list 如何更有效地将巨大的向量列表转换为矩阵？
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How to convert a huge list-of-vector to a matrix more efficiently?
提问by user1787675
I have a list of length 130,000 where each element is a character vector of length 110. I would like to convert this list to a matrix with dimension 1,430,000*10. How can I do it more efficiently?\ My code is :
我有一个长度为 130,000 的列表，其中每个元素都是一个长度为 110 的字符向量。我想将此列表转换为维度为 1,430,000*10 的矩阵。我怎样才能更有效地做到这一点？\ 我的代码是：
output=NULL
for(i in 1:length(z)) {
output=rbind(output,
matrix(z[[i]],ncol=10,byrow=TRUE))
}
回答by flodel
This should be equivalent to your current code, only a lot faster:
这应该等同于您当前的代码，只是要快得多：
output <- matrix(unlist(z), ncol = 10, byrow = TRUE)
回答by Ben Bolker
I thinkyou want
我想你想要
output <- do.call(rbind,lapply(z,matrix,ncol=10,byrow=TRUE))
i.e. combining @BlueMagister's use of do.call(rbind,...)with an lapplystatement to convert the individual list elements into 11*10 matrices ...
即结合@BlueMagister 的使用do.call(rbind,...)与lapply语句将单个列表元素转换为 11*10 矩阵......
Benchmarks (showing @flodel's unlistsolution is 5x faster than mine, and 230x faster than the original approach ...)
基准测试（显示@flodel 的unlist解决方案比我的解决方案快 5 倍，比原始方法快 230 倍......）
n <- 1000
z <- replicate(n,matrix(1:110,ncol=10,byrow=TRUE),simplify=FALSE)
library(rbenchmark)
origfn <- function(z) {
output <- NULL
for(i in 1:length(z))
output<- rbind(output,matrix(z[[i]],ncol=10,byrow=TRUE))
}
rbindfn <- function(z) do.call(rbind,lapply(z,matrix,ncol=10,byrow=TRUE))
unlistfn <- function(z) matrix(unlist(z), ncol = 10, byrow = TRUE)
## test replications elapsed relative user.self sys.self
## 1 origfn(z) 100 36.467 230.804 34.834 1.540
## 2 rbindfn(z) 100 0.713 4.513 0.708 0.012
## 3 unlistfn(z) 100 0.158 1.000 0.144 0.008
If this scales appropriately (i.e. you don't run into memory problems), the full problem would take about 130*0.2 seconds = 26 seconds on a comparable machine (I did this on a 2-year-old MacBook Pro).
如果这适当地扩展（即您没有遇到内存问题），完整的问题将需要大约 130*0.2 秒 = 26 秒在同类机器上（我在 2 岁的 MacBook Pro 上这样做）。
回答by Blue Magister
It would help to have sample information about your output. Recursively using rbindon bigger and bigger things is not recommended. My first guess at something that would help you:
获得有关输出的示例信息会有所帮助。不推荐rbind在越来越大的事物上递归使用。我第一个猜测对你有帮助的东西：
z <- list(1:3,4:6,7:9)
do.call(rbind,z)
See a related questionfor more efficiency, if needed.
如果需要，请参阅相关问题以提高效率。
回答by csta
You can also use,
您还可以使用，
output <- as.matrix(as.data.frame(z))
The memory usage is very similar to
内存使用情况非常类似于
output <- matrix(unlist(z), ncol = 10, byrow = TRUE)
Which can be verified, with mem_changed()from library(pryr).
可以验证，mem_changed()从library(pryr).
回答by Ahmed Gehad
you can use as.matrix as below:
您可以使用 as.matrix 如下：
output <- as.matrix(z)
