如何在给定时间戳的情况下从zoo/xts对象中删除行

Question

我很高兴使用这段代码:

z=lapply(filename_list, function(fname){
    read.zoo(file=fname,header=TRUE,sep = ",",tz = "")
    })
xts( do.call(rbind,z) )

直到Dirty Data在一个文件的末尾出现:

                        Open     High      Low    Close Volume
2011-09-20 21:00:00 1.370105 1.370105 1.370105 1.370105      1

这是在下一个文件的开头:

                        Open     High      Low  Close Volume
2011-09-20 21:00:00 1.370105 1.371045 1.369685 1.3702   2230

所以rbind.zoo抱怨重复.

我不能使用像:

 y <- x[ ! duplicated( index(x) ),  ]

因为它们在不同的动物园对象中,在列表中.我不能aggregate像这里建议的那样使用,因为它们是动物园对象的列表,而不是一个大的动物园对象.而且我无法得到一个重要的对象.第二十二条军规.

所以,当事情变得艰难时,强硬的黑客攻击一些for循环(原谅印刷品和停止,因为这还不是工作代码):

indexes <- do.call("c", unname(lapply(z, index)))
dups=duplicated(indexes)
if(any(dups)){
    duplicate_timestamps=indexes[dups]
    for(tix in 1:length(duplicate_timestamps)){
        t=duplicate_timestamps[tix]
        print("We have a duplicate:");print(t)
        for(zix in 1:length(z)){
            if(t %in% index(z[[zix]])){
                print(z[[zix]][t])
                if(z[[zix]][t]$Volume==1){
                    print("-->Deleting this one");
                    z[[zix]][t]=NULL    #<-- PROBLEM
                    }
                }
            }
        }
    stop("There are duplicate bars!!")
    }

我难以理解的是为动物园行分配NULL不会删除它(NextMethod中的错误("[< - "):替换的长度为零).好的,所以我做了一个过滤器副本,没有违规的项目...但我正在绊倒这些:

> z[[zix]][!t,]
Error in Ops.POSIXt(t) : unary '!' not defined for "POSIXt" objects

> z[[zix]][-t,]
Error in `-.POSIXt`(t) : unary '-' is not defined for "POSIXt" objects

PS虽然非常欢迎我在"动物园对象列表中复制行"的真正问题的高级解决方案,但这里的问题具体是关于如何在给定POSIXt索引对象的情况下从动物园对象中删除行.

---

一小部分测试数据:

list(structure(c(1.36864, 1.367045, 1.370105, 1.36928, 1.37039, 
1.370105, 1.36604, 1.36676, 1.370105, 1.367065, 1.37009, 1.370105, 
5498, 3244, 1), .Dim = c(3L, 5L), .Dimnames = list(NULL, c("Open", 
"High", "Low", "Close", "Volume")), index = structure(c(1316512800, 
1316516400, 1316520000), class = c("POSIXct", "POSIXt"), tzone = ""), class = "zoo"), 
    structure(c(1.370105, 1.370115, 1.36913, 1.371045, 1.37023, 
    1.37075, 1.369685, 1.36847, 1.367885, 1.3702, 1.36917, 1.37061, 
    2230, 2909, 2782), .Dim = c(3L, 5L), .Dimnames = list(NULL, 
        c("Open", "High", "Low", "Close", "Volume")), index = structure(c(1316520000, 
    1316523600, 1316527200), class = c("POSIXct", "POSIXt"), tzone = ""), class = "zoo"))

---

更新:感谢G. Grothendieck的行删除解决方案.在实际代码中,我遵循Joshua和GSee的建议来获取xts对象的列表而不是动物园对象列表.所以我的代码成了:

z=lapply(filename_list, function(fname){
    xts(read.zoo(file=fname,header=TRUE,sep = ",",tz = ""))
    })
x=do.call.rbind(z)

(顺便说一句,请注意电话do.call.rbind.这是因为rbind.xts有一些严重的内存问题.请参阅/sf/answers/842055651/)

然后我删除重复项作为后处理步骤:

dups=duplicated(index(x))
if(any(dups)){
    duplicate_timestamps=index(x)[dups]
    to_delete=x[ (index(x) %in% duplicate_timestamps) & x$Volume<=1]
    if(nrow(to_delete)>0){
        #Next line says all lines that are not in the duplicate_timestamp group
        #     OR are in the duplicate timestamps, but have a volume greater than 1.
        print("Will delete the volume=1 entry")
        x=x[ !(index(x) %in% duplicate_timestamps) | x$Volume>1]
        }else{
        stop("Duplicate timestamps, and we cannot easily remove them just based on low volume.")
        }
    }

Answer 1

如果z1和z2是你的动物园对象然后rbind同时去除任何重复z2:

rbind( z1, z2[ ! time(z2) %in% time(z1) ] )

关于删除具有指定时间的动物园对象中的点,上面已经说明了这一点,但一般来说,如果tt是要删除的时间向量:

z[ ! time(z) %in% tt ]

或者,如果我们知道有一个单一的元素tt,然后z[ time(z) != tt ].