用户定义的功能基于按类别分组的多个列

Question

This may be basic, but I've been trying to figure it out for days and haven't found an answer.

I am trying to calculate a new quantity based on two columns 'concentration' and 'area' grouped by 'catchment'. I've written a function to calculate the difference in concentration for each row and the row with the largest area normalized by proportion of area in that catchment, but it won't work with dplyr or aggregate (. It works fine with by, but then returns a list.

Ideally, I want to add a column onto the dataframe or replace the concentration column altogether. Here is the dataframe 'lev':

  area catchment concentration
1    1       Yup       2.00000
2   10       Yup      40.50000
3   25       Yup      50.82031
4   35       Yup      50.00000
5    1      Nope       1.00000
6   10      Nope       5.00000
7   25      Nope      40.08333
8   35      Nope      38.00000

Here is the function:

lever <- function(data=lev, x=data[,"concentration"], y=data[,"area"]){
N= which.max(y) 
L = (x - x[N]) * y/max(y)
return(L)}

And here is the desired result:

   area catchment concentration   leverage
1    1       Yup       2.00000 -1.3714286
2   10       Yup      40.50000 -2.7142857
3   25       Yup      50.82031  0.5859375
4   35       Yup      50.00000  0.0000000
5    1      Nope       1.00000 -1.0571429
6   10      Nope       5.00000 -9.4285714
7   25      Nope      40.08333  1.4880952
8   35      Nope      38.00000  0.0000000 

Using by , I can get two lists with the results for each catchment:

by(lev, lev$catchment, lever)

but I want to use the function on multiple columns categorized by several factors (eg date in addition to catchment) and I get

'incorrect number of dimensions'

errors with doBy and dplyr .

Answer 1

We can use tidyverse

library(tidyverse)
df1 %>% 
  group_by(catchment) %>%
  mutate(leverage = (concentration- concentration[which.max(area)]) * area/max(area))

---

Based on the description, if there are multiple columns as grouping variable, place those in the group_by , and the calculation can also be applied to multiple columns with mutate_each

Answer 2

Loading your data:

lev <- read.table(text = "area catchment concentration
    1       Yup       2.00000
   10       Yup      40.50000
   25       Yup      50.82031
   35       Yup      50.00000
    1      Nope       1.00000
   10      Nope       5.00000
   25      Nope      40.08333
   35      Nope      38.00000", 
   header=TRUE)

Grouped by catchment

library(dplyr)
lev %>% 
    group_by(catchment) %>% 
    mutate(N = which.max(area),
           L = (concentration - concentration[N]) * area/max(area))

# 
#    area catchment concentration     N          L
#   <int>    <fctr>         <dbl> <int>      <dbl>
# 1     1       Yup       2.00000     4 -1.3714286
# 2    10       Yup      40.50000     4 -2.7142857
# 3    25       Yup      50.82031     4  0.5859357
# 4    35       Yup      50.00000     4  0.0000000
# 5     1      Nope       1.00000     4 -1.0571429
# 6    10      Nope       5.00000     4 -9.4285714
# 7    25      Nope      40.08333     4  1.4880929
# 8    35      Nope      38.00000     4  0.0000000

Using your function

I modify your function so that it returns a data frame.

lever2 <- function(data, 
                   x = data[,"concentration"][[1]], 
                   y = data[,"area"][[1]]){
    # Use [[1]] to extract the vector only
    N <- which.max(y)
    L <- (x - x[N]) * y/max(y)
    # Put L back into the data frame 
    # so that we keep the concentration and area in the result
    data$L <- L
    return(data)
    }

The funtion can then be used with dplyr::group_by %>% do

lev %>% 
    group_by(catchment) %>% 
    do( lever2(.))

Answer 3

You can also use data.table to calculate this value:

library(data.table)
# convert to data.table
setDT(df)

df[, leverage := (concentration - concentration[which.max(area)]) * (area / max(area)),
   by=catchment]
df
   area catchment concentration   leverage
1:    1       Yup       2.00000 -1.3714286
2:   10       Yup      40.50000 -2.7142857
3:   25       Yup      50.82031  0.5859357
4:   35       Yup      50.00000  0.0000000
5:    1      Nope       1.00000 -1.0571429
6:   10      Nope       5.00000 -9.4285714
7:   25      Nope      40.08333  1.4880929
8:   35      Nope      38.00000  0.0000000

data

df <-
structure(list(area = c(1L, 10L, 25L, 35L, 1L, 10L, 25L, 35L), 
    catchment = structure(c(2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L), .Label = c("Nope", 
    "Yup"), class = "factor"), concentration = c(2, 40.5, 50.82031, 
    50, 1, 5, 40.08333, 38)), .Names = c("area", "catchment", 
"concentration"), class = "data.frame", row.names = c("1", "2", 
"3", "4", "5", "6", "7", "8"))