4 -38.3 -39.8 ...
$ :'data.frame': 790 obs. of 3 variables:
..$ K : num [1:790] 0 0.000237 0.000745 0.001252 0.00176 ...
..$ ShearStress : num [1:790] 0 94.8 347.7 633.8 1215.9 ...
..$ NormalStress: num [1:790] 0 -6 -12 -17 -28.4 ...
$ :'data.frame': 1060 obs. of 3 variables:
..$ K : num [1:1060] 0 0.00051 0.00102 0.00153 0.00204 ...
..$ ShearStress : num [1:1060] 0 44.2 70.4 100.3 133.3 ...
..$ NormalStress: num [1:1060] 0 0.1 0.18 -0.2 -1.2 ... ````
我们可以使用lapply在list"S"上循环，而不是单独执行此操作并创建多个对象
do.call(rbind, lapply(S, function(x) {
x1 <- approx(x[["K"]], x[["ShearStress"]], n = nrow(x))$y
data.frame(Mean = mean(x1, na.rm = TRUE), SD = sd(x1, na.rm = TRUE))
}))
如果我们需要元素式mean/sd，那么我们需要将输出的length更改为list的max长度，然后再更改rowMeans/rowSds
lst1 <- lapply(S, function(x) {
approx(x[["K"]], x[["ShearStress"]], n = nrow(x))$y
})
mx <- max(lengths(lst1))
m1 <- as.matrix(do.call(cbind, lapply(lst1, `length<-`, mx)))
library(matrixStats)
out <- data.frame(Mean = rowMeans(m1, na.rm = TRUE),
SD = rowSds(m1, na.rm = TRUE))