Question

这是我关于stackoverlow的第一个问题，如果我没有遵循正确的问题协议，请纠正我。

我正在尝试为三个时间点（时间1，时间2，时间3）收集的数据创建一些图表，这些图表相当于X1 ...，X2 ......和X3 ......列名称。图表也由数据框中的$ Group列分隔。

我创建图表没有问题，我只有很多变量（~170），我想比较时间1与时间2，时间2与时间3等等。所以我试图运行一个快捷方式来运行这个一种代码，而不是必须单独输出每一个。

如上所述，我创建了X1 ...... X2等变量名，表示记录变量的时间，即X1BCSTCAT =时间1; X2BCSTCAT =时间2; X3BCSTCAT =时间3.以下是我的数据的一小部分示例：

df <- structure(list(ID = structure(1:6, .Label = c("101","102","103","118","119","120"), class = "factor"), 
                   Group = structure(c(1L,1L,1L,2L,2L,2L), .Label = c("C8","TC"), class = "factor"), 
                   Wave = structure(c(1L, 2L, 3L, 4L, 1L, 2L), .Label = c("A","B","C","D"), class = "factor"), 
                   Yr = structure(c(1L, 2L, 1L, 2L, 1L, 2L), .Label = c("3","5"), class = c("ordered", "factor")), 
                   Age.Yr. = c(10.936,10.936, 9.311, 10.881, 10.683, 11.244), 
                   Training..hr. = c(10.667,10.333, 10.667, 10.333, 10.333, 10.333), 
                   X1BCSTCAT = c(-0.156,0.637,-1.133,0.637,2.189,1.229), 
                   X1BCSTCR = c(0.484,0.192, -1.309, 0.912, 1.902, 0.484), 
                   X1BCSTPR = c(-1.773,0.859, 0.859, 0.12, -1.111, 0.12), 
                   X2BCSTCAT = c(1.006, -0.379,-1.902, 0.444, 2.074, 1.006), 
                   X2BCSTCR = c(0.405, -0.457,-1.622, 1.368, 1.981, 0.168), 
                   X2BCSTPR = c(-0.511, -0.036,2.189, -0.036, -0.894, 0.949),
                   X3BCSTCAT = c(1.18, -1.399,-1.399, 1.18, 1.18, 1.18), 
                   X3BCSTCR = c(0.967, -1.622, -1.622,0.967, 0.967, 1.255), 
                   X3BCSTPR = c(-1.282, -1.282, 1.539,1.539, 0.792, 0.792)), 
              row.names = c(1L, 2L, 3L, 4L, 5L,8L), class = "data.frame")

下面是一些使用ggplot创建一个图表的工作代码，用于时间1对比一个变量的时间2数据：

library(ggplot2)

p <- ggplot(df, aes(x=df$X1BCSTCAT, y=df$X2BCSTCAT, shape = df$Group, color = df$Group)) + 
  geom_point() + geom_smooth(method=lm, aes(fill=df$Group), fullrange = TRUE) + 
  labs(title="BCSTCAT", x="Time 1", y = "Time 2") + 
  scale_color_manual(name = "Group",labels = c("C8","TC"),values = c("blue", "red")) +
  scale_shape_manual(name = "Group",labels = c("C8","TC"),values = c(16, 17)) +
  scale_fill_manual(name = "Group",labels = c("C8", "TC"),values = c("light blue", "pink"))

所以我真的想要创建一种快捷方式，其中R将循环并匹配变量名称X1 ... vs X2 ...依此类推并创建图形。我假设必须有一些方法来根据匹配的列号进行绘图，例如df [，7] vs df [，10]并迭代完成此过程或通过实际匹配名称进行绘图（其中变量名称的唯一差异是指示时间的数字）。

我之前已经使用lapply功能创建单个图表，但不知道从哪里开始尝试执行此操作。

Answer 1

使用tidyeval方法的解决方案。我们需要ggplot2 v3.0.0（记得重启你的R会话）

install.packages("ggplot2", dependencies = TRUE)

