使用R中的最小类来估算缺失值

Question

我是R的新手，需要帮助在我目前正在处理的数据集中的一列中输入缺失值。下图显示了我想要与少数列一起输入的缺失值。

我想使用之前的条目为客户填写最小数量的值，因为我认为这最适合我的情况和数据。例如，在图像中，我应该能够用1（最小值为1,5,2）填充缺失值。

在我的搜索过程中，我主要遇到的方法是使用给定类的均值，而不是最小值或最大值。

任何帮助或指示都会非常感激。

编辑：这是dput的输出。

structure(list(YEAR = c(2011L, 2012L, 2014L, 2015L, 2011L, 2012L
), CustomerId = c("00000063", "00000063", "00000063", "00000063", 
"00000065", "00000065"), MemberType = structure(c(2L, 2L, 2L, 
2L, 2L, 2L), .Label = c("GROUP", "INDIVIDUAL", "PARTNER"), class = "factor"), 
    MembershipTypeCode = structure(c(6L, 6L, 6L, 10L, 6L, 6L), .Label = c("EGROUP", 
    "EINDIV", "EINDIV2", "EPARTNER", "GROUP", "INDIV", "INDIV2", 
    "INDIV3", "PARTNER", "PLUS", "PLUS2", "PLUS20", "PLUS3", 
    "PLUSENTERPRI", "PLUSGROUP", "PLUSGROUP2", "PROF_ENTERPR", 
    "PROF_GROUP", "PROF_GROUP2", "PROF_INDIV", "PROF_INDIV2", 
    "PROF_INDIV3"), class = "factor"), MembershipPeriodBegin = structure(c(15279, 
    15677, 16071, 16436, 15006, 15371), class = "Date"), MembershipPeriodEnd = structure(c(15644, 
    16070, 16435, 16800, 15370, 15736), class = "Date"), ConsecutiveYearsAsMember = c(14L, 
    15L, 17L, 18L, 8L, 9L), AllocationUsage = c(0, 0, 0, 0, 0, 
    0), SetCOPPreference = structure(c(2L, 2L, 2L, 2L, 2L, 2L
    ), .Label = c("Y", "N"), class = "factor"), Purchase.Qty = c(2L, 
    5L, 1L, NA, 7L, 27L), Webcast.Registration = c(0L, 0L, 0L, 
    0L, 0L, 1L), Web.Visits = c(0L, 0L, 42L, 0L, 0L, 0L), Web.Page.Views = c(0L, 
    0L, 98L, 0L, 0L, 0L), Blog.Visits = c(0L, 0L, 3L, 0L, 0L, 
    0L), Blog.Page.Views = c(0L, 0L, 4L, 0L, 0L, 0L), Forum.Visits = c(0L, 
    0L, 45L, 0L, 0L, 0L), Forum.Page.Views = c(0L, 0L, 102L, 
    0L, 0L, 0L), ParatureTickets = c(0L, 0L, 0L, 0L, 0L, 0L), 
    ParatureChats = c(0L, 0L, 0L, 0L, 0L, 0L), Registered.for.Edu = c(0L, 
    0L, 0L, 0L, 0L, 0L), Attended.ICE = structure(c(2L, 2L, 2L, 
    2L, 2L, 2L), .Label = c("Y", "N"), class = "factor"), Attended.TK = structure(c(2L, 
    2L, 2L, 2L, 2L, 2L), .Label = c("Y", "N"), class = "factor"), 
    Frugal = structure(c(2L, 2L, 2L, 2L, 2L, 2L), .Label = c("Y", 
    "N"), class = "factor"), Chapter.Board = structure(c(2L, 
    2L, 2L, 2L, 2L, 2L), .Label = c("Y", "N"), class = "factor"), 
    Retained = structure(c(5L, 5L, 5L, 1L, 5L, 5L), .Label = c("Active", 
    "Awaiting Renewal", "Future Dated", "Lost", "Retained"), class = "factor"), 
    ProfileCompletion = c(60, 60, 60, 60, 60, 60), NumberofLogins = c(1L, 
    1L, 15L, 0L, 0L, 4L), Downloads = c(NA_integer_, NA_integer_, 
    NA_integer_, NA_integer_, NA_integer_, NA_integer_), ForumMember = structure(c(NA_integer_, 
    NA_integer_, NA_integer_, NA_integer_, NA_integer_, NA_integer_
    ), .Label = "N", class = "factor"), FreeUpgrade = structure(c(1L, 
    1L, 1L, 1L, 1L, 1L), .Label = c("Y", "N"), class = "factor")), .Names = c("YEAR", 
"CustomerId", "MemberType", "MembershipTypeCode", "MembershipPeriodBegin", 
"MembershipPeriodEnd", "ConsecutiveYearsAsMember", "AllocationUsage", 
"SetCOPPreference", "Purchase.Qty", "Webcast.Registration", "Web.Visits", 
"Web.Page.Views", "Blog.Visits", "Blog.Page.Views", "Forum.Visits", 
"Forum.Page.Views", "ParatureTickets", "ParatureChats", "Registered.for.Edu", 
"Attended.ICE", "Attended.TK", "Frugal", "Chapter.Board", "Retained", 
"ProfileCompletion", "NumberofLogins", "Downloads", "ForumMember", 
"FreeUpgrade"), row.names = c(NA, 6L), class = "data.frame")

谢谢，
PRATIK

Answer 1

我们可以将na.aggregate与FUN= min一起使用。我们将'data.frame'转换为'data.table'（setDT(df1)），按'CustomerID'分组，我们在'PurchaseQty'上应用na.aggregate并分配（:=）输出回到'PurchaseQty'。

library(data.table)
library(zoo)
setDT(df1)[, PurchaseQty := na.aggregate(PurchaseQty, FUN= min) , by = CustomerID]

数据

df1 <- data.frame(CustomerID= rep(1:2, each=4), PurchaseQty= c(4, 3, NA, 3, 1, 9, NA, 4))

Answer 2

由于您没有提供数据，这里有一个玩具示例，我将如何在基础R中执行此操作：

# simple sample data
data <- data.frame( a = rep( 10:12, each = 4 ), b = 12:1 )
data[ c( 3, 5, 12 ), 2 ] <- NA

# for each unique a value, get the row index with the min b value, 
# and write that min value to col b where b is NA
for( i in unique( data$a ) )
    data[ which( is.na( data$b ) & data$a == i  ), "b" ] <-
        min( data[ data$a == i, "b" ], na.rm = TRUE )

data
    a  b
1  10 12
2  10 11
3  10  9
4  10  9
5  11  5
6  11  7
7  11  6
8  11  5
9  12  4
10 12  3
11 12  2
12 12  2