R SQL查询列表中的数据帧
给定数据帧R SQL查询列表中的数据帧,r,list,dataframe,data.table,sqldf,R,List,Dataframe,Data.table,Sqldf,给定数据帧 df1 <- data.frame(CustomerId=c(1:6),Product=c(rep("Toaster",3),rep("Radio",3))) df2 <- data.frame(CustomerId=c(2,4,6),State=c(rep("Alabama",2),rep("Ohio",1))) df1如果将data.frames从列表复制到新环境,则可以使用envir参数来sqldf,或者通过命名列表的元素,并使用和 请注意以下几点: 我使用li
df1 <- data.frame(CustomerId=c(1:6),Product=c(rep("Toaster",3),rep("Radio",3)))
df2 <- data.frame(CustomerId=c(2,4,6),State=c(rep("Alabama",2),rep("Ohio",1)))
df1如果将data.frames从列表复制到新环境,则可以使用envir
参数来sqldf
,或者通过命名列表的元素,并使用和
请注意以下几点:
- 我使用
list
而不是c
创建dflist
注意区别
str(c(df1,df2))
##List of 4
## $ CustomerId: int [1:6] 1 2 3 4 5 6
## $ Product : Factor w/ 2 levels "Radio","Toaster": 2 2 2 1 1 1
## $ CustomerId: num [1:3] 2 4 6
## $ State : Factor w/ 2 levels "Alabama","Ohio": 1 1 2
str(list(df1,df2))
##List of 2
## $ :'data.frame': 6 obs. of 2 variables:
## ..$ CustomerId: int [1:6] 1 2 3 4 5 6
## ..$ Product : Factor w/ 2 levels "Radio","Toaster": 2 2 2 1 1 1
## $ :'data.frame': 3 obs. of 2 variables:
## ..$ CustomerId: num [1:3] 2 4 6
## ..$ State : Factor w/ 2 levels "Alabama","Ohio": 1 1 2
- 我已经调整了sql查询以反映data.frames中的名称(按照您的第二种方法)
命名数据
另一种使用proto
- 感谢@G.Grothendieck(参见评论
这将使用加载了sqldf
dflist <- list(a = df1, b = df2)
sqldf( "select a.CustomerId, a.Product, b.State from df1 a
inner join df2 b on b.CustomerId = a.CustomerId",
envir = as.proto(dflist))
我已经演示了如何使用assign
和mapply
来代替for循环
str(c(df1,df2))
##List of 4
## $ CustomerId: int [1:6] 1 2 3 4 5 6
## $ Product : Factor w/ 2 levels "Radio","Toaster": 2 2 2 1 1 1
## $ CustomerId: num [1:3] 2 4 6
## $ State : Factor w/ 2 levels "Alabama","Ohio": 1 1 2
str(list(df1,df2))
##List of 2
## $ :'data.frame': 6 obs. of 2 variables:
## ..$ CustomerId: int [1:6] 1 2 3 4 5 6
## ..$ Product : Factor w/ 2 levels "Radio","Toaster": 2 2 2 1 1 1
## $ :'data.frame': 3 obs. of 2 variables:
## ..$ CustomerId: num [1:3] 2 4 6
## ..$ State : Factor w/ 2 levels "Alabama","Ohio": 1 1 2
dflist <- list(df1,df2)
names(dflist) <- c('df1','df2')
# create a new environment
e <- new.env()
# assign the elements of dflist to this new environment
for(.x in names(dflist)){
assign(value = dflist[[.x]], x=.x, envir = e)
}
# this could also be done using mapply / lapply
# eg
# invisible(mapply(assign, value = dflist, x = names(dflist), MoreArgs =list(envir = e)))
# run the sql query
sqldf("select a.CustomerId, a.Product, b.State from df1 a
inner join df2 b on b.CustomerId = a.CustomerId", envir = e)
## CustomerId Product State
## 1 2 Toaster Alabama
## 2 4 Radio Alabama
## 3 6 Radio Ohio
# this is far simpler!!
with(dflist,sqldf("select a.CustomerId, a.Product, b.State from df1 a
inner join df2 b on b.CustomerId = a.CustomerId"))
dflist <- list(a = df1, b = df2)
sqldf( "select a.CustomerId, a.Product, b.State from df1 a
inner join df2 b on b.CustomerId = a.CustomerId",
envir = as.proto(dflist))
library(data.table)
dflist <- list(data.table(df1),data.table(df2))
names(dflist) <- c('df1','df2')
invisible(lapply(dflist, setkeyv, 'CustomerId'))
with(dflist, df1[df2])
## CustomerId Product State
## 1: 2 Toaster Alabama
## 2: 4 Radio Alabama
## 3: 6 Radio Ohio