首先，我们构建一个以列和组名称作为输入的函数。请注意使用rlang::sym，rlang::quo_name＆amp; !!。
然后为x-创建2个名称向量＆amp; y-值，以便我们可以使用purrr::map2同时循环播放它们。

library(rlang)
library(tidyverse)

df <- structure(list(ID = structure(1:6, .Label = c("101","102","103","118","119","120"), class = "factor"), 
                   Group = structure(c(1L,1L,1L,2L,2L,2L), .Label = c("C8","TC"), class = "factor"), 
                   Wave = structure(c(1L, 2L, 3L, 4L, 1L, 2L), .Label = c("A","B","C","D"), class = "factor"), 
                   Yr = structure(c(1L, 2L, 1L, 2L, 1L, 2L), .Label = c("3","5"), class = c("ordered", "factor")), 
                   Age.Yr. = c(10.936,10.936, 9.311, 10.881, 10.683, 11.244), 
                   Training..hr. = c(10.667,10.333, 10.667, 10.333, 10.333, 10.333), 
                   X1BCSTCAT = c(-0.156,0.637,-1.133,0.637,2.189,1.229), 
                   X1BCSTCR = c(0.484,0.192, -1.309, 0.912, 1.902, 0.484), 
                   X1BCSTPR = c(-1.773,0.859, 0.859, 0.12, -1.111, 0.12), 
                   X2BCSTCAT = c(1.006, -0.379,-1.902, 0.444, 2.074, 1.006), 
                   X2BCSTCR = c(0.405, -0.457,-1.622, 1.368, 1.981, 0.168), 
                   X2BCSTPR = c(-0.511, -0.036,2.189, -0.036, -0.894, 0.949),
                   X3BCSTCAT = c(1.18, -1.399,-1.399, 1.18, 1.18, 1.18), 
                   X3BCSTCR = c(0.967, -1.622, -1.622,0.967, 0.967, 1.255), 
                   X3BCSTPR = c(-1.282, -1.282, 1.539,1.539, 0.792, 0.792)), 
              row.names = c(1L, 2L, 3L, 4L, 5L,8L), class = "data.frame")

# define a function that accept strings as input
pair_plot <- function(x_var, y_var, group_var) {

  # convert strings to symbols
  x_var <- rlang::sym(x_var)
  y_var <- rlang::sym(y_var)
  group_var <- rlang::sym(group_var)

  # unquote symbols using !! 
  ggplot(df, aes(x = !! x_var, y = !! y_var, shape = !! group_var, color = !! group_var)) + 
    geom_point() + geom_smooth(method = lm, aes(fill = !! group_var), fullrange = TRUE) + 
    labs(title = "BCSTCAT", x = rlang::quo_name(x_var), y = rlang::quo_name(y_var)) +
    scale_color_manual(name = "Group", labels = c("C8", "TC"), values = c("blue", "red")) +
    scale_shape_manual(name = "Group", labels = c("C8", "TC"), values = c(16, 17)) +
    scale_fill_manual(name = "Group",  labels = c("C8", "TC"), values = c("light blue", "pink")) +
    theme_bw()
}

# Test if the new function works
pair_plot("X1BCSTCAT", "X2BCSTCAT", "Group")

# Create 2 parallel lists 
list_x <- colnames(df)[-c(1:6, (ncol(df)-2):(ncol(df)))]
list_x
#> [1] "X1BCSTCAT" "X1BCSTCR"  "X1BCSTPR"  "X2BCSTCAT" "X2BCSTCR"  "X2BCSTPR"

list_y <- lead(colnames(df)[-(1:6)], 3)[1:length(list_x)]
list_y
#> [1] "X2BCSTCAT" "X2BCSTCR"  "X2BCSTPR"  "X3BCSTCAT" "X3BCSTCR"  "X3BCSTPR"

# Loop through 2 lists simultaneously 
# Supply inputs to pair_plot function using purrr::map2
map2(list_x, list_y, ~ pair_plot(.x, .y, "Group"))

示例输出：

#> [[1]]

#> 
#> [[2]]

由reprex package（v0.2.0）创建于2018-05-24。

根据列名

1 个答案: