如何用中值填充NA？ [英] How to fill NA with median?

问题描述

范例资料：

set.seed(1)
df <- data.frame(years=sort(rep(2005:2010, 12)), 
                 months=1:12, 
                 value=c(rnorm(60),NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA))

head(df)
  years months      value
1  2005      1 -0.6264538
2  2005      2  0.1836433
3  2005      3 -0.8356286
4  2005      4  1.5952808
5  2005      5  0.3295078
6  2005      6 -0.8204684

请告诉我，我可以将df $值中的NA替换为其他月份的中位数吗？ value必须包含同一个月的所有先前值的中值。也就是说，如果当前月份是5月，value必须包含5月份以前所有值的中间值。

Tell me please, how i can replace NA in df$value to median of others months? "value" must contain the median of value of all previous values for the same month. That is, if current month is May, "value" must contain the median value for all previous values of the month of May.

推荐答案

p>或使用ave

df <- data.frame(years=sort(rep(2005:2010, 12)),
months=1:12,
value=c(rnorm(60),NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA))
df$value[is.na(df$value)] <- with(df, ave(value, months, 
   FUN = function(x) median(x, na.rm = TRUE)))[is.na(df$value)]

---

因为有这么多的答案让我们看看哪个是最快的。

---

Since there are so many answers let's see which is fastest.

plyr2 <- function(df){
  medDF <- ddply(df,.(months),summarize,median=median(value,na.rm=TRUE))
df$value[is.na(df$value)] <- medDF$median[match(df$months,medDF$months)][is.na(df$value)]
  df
}
library(plyr)
library(data.table)
DT <- data.table(df)
setkey(DT, months)


benchmark(ave = df$value[is.na(df$value)] <- 
  with(df, ave(value, months, 
               FUN = function(x) median(x, na.rm = TRUE)))[is.na(df$value)],
          tapply = df$value[61:72] <- 
            with(df, tapply(value, months, median, na.rm=TRUE)),
          sapply = df[61:72, 3] <- sapply(split(df[1:60, 3], df[1:60, 2]), median),
          plyr = ddply(df, .(months), transform, 
                       value=ifelse(is.na(value), median(value, na.rm=TRUE), value)),
          plyr2 = plyr2(df),
          data.table = DT[,value := ifelse(is.na(value), median(value, na.rm=TRUE), value), by=months],
          order = "elapsed")
        test replications elapsed relative user.self sys.self user.child sys.child
3     sapply          100   0.209 1.000000     0.196    0.000          0         0
1        ave          100   0.260 1.244019     0.244    0.000          0         0
6 data.table          100   0.271 1.296651     0.264    0.000          0         0
2     tapply          100   0.271 1.296651     0.256    0.000          0         0
5      plyr2          100   1.675 8.014354     1.612    0.004          0         0
4       plyr          100   2.075 9.928230     2.004    0.000          0         0

我敢打赌data.table是最快的。

I would have bet that data.table was the fastest.

[Matthew Dowle]最多花费0.02秒（2.075 / 100）。 data.table 认为不重要。尝试将复制设置为 1 ，并增加数据大小。或者定时最快的3次跑步也是一个常见的经验法则。在这些链接中进行更详细的讨论：

[ Matthew Dowle ] The task being timed here takes at most 0.02 seconds (2.075/100). data.table considers that insignificant. Try setting replications to 1 and increasing the data size, instead. Or timing the fastest of 3 runs is also a common rule of thumb. More verbose discussion in these links :

• 数据表不总是最快的证据
$ 平均化与其他列值对应的特定数据部分的列值的 b

• London R简报，2012年6月（幻灯片21标题为其他）


• 按组进行转换在极端情况下进行基准测试

• Evidence that data.table isn't always fastest
• Benchmarks in Averaging column values for specific sections of data corresponding to other column values
• London R presentation, June 2012 (slide 21 headed "Other")
• A transform by group benchmark in an extreme case

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