为多个变量绘制多个斜率图（使用 for 循环）（面临关键冗余问题）

Question

尽管我设法用假数据绘制了一个多斜率图（参见下面的可重现示例），但我在设法使代码适应我的真实数据时遇到了麻烦，并且由于密钥冗余而不断面临错误。

首先，一些上下文：我有一个包含许多“_x”和“_y”变量的数据集，它们是时间 1 和 2 的测量值 - 记录在一列中，因为每个条目都有时间 1 和时间 2 - 和我想为每个人绘制我的斜率，为每个变量（变量对）绘制一个图。

在一些帮助下，我设法为以下可重现示例中的一组变量执行了此操作，没有“_x”或“_y”列名。然而，当我尝试使用 selects 调整此代码时 - 为了只采用这些列而不是所有数据集 - 更改 colnames 以模仿示例，更改正则表达式等。我一直面临键冗余的错误。

"Error in spread():
! Each row of output must be identified by a unique combination of keys.
Keys are shared for 195 rows:"

我怀疑这是因为我的数据中确实有一些相同的值，但使用列 ID 应该不是问题，我不太明白我能做些什么来解决它.

foo 示例：

library(tidyverse)
Id <- rep(1:10)
a = c(5,10,15,12,13,25,12,13,11,9)
b = c(8,14,20,13,19,29,15,19,20,11)
c = c(10,14,20,1.5,9,21,13,21,11,10)
d = c(15,9,20,14,12,5,12,13,12,30)
group = as.factor( rep(1:2,each=5) )
data = data.frame(Id,a,b,c,d,group)

case_mapping <- data.frame(
  key  = c("a", "b", "c", "d"),
  key2 = c("x1", "x2", "y1", "y2")
)

data %>%
  gather(key, val, c(a:d)) %>%
  left_join(case_mapping, by = "key") %>%
  select(-key) %>% 
  extract(key2, into = c("key", "order"), "([a-z])([0-9])") %>%
  spread(key, val) %>%
  ggplot() +
  aes(x, y, group = Id, color = group) +  xlab("Age")+ #ggtitle(paste("Variable")+
  geom_point() +
  geom_line()

现在是我的数据示例。

library(tidyverse)
Id <- rep(1:10)
var1_x = c(5,10,15,12,13,25,12,13,11,9)
var2_x = c(8,14,20,13,19,29,NA,19,20,11) # just adding some nas.
var3_x = c(10,14,20,1.5,9,21,13,21,11,10)
var1_y = var1_x+3
var2_y = var2_x*2
var3_y = c(10,14,20,1.5,9,21,13,21,11,10) #same, just to see.
age1 = c(15,9,20,14,12,5,12,13,12,30)
age2 = c(18,19,24,16,15,9,16,19,14,37)
group = as.factor( rep(1:2,each=5) )
data = data.frame(Id,var1_x,var2_x,var3_x, var1_y,var2_y,var3_y,age1,age2,group)

现在，我是否应该创建一个 for 循环，以便我可以正确配对变量。首先我们创建两个字符串，colnames _x 和 _y

sub_x = colnames(data)[2:4] # sub x
sub_y = colnames(data)[5:7] # suby

现在我们应该可以实现 for 循环了。

for( i in 1:length(sub_x)) {
    
    # We define the matching keys.
    
    case_mapping <- data.frame(
    key  = c(sub_x[i],sub_y[i], "age1", "age2"),
    key2 = c("x1", "x2", "y1", "y2")
)
    
    # And now we should be able to plot this.
        
    data  %>% 
  gather(key, val, c(!!sym(sub_x[i]),    !!sym(sub_y[i]), age1,age2                   )) %>% 
  left_join(case_mapping, by = "key")  %>%
  select(-key) %>%
  extract(key2, into = c("key", "order"), "([a-z])([0-9])") %>%
  spread(key, val) %>%
  ggplot() +
  aes(x, y, group = Id, color = group) +
       xlab("Age")+
  geom_point() +
  geom_line()

