2

我有一个 xts 对象，我用它来“增加”周期以生成第二个 xts 对象。然后我将两个 xts 对象组合回更快的时期。如何设置 cbind() 以便替换 NA（见下文）并填充最新值？

require(xts)
data(sample_matrix)
x <- as.xts(sample_matrix)
x.wk <- to.weekly(x)
x2 <- cbind(x, x.wk[ , 4])
print(first(x2, "2 weeks"))

>head(x2)
                   Open     High      Low    Close  x.Close
    2007-01-02 50.03978 50.11778 49.95041 50.11778       NA
    2007-01-03 50.23050 50.42188 50.23050 50.39767       NA
    2007-01-04 50.42096 50.42096 50.26414 50.33236       NA
    2007-01-05 50.37347 50.37347 50.22103 50.33459       NA
    2007-01-06 50.24433 50.24433 50.11121 50.18112       NA
    2007-01-07 50.13211 50.21561 49.99185 49.99185 49.99185
    2007-01-08 50.03555 50.10363 49.96971 49.98806       NA
    2007-01-09 49.99489 49.99489 49.80454 49.91333       NA
    2007-01-10 49.91228 50.13053 49.91228 49.97246       NA
    2007-01-11 49.88529 50.23910 49.88529 50.23910       NA
    2007-01-12 50.21258 50.35980 50.17176 50.28519       NA
    2007-01-13 50.32385 50.48000 50.32385 50.41286       NA
    2007-01-14 50.46359 50.62395 50.46359 50.60145 50.60145

>head(x3)    # this is what I want instead
                   Open     High      Low    Close  x.Close
    2007-01-02 50.03978 50.11778 49.95041 50.11778       NA
    2007-01-03 50.23050 50.42188 50.23050 50.39767       NA
    2007-01-04 50.42096 50.42096 50.26414 50.33236       NA
    2007-01-05 50.37347 50.37347 50.22103 50.33459       NA
    2007-01-06 50.24433 50.24433 50.11121 50.18112       NA
    2007-01-07 50.13211 50.21561 49.99185 49.99185 49.99185
    2007-01-08 50.03555 50.10363 49.96971 49.98806 49.99185
    2007-01-09 49.99489 49.99489 49.80454 49.91333 49.99185
    2007-01-10 49.91228 50.13053 49.91228 49.97246 49.99185
    2007-01-11 49.88529 50.23910 49.88529 50.23910 49.99185
    2007-01-12 50.21258 50.35980 50.17176 50.28519 49.99185
    2007-01-13 50.32385 50.48000 50.32385 50.41286 49.99185
    2007-01-14 50.46359 50.62395 50.46359 50.60145 50.60145

谢谢！

4

2 回答 2

5

您只需要na.locf在合并两个对象后使用。

> head(na.locf(x2),15)
               Open     High      Low    Close  x.Close
2007-01-02 50.03978 50.11778 49.95041 50.11778       NA
2007-01-03 50.23050 50.42188 50.23050 50.39767       NA
2007-01-04 50.42096 50.42096 50.26414 50.33236       NA
2007-01-05 50.37347 50.37347 50.22103 50.33459       NA
2007-01-06 50.24433 50.24433 50.11121 50.18112       NA
2007-01-07 50.13211 50.21561 49.99185 49.99185 49.99185
2007-01-08 50.03555 50.10363 49.96971 49.98806 49.99185
2007-01-09 49.99489 49.99489 49.80454 49.91333 49.99185
2007-01-10 49.91228 50.13053 49.91228 49.97246 49.99185
2007-01-11 49.88529 50.23910 49.88529 50.23910 49.99185
2007-01-12 50.21258 50.35980 50.17176 50.28519 49.99185
2007-01-13 50.32385 50.48000 50.32385 50.41286 49.99185
2007-01-14 50.46359 50.62395 50.46359 50.60145 50.60145
2007-01-15 50.61724 50.68583 50.47359 50.48912 50.60145
2007-01-16 50.62024 50.73731 50.56627 50.67835 50.60145
于 2011-03-30T04:44:38.267 回答
2

这里有两个例子可以做到这一点。

第一种是一种函数式编程方式：

x2$x <- Reduce(function(x,y){
  if(is.na(y)) c(x,x[length(x)]) else c(x,y)},
  as.numeric(x2$x.Close)
  )

另一种是一种程序方式：

z <- x2$x.Close
z1 <- rle(c(is.na(z)))
z2 <- z1$lengths[z1$values==TRUE] +1
z3 <- z[!is.na(z)]

# in the case that first row is NA
if (is.na(z[1])) {
  z3 <- c(NA, z3)
  z2[1] <- z2[1]-1
  }

# in the case that the last row is not NA
if (!is.na(z[length(z)])) z2 <- c(z2, 1)
z4 <- rep(z3, z2)
x2$x <- z4
于 2011-03-30T02:15:49.313 回答