dplyr包可以用于条件变异吗？

Question

当突变是有条件的(取决于某些列值的值)时,是否可以使用mutate？

这个例子有助于显示我的意思.

structure(list(a = c(1, 3, 4, 6, 3, 2, 5, 1), b = c(1, 3, 4, 
2, 6, 7, 2, 6), c = c(6, 3, 6, 5, 3, 6, 5, 3), d = c(6, 2, 4, 
5, 3, 7, 2, 6), e = c(1, 2, 4, 5, 6, 7, 6, 3), f = c(2, 3, 4, 
2, 2, 7, 5, 2)), .Names = c("a", "b", "c", "d", "e", "f"), row.names = c(NA, 
8L), class = "data.frame")

  a b c d e f
1 1 1 6 6 1 2
2 3 3 3 2 2 3
3 4 4 6 4 4 4
4 6 2 5 5 5 2
5 3 6 3 3 6 2
6 2 7 6 7 7 7
7 5 2 5 2 6 5
8 1 6 3 6 3 2

我希望使用dplyr包找到解决我的问题的方法(是的,我知道这不是应该有效的代码,但我想它的目的很明确)用于创建新列g:

 library(dplyr)
 df <- mutate(df,
         if (a == 2 | a == 5 | a == 7 | (a == 1 & b == 4)){g = 2},
         if (a == 0 | a == 1 | a == 4 | a == 3 |  c == 4) {g = 3})

我正在寻找的代码的结果应该在这个特定的例子中有这个结果:

  a b c d e f  g
1 1 1 6 6 1 2  3
2 3 3 3 2 2 3  3
3 4 4 6 4 4 4  3
4 6 2 5 5 5 2 NA
5 3 6 3 3 6 2 NA
6 2 7 6 7 7 7  2
7 5 2 5 2 6 5  2
8 1 6 3 6 3 2  3

有没有人知道如何在dplyr中这样做？这个数据框只是一个例子,我正在处理的数据框要大得多.由于它的速度,我试图使用dplyr,但也许还有其他更好的方法来处理这个问题？

Answer 1

使用 ifelse

df %>%
  mutate(g = ifelse(a == 2 | a == 5 | a == 7 | (a == 1 & b == 4), 2,
               ifelse(a == 0 | a == 1 | a == 4 | a == 3 |  c == 4, 3, NA)))

添加 - if_else:注意,在dplyr 0.5中有一个if_else定义的函数,所以替代方法是替换ifelse为if_else; 但请注意,因为if_else比ifelse条件更严格(条件的两条腿必须具有相同的类型)所以NA在这种情况下必须更换NA_real_.

df %>%
  mutate(g = if_else(a == 2 | a == 5 | a == 7 | (a == 1 & b == 4), 2,
               if_else(a == 0 | a == 1 | a == 4 | a == 3 |  c == 4, 3, NA_real_)))

添加 - case_when自发布此问题后,dplyr已添加,case_when因此另一个替代方案是:

df %>% mutate(g = case_when(a == 2 | a == 5 | a == 7 | (a == 1 & b == 4) ~ 2,
                            a == 0 | a == 1 | a == 4 | a == 3 |  c == 4 ~ 3,
                            TRUE ~ NA_real_))

Answer 2

既然你要求其他更好的方法来处理这个问题,这里有另一种方法data.table:

require(data.table) ## 1.9.2+
setDT(df)
df[a %in% c(0,1,3,4) | c == 4, g := 3L]
df[a %in% c(2,5,7) | (a==1 & b==4), g := 2L]

请注意,条件语句的顺序是相反的,以便g正确获取.g即使在第二次任务期间也没有制作副本- 它已就地替换.

对于较大的数据,这将比使用嵌套 更好的性能if-else,因为它可以评估"是"和"否"的情况,并且嵌套可能更难以读取/维护恕我直言.

这是相对较大数据的基准:

# R version 3.1.0
require(data.table) ## 1.9.2
require(dplyr)
DT <- setDT(lapply(1:6, function(x) sample(7, 1e7, TRUE)))
setnames(DT, letters[1:6])
# > dim(DT) 
# [1] 10000000        6
DF <- as.data.frame(DT)

DT_fun <- function(DT) {
    DT[(a %in% c(0,1,3,4) | c == 4), g := 3L]
    DT[a %in% c(2,5,7) | (a==1 & b==4), g := 2L]
}

DPLYR_fun <- function(DF) {
    mutate(DF, g = ifelse(a %in% c(2,5,7) | (a==1 & b==4), 2L, 
            ifelse(a %in% c(0,1,3,4) | c==4, 3L, NA_integer_)))
}

BASE_fun <- function(DF) { # R v3.1.0
    transform(DF, g = ifelse(a %in% c(2,5,7) | (a==1 & b==4), 2L, 
            ifelse(a %in% c(0,1,3,4) | c==4, 3L, NA_integer_)))
}

system.time(ans1 <- DT_fun(DT))
#   user  system elapsed 
#  2.659   0.420   3.107 

system.time(ans2 <- DPLYR_fun(DF))
#   user  system elapsed 
# 11.822   1.075  12.976 

system.time(ans3 <- BASE_fun(DF))
#   user  system elapsed 
# 11.676   1.530  13.319 

identical(as.data.frame(ans1), as.data.frame(ans2))
# [1] TRUE

identical(as.data.frame(ans1), as.data.frame(ans3))
# [1] TRUE

不确定这是否是您要求的替代品,但我希望它有所帮助.