}

但这并没有给我任何结果，当我尝试调整它时，它会因收集而抛出错误。希望大家多多指教，明白我做错了什么。

PD：抱歉，如果我的语法不完全正确，但英语是我的第二语言。

编辑澄清：

我打算为每个变量绘制类似的东西 - 如果有一种方法可以指示每个斜率的 ID，那将非常好，这样我就不必从数据中查找它来查看他们对应的）

编辑 2 在 Tjebo 的帮助下，我有点“解决了它”，但我仍然需要通过 dplyr 自动化从提供的 data_long1 构建这个 data_long2。

data_long2 <- data.frame( Id = rep(data_long$Id,2), Group = rep(data_long$group,2), Var= rep(data_long$var,2)  , Valueage= c(data_long$age1,data_long$age2), Valuevar= c(data_long$x,data_long$y)                     )

ggplot(data_long2) +
  ## I've removed the grouping by ID, because there was only one observation per ID
  aes(Valueage, Valuevar, color=Id) +  
  geom_point() +
  geom_line(aes(group = Id))+
  # geom_line() +
  ## you can for example facet by your new variable column
  facet_grid(~Var)
#> Warning: Removed 1 rows containing missing values (geom_point).

并更改颜色组

Answer 1

我认为你可能把事情复杂化了。据我了解，您很难重塑数据然后绘制所有变量，对吗？

下面是一种使用 new-ish pivot_longer 进行重塑的方法（它具有惊人的功能，尤其是在“多次聚会”方面），然后分面而不是循环。

更新

你基本上需要旋转更长的时间两次

library(tidyverse)

Id <- rep(1:10)
var1_x = c(5,10,15,12,13,25,12,13,11,9)
var2_x = c(8,14,20,13,19,29,NA,19,20,11) # just adding some nas.
var3_x = c(10,14,20,1.5,9,21,13,21,11,10)
var1_y = var1_x+3
var2_y = var2_x*2
var3_y = c(10,14,20,1.5,9,21,13,21,11,10) #same, just to see.
age1 = c(15,9,20,14,12,5,12,13,12,30)
age2 = c(18,19,24,16,15,9,16,19,14,37)
group = as.factor( rep(1:2,each=5) )
data = data.frame(Id,var1_x,var2_x,var3_x, var1_y,var2_y,var3_y,age1,age2,group)

data_long <-
  data %>%
  ## make use of the cool pivot_longer function
  pivot_longer(cols = matches("_[x|y]"), 
               names_to = c("var", ".value"), 
               names_pattern = "(.*)_(.*)") %>%
  ## now make even longer! all y (currently confusingly called x and y) belong into one column 
  ## and all x (currently called age1 and age2) in another column 
  ## this is easier with a similar pattern in both, therefore renaming
  ## note the .value name is switched when compared with the first pivoting
  rename(y1= x, y2 = y) %>%
  pivot_longer(
    matches(".*([1-2])"),
    names_to = c(".value", "set"),
    names_pattern = "(.+)([0-9+])"
  )

ggplot(data_long) +
  ## I've removed the grouping by ID, because there was only one observation per ID
  aes(age, y, color = as.character(Id)) +  
  geom_point() +
  geom_line() +
  ## you can for example facet by your new variable column
  facet_grid(~var)
#> Warning: Removed 2 rows containing missing values (geom_point).

在循环中分别创建每个绘图：

## split by your new variable and run a loop to create a list of plots 
ls_p <- lapply(split(data_long, data_long$var), function(.x){
  ggplot(.x) +
    ## I've removed the grouping by ID, because there was only one observation per ID
    aes(age, y, color = as.character(Id)) +  
    geom_point() +
    geom_line() +
    ## you can for example facet by your new variable column
    facet_grid(~var)
} )

## you can then either print them separately or all together, e.g. with patchwork
patchwork::wrap_plots(ls_p) + patchwork::plot_layout(ncol = 1)
#> Warning: Removed 2 rows containing missing values (geom_point).
#> Warning: Removed 2 row(s) containing missing values (geom_path).

^{由 reprex package (v2.0.1)}

于 2022-05-31 创建

为多个变量绘制多个斜率图（使用 for 循环）（面临关键冗余问题）

Plotting multiple slope plot's for multiple variables (With a for loop) (Facing issues with key redundancy)

r

ggplot2

dplyr

tidyverse