从数据框中每三行绘制一次

Question

我想根据我的数据制作一些图。不幸的是，很难预测我会生成多少图，因为它取决于数据并且可能会有所不同。这就是我想让它易于调节的原因。但是，它通常是每次 3 行组中的一个图。

所以，我想从1:3、4-6、7-9 等行绘制。

这是数据：

> dput(DF_final)
structure(list(AC = c(0.0031682160632777, 0.00228591145206846, 
0.00142094444568728, 0.000661218113472149, 0.0010078157353918, 
0.000400289437089513, 40.4634784175177, 40.5055070858594, 0.0183737773741582
), SD = c(0.00250647379467532, 0.0013244185401148, 0.000469332241199189, 
0.000294558308707343, 0.000385553400676202, 0.000104447914881357, 
11.0693842400794, 8.78768774254084, 0.00696532251341454), ln_AC = c(-5.75458660556339, 
-6.08099044923792, -6.556433525855, -7.32142679754668, -6.89996992823399, 
-7.8233226797995, 3.70039979980691, 3.70143794229703, -3.99683077355773
), ln_SD = c(-5.98887837626238, -6.62678175351058, -7.66419963690747, 
-8.13003358225542, -7.86083085139947, -9.16682203300101, 2.40418312097106, 
2.17335162163583, -4.96681136795312), Percent_AC = c(126.401324043689, 
172.597361244303, 302.758754023937, 224.477834753288, 261.394591157605, 
383.243109777925, 365.544076706723, 460.934756361151, 263.789326894369
), Percent_SD = c(100, 100, 100, 100, 100, 100, 100, 100, 100
), TP = c(0, 40, 80, 0, 40, 80, 0, 40, 80)), row.names = c("Tim_0", 
"Tim_40", "Tim_80", "Jack_0", "Jack_40", "Jack_80", "Tom_0", 
"Tom_40", "Tom_80"), class = "data.frame")

ln_AC 列应设置为 Y 轴，TP 列应设置为 X 轴。首先，我希望将所有这些都放在彼此相邻的单独图表上（请记住，在某些时候绘图的数量可能会很高），如果可能的话，所有内容都在同一个图表上。它应该是带有趋势线的点图。

是否也可以从线性回归得到一个斜率，SD斜率，R^2？

我设法为单个图做到了，但回归线看起来很奇怪......

以下代码用于生成此图和回归线。

fit <- lm(DF_final$ln_AC~DF_final$TP, data=DF_final)
plot(DF_final[1:3,7], DF_final[1:3,3], type = "p", ylim = c(-10,0), xlim=c(0,100), col = "red")
lines(DF_final$TP, fitted(fit), col="blue")

Answer 1

在 base R 中（没有这么多包），你可以这样做：

# splits every 3 rows
DF = split(DF_final,gsub("_[^ ]*","",rownames(DF_final) ))
# you can also do
# DF = split(DF_final,(1:nrow(DF_final) - 1) %/%3 ))

存储您的值：

slopes =  vector("numeric",3)
names(slopes) = names(DF)
rsq = vector("numeric",3)
names(rsq) = names(DF)

绘制：

par(mfrow=c(1,3))
for(i in names(DF)){
fit <- lm(ln_AC~TP, data=DF[[i]])
plot(DF[[i]]$TP, DF[[i]]$ln_AC, type = "p", col = "red",main=i)
abline(fit, col="blue")
slopes[i]=round(fit$coefficients[2],digits=2)
rsq[i]=round(summary(fit)$r.squared,digits=2)
mtext(side=1,paste("slope=",slopes[i],"\nrsq=",rsq[i]),
padj=-2,cex=0.7)
}

还有你的价值观：

slopes
 Jack   Tim   Tom 
-0.01 -0.01 -0.10 
rsq
Jack  Tim  Tom 
0.29 0.99 0.75 

Answer 2

如果我理解正确，您希望每张图进行 3 次观察的原因是因为您有不同的人 (Jack,Tim,Tom) 。是这样吗？ 如果你不想担心这个数字，你可以这样做

# move rownames to column  
data$person <- rownames(data)
data$person <- gsub("\\_.*","",data$person)  # remove TP from names

# better to use library(data.table) for this step
data <- melt(data,id.vars=c("person","TP","ln_AC"))

ggplot(data,aes(x=TP, y=ln_AC)) + geom_point() +
     geom_smooth(method = "lm") + facet_grid(~person)

