更改R中两个不同列值之间的名称位置

Question

我有一个data.frame，它显示水质与土地利用参数之间的相关性。在这个data.frame中，您有列xName、yName、corr和p.value。然而，在列xName中，yName并不是一个单独的专栏，仅仅是为了质量和单独使用。

例如，我只想要xName栏中的水质数据，yName栏中只有土地利用数据，因为顺序不会改变corr和p.value的值。

最初，我考虑用mutate创建新列，其中显示xName和yName是否是土地使用和水质，然后使用case_when和recode转换可能的名称。

但是我无法继续前进，因为它返回一个错误并且不更改名称。

df=read.table(text="xName   yName   corr    p.value classe_tipox2   classe_tipoy2
IQA_mean    soil    0.639727697 0.025073852 water   land
OD_mean veg 0.60989011  0.03029001  water   land
OD_mean grass   -0.576923077    0.042537186 water   land
soil    N_Total_mean    0.604577823 0.037305053 land    water
crop    N_Total_mean    0.695600561 0.012007646 land    water
crop    P_Total_mean    -0.624544589    0.029931227 land    water
DBO_mean    veg 0.797202797 0.003161252 water   land
DBO_mean    city    0.756757445 0.004382975 water   land
DBO_mean    grass   -0.825174825    0.001718596 water   land
veg P_Total_mean    -0.587412587    0.048844858 land    water
P_Total_mean    grass   0.629370629 0.03239475  water   land", sep="", header=TRUE)%>%
   
  #change names positions
    mutate(xName=case_when(
    classe_tipox2=='land' & classe_tipoy2=='water' ~ recode(xName=yName )
  ))

我希望当xName是土地使用类型时，xName被重命名为yName。

Answer 1

另一种选择(虽然有点冗长)。首先，我将数据放入长格式，然后使用ifelse语句“修复”xName和yName，然后将数据转回宽格式。

library(tidyverse)

df %>%
  pivot_longer(c(xName, yName)) %>%
  mutate(
    name = ifelse(
      name == "xName" &
        classe_tipox2 == 'land',
      "yName",
      ifelse(
        name == "yName" &
          classe_tipox2 == 'land',
        "xName",
        name
      )
    )
  ) %>%
  pivot_wider(names_from = name, values_from = value) %>% 
  select(5:6, 1:4) %>% 
  mutate(classe_tipox2 = "water", classe_tipoy2 = "land")

输出

# A tibble: 11 × 6
   xName        yName   corr p.value classe_tipox2 classe_tipoy2
   <chr>        <chr>  <dbl>   <dbl> <chr>         <chr>        
 1 IQA_mean     soil   0.640 0.0251  water         land         
 2 OD_mean      veg    0.610 0.0303  water         land         
 3 OD_mean      grass -0.577 0.0425  water         land         
 4 N_Total_mean soil   0.605 0.0373  water         land         
 5 N_Total_mean crop   0.696 0.0120  water         land         
 6 P_Total_mean crop  -0.625 0.0299  water         land         
 7 DBO_mean     veg    0.797 0.00316 water         land         
 8 DBO_mean     city   0.757 0.00438 water         land         
 9 DBO_mean     grass -0.825 0.00172 water         land         
10 P_Total_mean veg   -0.587 0.0488  water         land         
11 P_Total_mean grass  0.629 0.0324  water         land  

Answer 2

一种基R方法

# get the conditional logic
chose <- df$classe_tipox2 == "land"

# then select the columns and bind with the rest of the data
cbind(setNames(data.frame(t(sapply(seq_along(chose), function(x) 
  'if'(chose[x], c(df$yName[x],df$xName[x]), c(df$xName[x],df$yName[x]))))), 
  colnames(df[,1:2])), df[,-c(1,2)])

          xName yName       corr     p.value classe_tipox2 classe_tipoy2
1      IQA_mean  soil  0.6397277 0.025073852         water          land
2       OD_mean   veg  0.6098901 0.030290010         water          land
3       OD_mean grass -0.5769231 0.042537186         water          land
4  N_Total_mean  soil  0.6045778 0.037305053          land         water
5  N_Total_mean  crop  0.6956006 0.012007646          land         water
6  P_Total_mean  crop -0.6245446 0.029931227          land         water
7      DBO_mean   veg  0.7972028 0.003161252         water          land
8      DBO_mean  city  0.7567574 0.004382975         water          land
9      DBO_mean grass -0.8251748 0.001718596         water          land
10 P_Total_mean   veg -0.5874126 0.048844858          land         water
11 P_Total_mean grass  0.6293706 0.032394750         water          land