如何使用sqldf循环数据帧? [英] how to loop the dataframe using sqldf?
问题描述
第一个代码:
样本数据:
vector1 <- data.frame("name"="a","age"=10,"gender"="m")
vector2 <- data.frame("name"="b","age"=33,"gender"="m")
vector3 <- data.frame("name"="b","age"=58,"gender"="f")
list <- list(vector1,vector2,vector3)
sql <- list()
for(i in 1:length(list)){
print(list[[1]]) # access dataframe
sql[[i]]<-
sqldf(paste0("select name,gender,count(name) from ",list[[i]]," group by gender "))
}
如何使用sqldf函数正确循环数据帧?我已经尝试在sqldf函数中使用list [[1]]或list [1]进行测试,但它不会返回任何此类表或语法错误.在循环功能中,我可以访问数据帧.可以使用这种格式吗?
How to loop the data frame correctly using sqldf function? I have tried list[[1]] or list[1] in the sqldf function to do the test but it will return no such table or syntax error. In the loop function, I can access the data frame. Is it possible to use this format?
print(str(list))
List of 3
$ :'data.frame': 1 obs. of 3 variables:
..$ name : Factor w/ 1 level "a": 1
..$ age : num 10
..$ gender: Factor w/ 1 level "m": 1
$ :'data.frame': 1 obs. of 3 variables:
..$ name : Factor w/ 1 level "b": 1
..$ age : num 33
..$ gender: Factor w/ 1 level "m": 1
$ :'data.frame': 1 obs. of 3 variables:
..$ name : Factor w/ 1 level "b": 1
..$ age : num 58
..$ gender: Factor w/ 1 level "f": 1
NULL
第二:
这是我的期望.
f<- lapply(list, function(dataframe) {
sql <-
sqldf("select name,gender,count(name) from dataframe group by gender ")
})
print(f)
这是输出.
> print(f)
[[1]]
name gender count(name)
1 a m 1
[[2]]
name gender count(name)
1 b m 1
[[3]]
name gender count(name)
1 b f 1
是否可以使用第一个代码访问列表?当我想使用粘贴功能访问列表中的每个数据框时,如何解决此问题.
Is it possible to use the first code to access the list? How to fix it when I want to use paste function to access each data frame in a list.
推荐答案
sqldf::sqldf
是指环境中存在的对象.因此,只需创建DF <- list[[i]]
并在查询中使用此名称即可.
sqldf::sqldf
refers to the object existing in the environment. So, just create DF <- list[[i]]
and use this name in query.
for(i in 1:length(list)){
print(list[[1]]) # access dataframe
DF <- list[[i]]
sql[[i]]<- sqldf("select name,gender,count(name) from DF group by gender ")
}
print(sql)
# [[1]]
# name gender count(name)
# 1 a m 1
#
# [[2]]
# name gender count(name)
# 1 b m 1
#
# [[3]]
# name gender count(name)
# 1 b f 1
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