LucaS's questions

我已经在这个问题上纠结了好几个小时，这有点令人沮丧。基本上，我想安排一些数据，以便 NA 基于分组结构首先出现。我可以部分实现，但我尝试的任何方法都无法让我得到想要的结果。

使用此代码，

df <-  df |> 
  group_by(AESOC, AEPT) |> 
  arrange(!is.na(AEPT), !is.na(Severity), .by_group = TRUE)

我已经能够实现图中所示的效果。

但是我仍想进一步安排，以便第 9-12 行出现在第 1 行之前，第 25-28 行出现在第 13 行之前（即在 AESOC 和 AEPT 确定的组的最开始处）。

这些小数据包括在这里：

df <-  structure(list(AESOC = c("Blood and lymphatic system disorders", 
"Blood and lymphatic system disorders", "Blood and lymphatic system disorders", 
"Blood and lymphatic system disorders", "Blood and lymphatic system disorders", 
"Blood and lymphatic system disorders", "Blood and lymphatic system disorders", 
"Blood and lymphatic system disorders", "Blood and lymphatic system disorders", 
"Blood and lymphatic system disorders", "Blood and lymphatic system disorders", 
"Blood and lymphatic system disorders", "Cardiac disorders", 
"Cardiac disorders", "Cardiac disorders", "Cardiac disorders", 
"Cardiac disorders", "Cardiac disorders", "Cardiac disorders", 
"Cardiac disorders", "Cardiac disorders", "Cardiac disorders", 
"Cardiac disorders", "Cardiac disorders", "Cardiac disorders", 
"Cardiac disorders", "Cardiac disorders", "Cardiac disorders"
), AEPT = c("    Anaemia", "    Anaemia", "    Anaemia", "    Anaemia", 
"    Lymphopenia", "    Lymphopenia", "    Lymphopenia", "    Lymphopenia", 
NA, NA, NA, NA, "    Dizziness", "    Dizziness", "    Dizziness", 
"    Dizziness", "    Palpitations", "    Palpitations", "    Palpitations", 
"    Palpitations", "    Presyncope", "    Presyncope", "    Presyncope", 
"    Presyncope", NA, NA, NA, NA), Severity = c("        mild", 
"        moderate", "        severe", NA, "        mild", "        moderate", 
"        severe", NA, "    mild", "    moderate", "    severe", 
NA, "        mild", "        moderate", "        severe", NA, 
"        mild", "        moderate", "        severe", NA, "        moderate", 
"        mild", "        severe", NA, "    moderate", "    mild", 
"    severe", NA)), row.names = c(NA, -28L), class = c("tbl_df", 
"tbl", "data.frame"))

任何帮助都将不胜感激。

我正在尝试使用传递给 的函数重命名多个 df 的列purrr::map，但显然我做错了，因为我收到错误。有人能告诉我怎么做吗？

# Create new df rows
newrow1 <-  data.frame(t(c("a", rep("", 6))))
newrow2 <-  data.frame(t(c("b", rep("", 6))))
newrow3 <-  data.frame(t(c("c", rep("", 6))))
newrow4 <-  data.frame(t(c("d", rep("", 6))))
newrow5 <-  data.frame(t(c("e", rep("", 6))))
newrow6 <-  data.frame(t(c("f", rep("", 6))))
newrow7 <-  data.frame(t(c("g", rep("", 6))))
newrow8 <-  data.frame(t(c("h", rep("", 6))))
newrow9 <-  data.frame(t(c("i", rep("", 6))))
# Function to rename dfs
changenames <- function(x) {
  names(x) <- c("row","n_1","n_2","n_3","pct_1","pct_2","pct_3")
  return(x)
}
df_names <-  paste0("newrow", 1:9)
purrr::map(df_names, changenames)

我正在尝试学习一些基本的 purrr。以这个玩具示例为例 - 我如何将“rank”附加到使用 map_df 创建的当前变量名称中？

library(tidyverse)
# generate some random data
dat <- tibble(
  x_1 = runif(10),
  x_2 = runif(10),
  x_3 = runif(10)
)
# find the ranks
dat <-  cbind(dat, dat |> map_df(rank))

目前生产的产品如下：

> dat
          x_1        x_2       x_3 x_1 x_2 x_3
1  0.37593751 0.20597008 0.4826491   6   3   4
2  0.01805845 0.05432019 0.6732073   1   1   7
3  0.39243407 0.47535664 0.9920948   7   7  10
4  0.06361097 0.21624621 0.4309171   2   4   3
5  0.53391717 0.28778664 0.9323643   9   6   9
6  0.24612385 0.83270359 0.2719036   4  10   1
7  0.47170853 0.26261303 0.3418129   8   5   2
8  0.33821497 0.57270648 0.7606502   5   9   8
9  0.85717819 0.48097921 0.4848689  10   8   5
10 0.11503862 0.07063651 0.6391102   3   2   6

有人能帮我看看我哪里出错了吗？我认为在管道函数中使用花括号时，您可以调用变量而不必将其括在“”（即作为字符串）。那么我的函数哪里出错了：

library(tidyverse)
dat <-  structure(list(id = 1:6, 
                       var_1 = c("1,1,-11000", "1,1,1", "0,0,0", "1,1,0", "1,1,1", "1,1,1"), 
                       var_2 = c("0,0,-13000", "0,0,0", "-13000,-13000,-13000", "6,4,-13000", "0,0,0", "0,0,0"), 
                       var_3 = c("24,7,-13000", "0,0,0", "-13000,-13000,-13000", "0,0,-13000", "0,0,0", "0,0,0")), 
                  row.names = 1:6, class = "data.frame")

# Separate to wide and convert to long in one step
split_to_long <-  function(col){
  i <-  substr({{col}}, 5, 5)
  temp <-  dat |>
    select("id", {{col}}) |>
    separate_wider_delim({{col}}, ",", too_few = "align_start",
                         names = c(paste0({{col}},"_1"),
                                   paste0({{col}},"_2"),
                                   paste0({{col}},"_3"))) |>
    pivot_longer(2:4,
                 names_to = "visit",
                 values_to = paste0("var_", i),
                 names_prefix = paste0("var_", i, "_"))
  temp
}

# This works
split_to_long("var_1")

# But this doesn't
split_to_long(var_1)

这来自我发布的另一个问题： 如何用 purrr::map 替换这个 for 循环

我有一个相当简单的函数，其中包含一个 for 循环，将特定编码的缺失值转换为 NA：

make_na <- function(value, df){
  for(thisvar in names(df)){
    df[[thisvar]][df[[thisvar]] == value] <- NA
    num_NA <- length(df[[thisvar]][is.na(df[[thisvar]])])
    print(paste0(thisvar, " has ", num_NA, " NA's"))
  }  
  return(df)
} 

运行该函数可以正确打印修改后的数据框，但我不知道如何将数据框放回工作区。我也尝试new_dat <- return(df)将数据框分配给一个对象，但那也不起作用。显然我遗漏了一些简单的东西。