我有一个复杂的问题，我不知道该怎么办。我有两个名为df1的数据框：
structure(list(State = structure(1:2, .Label = c("Aaa", "Dd"), class = "factor"),
City = structure(1:2, .Label = c("bb", "e"), class = "factor"),
Type1 = c(NA, NA), Type2 = c(NA, NA)), .Names = c("State",
"City", "Type1", "Type2"), class = "data.frame", row.names = c(NA,
-2L))
和df2：
structure(list(state = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 2L
), .Label = c("Aaa", "Dd"), class = "factor"), city = structure(c(1L,
2L, 3L, 4L, 4L, 5L, 6L), .Label = c("bb", "ccc", "ddd", "fff",
"ggg", "hh"), class = "factor"), type = structure(c(1L, 2L, 2L,
2L, 2L, 2L, 3L), .Label = c("Type 1", "Type 2", "Type 4"), class = "factor"),
value = 1:7), .Names = c("state", "city", "type", "value"
), class = "data.frame", row.names = c(NA, -7L))
Dataframe df1如下所示：
State City Type1 Type2
Aaa bb NA NA
Dd e NA NA
和dataframe df2如下所示：
state city type value
Aaa bb Type 1 1
Aaa ccc Type 2 2
Aaa ddd Type 2 3
Dd fff Type 2 4
Dd fff Type 2 5
Dd ggg Type 2 6
Dd hh Type 4 7
对于NA中的df1，我需要根据以下规则从df2查找值：
1）如果只有单个实例，其中State = state和City = city对于给定type在df2中，将value插入相应的df1列Type1或Type2
2）如果多个实例，其中State = state和City = city对于给定的type，我需要对所有value求平均值并将其插入df1
3）如果给定State 无实例，其中state = City和city = type，我需要获取此state的所有type的平均值并插入df1
4）如果给定的State 无实例其中state = type，则该值应保留NA { {1}}
只是为了澄清 - 基本上我想尽可能将值df1和Type1平均为“已解决”。换句话说，我希望在可能的情况下使用Type2级别的平均值，但如果不可能，那么我想使用City级别平均值。不过，我想要回复State中概述的原始State和City的平均值（即使df1平均值都可用
我知道这很复杂！我想要的结果是
State这是一个数据框，如：
structure(list(State = structure(1:2, .Label = c("Aaa", "Dd"), class = "factor"),
City = structure(1:2, .Label = c("bb", "e"), class = "factor"),
Type1 = c(1L, NA), Type2 = c(2.5, 5)), .Names = c("State",
"City", "Type1", "Type2"), class = "data.frame", row.names = c(NA,
-2L))
我甚至不知道从哪里开始解决这个问题。我的第一个想法是，我需要使用State City Type1 Type2
Aaa bb 1 2.5
Dd e NA 5.0
来重塑acast。例如，我可以使用
df2它将数据重新整形为更接近acast(df2, state+city+value~type)
，但随后我松开了一些我需要保留的列（这些列被压缩到rowname中）。我甚至不知道如何开始搜索df1和City的挑战，然后根据这些结果进行平均。
有人能指出我正确的方向吗？
编辑（2015年1月）：我在下面的特洛伊回答下面添加了一条新评论，询问是否有一种简单的方法可以添加一个列，用于确定计算均值的级别（城市或州） 。我找到了一个解决方案，虽然可能有更好的方法，但它对我有用。希望这有助于某人！
State然后
getlevel<-function(state,city,type){
m<-means[means$state==state & means$city==city & means$type==type, "mean"]
sm<-state_means[state_means$state==state & state_means$type==type, "mean"]
ifelse(length(m)>0,"city","state")
}
答案 0 :(得分：1)
编辑 - 抱歉误解了这些问题：以下是符合您条件的更正代码：
require(plyr)
means<-ddply(df2,.(state,city,type),summarize,mean=mean(value))
state_means<-ddply(df2,.(state,type),summarize,mean=mean(value))
getval<-function(state,city,type){
m<-means[means$state==state & means$city==city & means$type==type, "mean"]
sm<-state_means[state_means$state==state & state_means$type==type, "mean"]
ifelse(length(m)>0,m,sm)
}
## this gives you the new df1
ddply(df1,.(State,City),transform,Type1=getval(as.character(State),as.character(City),"Type 1"),Type2=getval(as.character(State),as.character(City),"Type 2"))
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 以前的答案（不完整）
这有点困难，因为你的结构调用对于df2不能正常工作，你的示例数据集不会给你预期结果中的所有数据，但我认为你想要的是：
require(plyr)
means<-ddply(df2,.(state,city,type),summarize,mean=mean(value))
getval<-function(state,city,type){means[means$state==state & means$city==city & means$type==type, "mean"]}
## this gives you the new df1
ddply(df1,.(State,City),transform,Type1=getval(as.character(State),as.character(City),"Type 1"),Type2=getval(as.character(State),as.character(City),"Type 2"))
############################################################################X
## what's happening in detail:
require(plyr) # calls the plyr library
means<-ddply(df2, # base on df2
.(state,city,type), # summarize by combination of city/state/type
summarize, # tells plyr to summarize rather than transform
mean=mean(value)) # show one column at each summary level, called 'mean', the average val
getval<-function(state,city,type){ # create function called getval, takes 3 parameters
means[means$state==state & # first part of [X,]
means$city==city & # selects the row that matches all criteria
means$type==type,
"mean"]} # and [,X] the column relating to the type
getval("Aaa","bb","Type 2")
# this gives you the new df2
ddply(df1, # base on df1
.(State,City), # summarize by State & City
transform, # tell plyr to transform existingn set rather than roll up
Type1=getval(as.character(State),as.character(City),"Type 1"), # call getval() for Type 1
Type2=getval(as.character(State),as.character(City),"Type 2")) # and for Type 2
为您提供以下内容（不是您的预期结果，但数据隐含的内容）
State City Type1 Type2
1 Aaa bb 1 NA
2 Dd e NA NA
答案 1 :(得分：0)
首先重塑df2中的数据，然后使用data.table的密钥合适地合并数据：
library(data.table)
library(reshape2)
dt1 <- as.data.table(df1)
dt2 <- as.data.table(df2)
Type列
dt2.casted <- reshape2::dcast(dt2, state + city ~ type
, fill=NA_real_
, fun.aggregate=mean, na.rm=TRUE)
dt2.casted <- as.data.table(dt2.casted)
keys以便合并setkey(dt2.casted, state, city)
setkey(dt1, State, City)
NA s dt1[dt2.casted][, lapply(.SD, mean, na.rm=TRUE), by=State, .SDcols=grep("Type", names(dt2.casted), value=TRUE)]
State Type 1 Type 2 Type 4
1: Aaa 1 2.50 NaN
2: Dd NaN 5.25 7
dt2.casted <- reshape2::dcast(dt2, state ~ type
, fill=NA_real_
, fun.aggregate=mean, na.rm=TRUE)
dt2.casted <- as.data.table(dt2.casted)
setkey(dt2.casted, state)
setkey(dt1, State)
dt1[dt2.casted][, lapply(.SD, mean, na.rm=TRUE)
, by=list(State, City)
, .SDcols=grep("Type"
, names(dt2.casted), value=TRUE)
]
State City Type 1 Type 2 Type 4
1: Aaa bb 1 2.5 NaN
2: Dd e NaN 5.0 7