Answer 3

dplyr现在有一个case_when提供矢量化if 的函数.语法有点奇怪,mosaic:::derivedFactor因为你不能以标准的dplyr方式访问变量,并且需要声明NA的模式,但它要快得多mosaic:::derivedFactor.

df %>%
mutate(g = case_when(a %in% c(2,5,7) | (a==1 & b==4) ~ 2L, 
                     a %in% c(0,1,3,4) | c == 4 ~ 3L, 
                     TRUE~as.integer(NA)))

编辑:如果你在dplyr::case_when()包的0.7.0版本之前使用,那么你需要在变量名前加上' .$'(例如.$a == 1在里面写case_when).

基准:对于基准(重用Arun的帖子中的功能)和减少样本量:

require(data.table) 
require(mosaic) 
require(dplyr)
require(microbenchmark)

DT <- setDT(lapply(1:6, function(x) sample(7, 10000, TRUE)))
setnames(DT, letters[1:6])
DF <- as.data.frame(DT)

DPLYR_case_when <- function(DF) {
  DF %>%
  mutate(g = case_when(a %in% c(2,5,7) | (a==1 & b==4) ~ 2L, 
                       a %in% c(0,1,3,4) | c==4 ~ 3L, 
                       TRUE~as.integer(NA)))
}

DT_fun <- function(DT) {
  DT[(a %in% c(0,1,3,4) | c == 4), g := 3L]
  DT[a %in% c(2,5,7) | (a==1 & b==4), g := 2L]
}

DPLYR_fun <- function(DF) {
  mutate(DF, g = ifelse(a %in% c(2,5,7) | (a==1 & b==4), 2L, 
                    ifelse(a %in% c(0,1,3,4) | c==4, 3L, NA_integer_)))
}

mosa_fun <- function(DF) {
  mutate(DF, g = derivedFactor(
    "2" = (a == 2 | a == 5 | a == 7 | (a == 1 & b == 4)),
    "3" = (a == 0 | a == 1 | a == 4 | a == 3 |  c == 4),
    .method = "first",
    .default = NA
  ))
}

microbenchmark(
  DT_fun(DT),
  DPLYR_fun(DF),
  DPLYR_case_when(DF),
  mosa_fun(DF),
  times=20
)

这给出了:

            expr        min         lq       mean     median         uq        max neval
         DT_fun(DT)   1.503589   1.626971   2.054825   1.755860   2.292157   3.426192    20
      DPLYR_fun(DF)   2.420798   2.596476   3.617092   3.484567   4.184260   6.235367    20
DPLYR_case_when(DF)   2.153481   2.252134   6.124249   2.365763   3.119575  72.344114    20
       mosa_fun(DF) 396.344113 407.649356 413.743179 412.412634 416.515742 459.974969    20

Answer 4

包中的derivedFactor函数mosaic似乎旨在处理这个问题.使用此示例,它看起来像:

library(dplyr)
library(mosaic)
df <- mutate(df, g = derivedFactor(
     "2" = (a == 2 | a == 5 | a == 7 | (a == 1 & b == 4)),
     "3" = (a == 0 | a == 1 | a == 4 | a == 3 |  c == 4),
     .method = "first",
     .default = NA
     ))

(如果您希望结果是数字而不是因子,则可以包含derivedFactor在as.numeric呼叫中.)

derivedFactor 也可以用于任意数量的条件.

Answer 5

case_when 在以下情况下,现在是SQL风格案例的一个非常干净的实现:

structure(list(a = c(1, 3, 4, 6, 3, 2, 5, 1), b = c(1, 3, 4, 
2, 6, 7, 2, 6), c = c(6, 3, 6, 5, 3, 6, 5, 3), d = c(6, 2, 4, 
5, 3, 7, 2, 6), e = c(1, 2, 4, 5, 6, 7, 6, 3), f = c(2, 3, 4, 
2, 2, 7, 5, 2)), .Names = c("a", "b", "c", "d", "e", "f"), row.names = c(NA, 
8L), class = "data.frame") -> df


df %>% 
    mutate( g = case_when(
                a == 2 | a == 5 | a == 7 | (a == 1 & b == 4 )     ~   2,
                a == 0 | a == 1 | a == 4 |  a == 3 | c == 4       ~   3
))

使用dplyr 0.7.4

手册:http://dplyr.tidyverse.org/reference/case_when.html