可以的,而最简单的方式可能是通过 write.zoo
:
R> write.zoo
function (x, file = "", index.name = "Index", row.names = FALSE,
col.names = NULL, ...)
{
if (is.null(col.names))
col.names <- !is.null(colnames(x))
dx <- as.data.frame(x)
stopifnot(all(names(dx) != index.name))
dx[[index.name]] <- index(x)
dx <- dx[, c(ncol(dx), 1:(ncol(dx) - 1))]
write.table(dx, file = file, row.names = row.names, col.names = col.names,
...)
}
<environment: namespace:zoo>
R>
这里是一个完整的示例代码:
R> mat <- matrix(rnorm(20),5,4, dimnames=list(NULL, LETTERS[1:4]))
R> mat
A B C D
[1,] -2.5304768 0.5454043 0.754670 0.330617
[2,] -0.5199045 0.3943289 -1.271524 -2.243113
[3,] -0.0996277 -0.0513063 -0.846310 -0.140727
[4,] 0.3819981 0.5230709 1.131108 2.398311
[5,] 1.4366976 -1.7750772 0.193936 1.047754
R> xmat <- xts(mat, order.by=Sys.Date() + seq(-4,0))
R> xmat
A B C D
2012-01-19 -2.5304768 0.5454043 0.754670 0.330617
2012-01-20 -0.5199045 0.3943289 -1.271524 -2.243113
2012-01-21 -0.0996277 -0.0513063 -0.846310 -0.140727
2012-01-22 0.3819981 0.5230709 1.131108 2.398311
2012-01-23 1.4366976 -1.7750772 0.193936 1.047754
现在我们已经有了数据,只需要把它写出来就可以了:
R> write.zoo(xmat, file="/tmp/demo.csv", sep=",")
R> system("cat /tmp/demo.csv")
"Index","A","B","C","D"
2012-01-19,-2.53047680387774,0.545404313269755,0.754669841541681,0.330616876246245
2012-01-20,-0.519904544868541,0.394328857686792,-1.27152367237311,-2.24311276135881
2012-01-21,-0.0996276931028331,-0.0513062656752562,-0.846309564748021,-0.14072731914499
2012-01-22,0.381998053276389,0.523070920853495,1.13110826400249,2.39831100812159
2012-01-23,1.43669757366164,-1.77507724264279,0.193935657150967,1.04775355172344
R>
2012年1月25日修改 使用row.names=FALSE
而不是TRUE
来防止重复的行名。由于row.names=FALSE
是默认设置,因此从调用中删除它。
> xmat[,1] A 2012-01-21 -1.4232098 2012-01-22 0.1456240 2012-01-23 -0.6625430 2012-01-24 -0.3947322 2012-01-25 0.5433947
时,它会再次返回日期,而我只想得到 A 中的值。这可能吗? - user1155299row.names=FALSE
这是默认设置。我已经相应地编辑了我的帖子。 - Dirk Eddelbuettel