按组提取对应于变量最小值的行

Posted 2021-03-26

我希望（1）通过一个变量（State）对数据进行分组，（2）在每个组中找到另一个变量的最小值行（Employees），以及（3）提取整行。

（1）和（2）是简单的单行，我觉得（3）也应该是，但我不能得到它。

这是一个示例数据集：

> data
  State Company Employees
1    AK       A        82
2    AK       B       104
3    AK       C        37
4    AK       D        24
5    RI       E        19
6    RI       F       118
7    RI       G        88
8    RI       H        42

data <- structure(list(State = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 
        2L), .Label = c("AK", "RI"), class = "factor"), Company = structure(1:8, .Label = c("A", 
        "B", "C", "D", "E", "F", "G", "H"), class = "factor"), Employees = c(82L, 
        104L, 37L, 24L, 19L, 118L, 88L, 42L)), .Names = c("State", "Company", 
        "Employees"), class = "data.frame", row.names = c(NA, -8L))

使用min按组计算aggregate很容易：

> aggregate(Employees ~ State, data, function(x) min(x))
  State Employees
1    AK        24
2    RI        19

...或data.table：

> library(data.table)
> DT <- data.table(data)
> DT[ , list(Employees = min(Employees)), by = State]
   State Employees
1:    AK        24
2:    RI        19

但是如何提取对应于这些min值的整行，即在结果中还包括Company？

答案

稍微优雅一点：

library(data.table)
DT[ , .SD[which.min(Employees)], by = State]

   State Company Employees
1:    AK       D        24
2:    RI       E        19

---

略微不如使用.SD优雅，但速度更快（对于有许多组的数据）：

DT[DT[ , .I[which.min(Employees)], by = State]$V1]

另外，如果您的数据集具有多个相同的最小值并且您想要将它们全部分组，则只需将表达式which.min(Employees)替换为Employees == min(Employees)。

另见Subset by group with data.table。

另一答案

一个dplyr解决方案：

library(dplyr)    
data %>% 
    group_by(State) %>% 
    slice(which.min(Employees))

另一答案

由于这是谷歌的热门话题，我想我会添加一些我觉得有用的其他选项。这个想法基本上是由Employees安排一次，然后根据State采取独特的

要么使用data.table

library(data.table)
unique(setDT(data)[order(Employees)], by = "State")
#    State Company Employees
# 1:    RI       E        19
# 2:    AK       D        24

---

或者，我们也可以先订购然后再订购.SD。这两个操作都在resent data.table版本和order优化，似乎触发data.table:::forderv，而.SD[1L]触发Gforce

setDT(data)[order(Employees), .SD[1L], by = State, verbose = TRUE] # <- Added verbose
# order optimisation is on, i changed from 'order(...)' to 'forder(DT, ...)'.
# i clause present and columns used in by detected, only these subset: State 
# Finding groups using forderv ... 0 sec
# Finding group sizes from the positions (can be avoided to save RAM) ... 0 sec
# Getting back original order ... 0 sec
# lapply optimization changed j from '.SD[1L]' to 'list(Company[1L], Employees[1L])'
# GForce optimized j to 'list(`g[`(Company, 1L), `g[`(Employees, 1L))'
# Making each group and running j (GForce TRUE) ... 0 secs
#    State Company Employees
# 1:    RI       E        19
# 2:    AK       D        24

---

或者dplyr

library(dplyr)
data %>% 
  arrange(Employees) %>% 
  distinct(State, .keep_all = TRUE)
#   State Company Employees
# 1    RI       E        19
# 2    AK       D        24

---

从@Khashaas借来的另一个有趣的想法很棒的答案（为了处理多个匹配而对mult = "first"进行了一些小修改）是首先找到每组的最小值，然后执行二进制连接。这样做的好处是可以利用data.tables gmin函数（跳过评估开销）和二进制连接功能

tmp <- setDT(data)[, .(Employees = min(Employees)), by = State]
data[tmp, on = .(State, Employees), mult = "first"]
#    State Company Employees
# 1:    AK       D        24
# 2:    RI       E        19

