用以前的值替换缺失值[重复]答案

【问题标题】：Replace missing value with previous value [duplicate]用以前的值替换缺失值[重复]
【发布时间】：2013-01-17 07:06:06
【问题描述】：

Event,Time,Bid,Offer
Quote,0.458338,9.77,9.78
Order,0.458338,NA,NA
Order,0.458338,NA,NA
Order,0.458338,NA,NA
Quote,0.458363,9.78,9.79
Order,0.458364,NA,NA

我有一个这样的数据框我想编写一个高效的代码来用以前的报价报价和报价填写NA，时间是排序的，只有报价包含报价和报价字段（最好是矢量化）

原来如此

Event,Time,Bid,Offer
Quote,0.458338,9.77,9.78
Order,0.458338,9.77,9.78
Order,0.458338,9.77,9.78
Order,0.458338,9.77,9.78
Quote,0.458363,9.78,9.79
Order,0.458364,9.78,9.79

谢谢

【问题讨论】：

标签： r zoo

【解决方案1】：

试试这个：

# Last Observation Move Forward
na.lomf <- function(object, na.rm = F) {
    na.lomf.0 <- function(object) {
        idx <- which(!is.na(object))
        if (is.na(object[1])) idx <- c(1, idx)
        rep.int(object[idx], diff(c(idx, length(object) + 1)))
    }    
    dimLen <- length(dim(object))
    object <- if (dimLen == 0) na.lomf.0(object) else apply(object, dimLen, na.lomf.0)
    if (na.rm) na.trim(object, sides = "left", is.na = "all") else object
}

【讨论】：

【解决方案2】：

zoo 包中的na.locf() 函数是您的朋友。 locf 代表“最后一个结转”。使用您的数据：

dat <- read.table(text = "Event,Time,Bid,Offer
Quote,0.458338,9.77,9.78
Order,0.458338,NA,NA
Order,0.458338,NA,NA
Order,0.458338,NA,NA
Quote,0.458363,9.78,9.79
Order,0.458364,NA,NA
", header = TRUE, sep = ",")

require(zoo)

dat2 <- transform(dat, Bid = na.locf(Bid), Offer = na.locf(Offer))

生产。

> dat2
  Event     Time  Bid Offer
1 Quote 0.458338 9.77  9.78
2 Order 0.458338 9.77  9.78
3 Order 0.458338 9.77  9.78
4 Order 0.458338 9.77  9.78
5 Quote 0.458363 9.78  9.79
6 Order 0.458364 9.78  9.79

【讨论】：