抱歉,我是R新手,例如,我有一个具有树的高度和树冠密度的数据集:
allprojects {
// ...
configurations.all {
resolutionStrategy {
force 'com.google.firebase:firebase-common:17.0.0'
force 'com.google.android.gms:play-services-basement:16.2.0'
force 'com.google.firebase:firebase-iid:16.0.0'
force 'com.google.firebase:firebase-auth:17.0.0'
}
}
}
我想将“ h_100”重新分组为2m个间隔(从2m最小到最大30m),然后我想计算每个间隔的平均i_cd值和四分位数范围,以便我可以用最小二乘回归绘制它们。我用来获取均值的代码有问题。这是我到目前为止的内容:
i_h100 i_cd
2.89 0.0198
2.88 0.0198
17.53 0.658
27.23 0.347
预先感谢您的任何建议。
答案 0 :(得分:2)
使用aggregate()
计算分组均值。
# Some example data
set.seed(1)
i_h100 <- round(runif(100, 2, 30), 2)
i_cd <- rexp(100, 1/i_h100)
mydata <- data.frame(i_cd, i_h100)
# Grouping i_h100
mydata$i_h100_2m <- cut(mydata$i_h100, seq(2, 30, by=2))
head(mydata)
# i_cd i_h100 i_h100_2m
# 1 2.918093 9.43 (8,10]
# 2 13.735728 12.42 (12,14]
# 3 13.966347 18.04 (18,20]
# 4 2.459760 27.43 (26,28]
# 5 8.477551 7.65 (6,8]
# 6 6.713224 27.15 (26,28]
# Calculate groupwise means of i_cd
i_cd_2m_mean <- aggregate(i_cd ~ i_h100_2m, mydata, mean)
# And IQR
i_cd_2m_iqr <- aggregate(i_cd ~ i_h100_2m, mydata, IQR)
upper <- i_cd_2m_mean[,2]+(i_cd_2m_iqr[,2]/2)
lower <- i_cd_2m_mean[,2]-(i_cd_2m_iqr[,2]/2)
# Plotting the result
plot.default(i_cd_2m_mean, xaxt="n", ylim=range(c(upper, lower)),
main="Groupwise means \U00B1 0.5 IQR", type="n")
points(upper, pch=2, col="lightblue", lwd=1.5)
points(lower, pch=6, col="pink", lwd=1.5)
points(i_cd_2m_mean, pch=16)
axis(1, i_cd_2m[,1], as.character(i_cd_2m[,1]), cex.axis=0.6, las=2)
答案 1 :(得分:1)
这是一个解决方案,
library(reshape2)
library(dplyr)
mydata <- data_frame(i_h100=c(2.89,2.88,17.53,27.23),i_cd=c(0.0198,0.0198,0.658,0.347))
height <- mydata$i_h100
breaks <- seq(2,30,by=2) #2m intervals
height.cut <- cut(height, breaks, right=TRUE)
mydata$height.cut <- height.cut
mean_i_h100 <- mydata %>% group_by(height.cut) %>% summarise(mean_i_h100 = mean(i_h100))
一些评论:
mean
变量更改为mean_i_h100
mean_i_h100 <- summarise(group_by(mydata,height.cut),mean_i_h100 = mean(i_h100))
library
安装的两个软件包