按组提取与变量最小值对应的行

73

我希望（1）将数据按一个变量（）分组State，（2）在每个组中找到另一个变量（Employees）的最小值行，以及（3）提取整行。

（1）和（2）是简单的一线客，我觉得（3）也是，但我无法理解。

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))

min按组计算很容易，使用aggregate：

> 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，如何提取与这些min值相对应的整行，也就是包括Company在结果中？

r  dplyr  data.table  aggregate 
— 埃德·斯温德勒斯
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Answers:

64

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

略低于使用.SD，但略快（对于具有多个组的数据）：

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

此外，只需更换的表达which.min(Employees)与Employees == min(Employees)，如果你的数据集有多个相同的最小值和你想子集所有的人。

另请参见使用data.table按组对应于最大值的子集行。

— 塞诺尔
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如果该数字应返回最小值，但基于使用该数字的频率，该如何获得呢？
— Abhishek Singh
57

这里是一个dplyr解决方案（请注意，我不是普通用户）：

library(dplyr)    
data %>% 
    group_by(State) %>% 
    slice(which.min(Employees))
— 研究
source
1
使用dplyr 1.0.0，可以通过以下方式实现相同的功能：group_by(data, State) %>% slice_min(order_by = Employees)。
— jazzurro '20年
32

由于这是Google的热门产品，我想我会添加一些我认为有用的附加选项。这个想法基本上是安排一次Employees，然后按照State

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，我们也可以先排序然后子集.SD。这两个操作都在resent data.table版本中进行了优化，并且order似乎是触发器data.table:::forderv，而.SD[1L]触发器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

要么 dplyr

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

从@Khashaas很棒的答案中借用的另一个有趣的想法（对进行了微小的修改mult = "first"以处理多个匹配项）是首先找到每个组的最小值，然后执行二进制联接。这样做的优点是利用了data.tablesgmin函数（跳过了评估开销）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
— 大卫·阿伦堡
source
做得很好。出于完整性/好奇心，我提供了正确的plyr解决方案。随时将其包含在基准测试中……我不希望它能够承受data.table。
— C8H10N4O2
1
@ C8H10N4O2更新。
— David Arenburg
8

基本功能by通常对于处理data.frames中的块数据很有用。例如

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

它确实返回列表中的数据，但是您可以使用以下方法将其折叠 

do.call(rbind, by(data, data$State, function(x) x[which.min(x$Employees), ] ))
— 弗里克先生
source
3

更正的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

感谢@ joel.wilson

— C8H10N4O2
source
1

在base中，您可以ave用来获取min每个组并将其与进行比较，Employees并获得一个逻辑向量来将子集化data.frame。

data[data$Employees == ave(data$Employees, data$State, FUN=min),]
#  State Company Employees
#4    AK       D        24
#5    RI       E        19

或在功能中进行比较。

data[as.logical(ave(data$Employees, data$State, FUN=function(x) x==min(x))),]
#data[ave(data$Employees, data$State, FUN=function(x) x==min(x))==1,] #Variant
#  State Company Employees
#4    AK       D        24
#5    RI       E        19
— K
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