Replace missing values (NA) with most recent non-NA by group

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南旧
南旧 2020-11-22 05:42

I would like to solve the following problem with dplyr. Preferable with one of the window-functions. I have a data frame with houses and buying prices. The following is an e

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  •  悲&欢浪女
    2020-11-22 05:58

    These all use na.locf from the zoo package. Also note that na.locf0 (also defined in zoo) is like na.locf except it defaults to na.rm = FALSE and requires a single vector argument. na.locf2 defined in the first solution is also used in some of the others.

    dplyr

    library(dplyr)
    library(zoo)
    
    na.locf2 <- function(x) na.locf(x, na.rm = FALSE)
    df %>% group_by(houseID) %>% do(na.locf2(.)) %>% ungroup
    

    giving:

    Source: local data frame [15 x 3]
    Groups: houseID
    
       houseID year price
    1        1 1995    NA
    2        1 1996   100
    3        1 1997   100
    4        1 1998   120
    5        1 1999   120
    6        2 1995    NA
    7        2 1996    NA
    8        2 1997    NA
    9        2 1998    30
    10       2 1999    30
    11       3 1995    NA
    12       3 1996    44
    13       3 1997    44
    14       3 1998    44
    15       3 1999    44
    

    A variation of this is:

    df %>% group_by(houseID) %>% mutate(price = na.locf0(price)) %>% ungroup
    

    Other solutions below give output which is quite similar so we won't repeat it except where the format differs substantially.

    Another possibility is to combine the by solution (shown further below) with dplyr:

    df %>% by(df$houseID, na.locf2) %>% bind_rows
    

    by

    library(zoo)
    
    do.call(rbind, by(df, df$houseID, na.locf2))
    

    ave

    library(zoo)
    
    transform(df, price = ave(price, houseID, FUN = na.locf0))
    

    data.table

    library(data.table)
    library(zoo)
    
    data.table(df)[, na.locf2(.SD), by = houseID]
    

    zoo This solution uses zoo alone. It returns a wide rather than long result:

    library(zoo)
    
    z <- read.zoo(df, index = 2, split = 1, FUN = identity)
    na.locf2(z)
    

    giving:

           1  2  3
    1995  NA NA NA
    1996 100 NA 44
    1997 100 NA 44
    1998 120 30 44
    1999 120 30 44
    

    This solution could be combined with dplyr like this:

    library(dplyr)
    library(zoo)
    
    df %>% read.zoo(index = 2, split = 1, FUN = identity) %>% na.locf2
    

    input

    Here is the input used for the examples above:

    df <- structure(list(houseID = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 
      2L, 3L, 3L, 3L, 3L, 3L), year = c(1995L, 1996L, 1997L, 1998L, 
      1999L, 1995L, 1996L, 1997L, 1998L, 1999L, 1995L, 1996L, 1997L, 
      1998L, 1999L), price = c(NA, 100L, NA, 120L, NA, NA, NA, NA, 
      30L, NA, NA, 44L, NA, NA, NA)), .Names = c("houseID", "year", 
      "price"), class = "data.frame", row.names = c(NA, -15L))
    

    REVISED Re-arranged and added more solutions. Revised dplyr/zoo solution to conform to latest changes dplyr. Applied fixed and factored out na.locf2 from all solutions.

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