---

一些基准

library(data.table)
library(dplyr)
library(plyr)
library(stringi)
library(microbenchmark)

set.seed(123)
N <- 1e6
data <- data.frame(State = stri_rand_strings(N, 2, '[A-Z]'),
                   Employees = sample(N*10, N, replace = TRUE))
DT <- copy(data)
setDT(DT)
DT2 <- copy(DT)
str(DT)
str(DT2)

microbenchmark("(data.table) .SD[which.min]: " = DT[ , .SD[which.min(Employees)], by = State],
               "(data.table) .I[which.min]: " = DT[DT[ , .I[which.min(Employees)], by = State]$V1],
               "(data.table) order/unique: " = unique(DT[order(Employees)], by = "State"),
               "(data.table) order/.SD[1L]: " = DT[order(Employees), .SD[1L], by = State],
               "(data.table) self join (on):" = {
                 tmp <- DT[, .(Employees = min(Employees)), by = State]
                 DT[tmp, on = .(State, Employees), mult = "first"]},
               "(data.table) self join (setkey):" = {
                 tmp <- DT2[, .(Employees = min(Employees)), by = State] 
                 setkey(tmp, State, Employees)
                 setkey(DT2, State, Employees)
                 DT2[tmp, mult = "first"]},
               "(dplyr) slice(which.min): " = data %>% group_by(State) %>% slice(which.min(Employees)),
               "(dplyr) arrange/distinct: " = data %>% arrange(Employees) %>% distinct(State, .keep_all = TRUE),
               "(dplyr) arrange/group_by/slice: " = data %>% arrange(Employees) %>% group_by(State) %>% slice(1),
               "(plyr) ddply/which.min: " = ddply(data, .(State), function(x) x[which.min(x$Employees),]),
               "(base) by: " = do.call(rbind, by(data, data$State, function(x) x[which.min(x$Employees), ])))


# Unit: milliseconds
#                             expr        min         lq       mean     median         uq       max neval      cld
#    (data.table) .SD[which.min]:   119.66086  125.49202  145.57369  129.61172  152.02872  267.5713   100    d    
#     (data.table) .I[which.min]:    12.84948   13.66673   19.51432   13.97584   15.17900  109.5438   100 a       
#      (data.table) order/unique:    52.91915   54.63989   64.39212   59.15254   61.71133  177.1248   100  b      
#     (data.table) order/.SD[1L]:    51.41872   53.22794   58.17123   55.00228   59.00966  145.0341   100  b      
#     (data.table) self join (on):   44.37256   45.67364   50.32378   46.24578   50.69411  137.4724   100  b      
# (data.table) self join (setkey):   14.30543   15.28924   18.63739   15.58667   16.01017  106.0069   100 a       
#       (dplyr) slice(which.min):    82.60453   83.64146   94.06307   84.82078   90.09772  186.0848   100   c     
#       (dplyr) arrange/distinct:   344.81603  360.09167  385.52661  379.55676  395.29463  491.3893   100     e   
# (dplyr) arrange/group_by/slice:   367.95924  383.52719  414.99081  397.93646  425.92478  557.9553   100      f  
#         (plyr) ddply/which.min:   506.55354  530.22569  568.99493  552.65068  601.04582  727.9248   100       g 
#                      (base) by:  1220.38286 1291.70601 1340.56985 1344.86291 1382.38067 1512.5377   100        h

另一答案

基函数by通常用于处理data.frames中的块数据。例如

by(data, data$State, function(x) x[which.min(x$Employees), ] )

它确实返回列表中的数据，但您可以使用它来折叠它

do.call(rbind, by(data, data$State, function(x) x[which.min(x$Employees), ] ))

另一答案

更正了plyr解决方案：

ddply(df, .(State), function(x) x[which.min(x$Employees),])
#   State Company Employees
# 1    AK       D        24
# 2    RI       E        19

thanks to @joel.wilson