list 如何合并R中列表的所有元素？
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How to merge all elements of list in R?
提问by user2109248
I have a list containing data frames as its elements in R.
我有一个列表，其中包含数据框作为 R 中的元素。
Example:
例子：
df1 <- data.frame("names"=c("John","Sam","Dave"),"age"=c(21,22,25))
df2 <- data.frame("names"=c("John","Sam"),"score"=c(22,25))
df3 <- data.frame("names"=c("John","Sam","Dave"),"country"=c("US","SA","NZ"))
mylist <- list(df1,df2,df3)
Is it possible to merge all the elements of mylist together without using a loop?
是否可以在不使用循环的情况下将 mylist 的所有元素合并在一起？
My desired output for this example is:
这个例子我想要的输出是：
names age score country
1 John 21 22 US
2 Sam 22 25 SA
The list in this example has only three elements; however, I am looking for a solution that can handle an arbitrary number of elements.
这个例子中的列表只有三个元素；但是，我正在寻找一种可以处理任意数量元素的解决方案。
回答by agstudy
You can use Reduce, one liner solution:
您可以使用Reduce一种衬垫解决方案：
Reduce(merge,mylist)
names age score country
1 John 21 22 US
2 Sam 22 25 SA
回答by alexwhan
Quick and dirty example:
快速而肮脏的例子：
merge(merge(df1, df2),df3)
EDIT- Very similar question here:Simultaneously merge multiple data.frames in a list
编辑- 非常相似的问题：在列表中同时合并多个 data.frames
solution:
解决方案：
merged.data.frame = Reduce(function(...) merge(..., all=F), my.list)
Disclaimer - All I changed from @Charles answer was to make merge(..., all=F)rather than T- this way it gives your desired output.
免责声明 - 我从 @Charles 答案中更改的所有内容都是 makemerge(..., all=F)而不是T- 这样它就可以提供您想要的输出。
回答by Aaron left Stack Overflow
Just to show it could be done another way...
只是为了表明它可以以另一种方式完成......
mymerge <- function(mylist) {
names(mylist) <- sapply(mylist, function(x) names(x)[2])
ns <- unique(unlist(lapply(mylist, function(x) levels(x$names))))
as.data.frame(c(list(names=ns), lapply(mylist, function(x)
{x[match(ns, x$names),2]})))
}
> mymerge(mylist)
names age score country
1 Dave 25 NA NZ
2 John 21 22 US
3 Sam 22 25 SA
One could easily adapt to remove rows with missing values, or perhaps just remove afterwards with complete.cases.
人们可以很容易地适应删除缺少值的行，或者可能只是在之后使用complete.cases.
To show that it's faster, we'll make up a bigger data set; 100 variables and 25 names.
为了证明它更快，我们将组成一个更大的数据集；100 个变量和 25 个名称。
set.seed(5)
vs <- paste0("V", 1:100)
mylist <- lapply(vs, function(v) {
x <- data.frame(names=LETTERS[1:25], round(runif(25, 0,100)))
names(x)[2] <- v
x
})
> microbenchmark(Reduce(merge, mylist), myf(mylist))
Unit: milliseconds
expr min lq median uq max
1 myf(mylist) 12.81371 13.19746 13.36571 14.40093 33.90468
2 Reduce(merge, mylist) 199.23714 206.28608 207.30247 208.44939 226.05980
回答by alap
Have you tried this function?
你试过这个功能吗？
http://rss.acs.unt.edu/Rdoc/library/gtools/html/smartbind.html
http://rss.acs.unt.edu/Rdoc/library/gtools/html/smartbind.html
library(gtools)
df1 <- data.frame(list(A=1:10), B=LETTERS[1:10], C=rnorm(10) )
df2 <- data.frame(A=11:20, D=rnorm(10), E=letters[1:10] )
df3 <- df1
out <- smartbind( mylist <- list(df1,df2,df3))