这会产生类似@giocomai 的图，但如果您的数据中有 4、5、6 或任何人，它也可以工作。

---- 编辑

如果你想添加 R2 值，你可以这样做。请注意，它可能不是最好和优雅的解决方案，但它确实有效。

data <- data.frame(...)
data$person <- rownames(data)
data$person <- gsub("\\_.*","",data$person)  

# run lm for all persons and save them in a data.frame
nomi <- unique(data$person)
#lmStats <- data.frame()
lmStats <- sapply(nomi, 
   function(ita){
      model <- lm(ln_AC~TP,data= data[which(data$person == ita),])
      lmStat <- summary(model)
      # I only save r2, but you can get all the statistics you need
      lmRow <- data.frame("r2" = lmStat$r.squared )
      #lmStats <- rbind(lmStats,lmRow)
   }
)
lmStats <- do.call(rbind,lmStats)

# format the output,and create a dataframe we will use to annotate facet_grid
lmStats <- as.data.frame(lmStats)
rownames(lmStats) <- gsub("\\..*","",rownames(lmStats))
lmStats$person <- rownames(lmStats)
colnames(lmStats)[1] <- "r2"
lmStats$r2 <- round(lmStats$r2,2)
lmStats$TP <- 40
lmStats$ln_AC <- 0
lmStats$lab <- paste0("r2= ",lmStats$r2)

# melt and add r2 column to the data (not necessary, but I like to have everything I plot in teh data)
data <- melt(data,id.vars=c("person","TP","ln_AC"))
data$r2 <- lmStats[match(data$person,rownames(lmStats)),1] 


ggplot(data,aes(x=TP, y=ln_AC)) + geom_point() +
   geom_smooth(method = "lm") + facet_grid(~person) +
   geom_text(data=lmStats,label=lmStats$lab)

一种更简单的方法（更少的步骤）是使用facet_grid(~r2)，这样您就可以在标题中使用 R.square 值。

Answer 3

如果我正确理解您的意思，假设您将始终对每个图进行三个观察，那么您的主要问题将是创建一个分类变量来将它们分开。这是实现它的一种方法。根据您喜欢的布局，您可能需要检查facet_wrap 而不是facet_grid。

library("dplyr")
library("ggplot2")
DF_final <- structure(list(AC = c(0.0031682160632777, 0.00228591145206846, 
                      0.00142094444568728, 0.000661218113472149, 0.0010078157353918, 
                      0.000400289437089513, 40.4634784175177, 40.5055070858594, 0.0183737773741582
), SD = c(0.00250647379467532, 0.0013244185401148, 0.000469332241199189, 
          0.000294558308707343, 0.000385553400676202, 0.000104447914881357, 
          11.0693842400794, 8.78768774254084, 0.00696532251341454), ln_AC = c(-5.75458660556339, 
                                                                              -6.08099044923792, -6.556433525855, -7.32142679754668, -6.89996992823399, 
                                                                              -7.8233226797995, 3.70039979980691, 3.70143794229703, -3.99683077355773
          ), ln_SD = c(-5.98887837626238, -6.62678175351058, -7.66419963690747, 
                       -8.13003358225542, -7.86083085139947, -9.16682203300101, 2.40418312097106, 
                       2.17335162163583, -4.96681136795312), Percent_AC = c(126.401324043689, 
                                                                            172.597361244303, 302.758754023937, 224.477834753288, 261.394591157605, 
                                                                            383.243109777925, 365.544076706723, 460.934756361151, 263.789326894369
                       ), Percent_SD = c(100, 100, 100, 100, 100, 100, 100, 100, 100
                       ), TP = c(0, 40, 80, 0, 40, 80, 0, 40, 80)), row.names = c("Tim_0", 
                                                                                  "Tim_40", "Tim_80", "Jack_0", "Jack_40", "Jack_80", "Tom_0", 
                                                                                  "Tom_40", "Tom_80"), class = "data.frame")
DF_final %>% 
  mutate(id = as.character(sapply(1:(nrow(DF_final)/3), rep, 3))) %>% 
  ggplot(aes(x=TP, y=ln_AC)) +
  geom_point() +
  geom_smooth(method = "lm") +
  facet_grid(~id)

^{由reprex package (v0.3.0) 于 2020 年 2 月 6 日创建}