Question

我有几百个重复的主键和与这些键关联的日期。日期可能有也可能没有缺少条目，但需要缺少条目需要用max（日期）替换。

#Create Proxy dataframe
df <- tibble(
  key = c("a", "a", "b", "b", "c", "c", "d", "d", "e", "e", "f", "f", "h", "h", "i","i", "j", "j", "k", "k", "l", "l", "m", "m"),
  date1 = c("NA", "2017-02-13", "NA", "2017-04-14", "2017-05-18", "2017-05-18", "NA", "2018-01-07",
          "2017-09-24", "2017-09-25", "NA", "2017-09-29", "NA", "2017-08-13", "NA", "2017-04-29",
          "NA", "2018-01-28", "NA", "2017-10-08", "NA", "2017-01-10", "NA", "2017-11-01")
)
df$date1 <- as.Date(df$date1, format = "%Y-%m-%d")

请注意

-key＆＃34; a＆＃34;有一个缺少的日期，需要用唯一可用的日期替换 -key＆＃34; c＆＃34;没有任何遗漏日期
-key＆＃34; e＆＃34;有两个不同的日期，但最后一个日期需要记录

 df
    # A tibble: 24 x 2
         key      date1
       <chr>     <date>
     1     a         NA
     2     a 2017-02-13
     3     b         NA
     4     b 2017-04-14
     5     c 2017-05-18
     6     c 2017-05-18
     7     d         NA
     8     d 2018-01-07
     9     e 2017-09-24
    10     e 2017-09-25
    # ... with 14 more rows

我尝试过的解决方案不起作用：

library(lubridate)
df$date <- with(df$date, as.Date(ifelse(is.na(df$date), orderDate, df$date), origin = "1970-01-01"))

library(dplyr)
df %>% group_by(key) %>%
        mutate(date = (date, NA, df$date)) %>%
        as.data.frame

任何帮助将不胜感激！谢谢！

Answer 1

假设您只想在+------+------------+--+ | id | _c1 | +------+------------+--+ | abc | [1,2,2,1] | +------+------------+--+为NA时替换为每个组中的max()值，这将有效。请注意，您需要指定date1，因为na.rm = TRUE会返回NA，而不是1。

max(NA, 1)

Answer 2

有一种替代方法比Mako212's dplyr solution快得多。它在加入时使用更新，将NA值替换为每个max(date1)组的key：

library(data.table) DT <- as.data.table(df) tmp <- DT[, .(date1 = as.Date(NA), max(date1, na.rm = TRUE)), by = key] DT[tmp, on = .(key, date1), date1 := V2][]

key date1 1: a 2017-02-13 2: a 2017-02-13 3: b 2017-04-14 4: b 2017-04-14 5: c 2017-05-18 6: c 2017-05-18 7: d 2018-01-07 8: d 2018-01-07 9: e 2017-09-24 10: e 2017-09-25 11: f 2017-09-29 12: f 2017-09-29 13: h 2017-08-13 14: h 2017-08-13 15: i 2017-04-29 16: i 2017-04-29 17: j 2018-01-28 18: j 2018-01-28 19: k 2017-10-08 20: k 2017-10-08 21: l 2017-01-10 22: l 2017-01-10 23: m 2017-11-01 24: m 2017-11-01 key date1

请注意，只有date1为NA的行被替换为到位，即不复制整个数据对象。

tmp包含每个key组的replacemnet值：

key date1 V2 1: a <NA> 2017-02-13 2: b <NA> 2017-04-14 3: c <NA> 2017-05-18 4: d <NA> 2018-01-07 5: e <NA> 2017-09-25 6: f <NA> 2017-09-29 7: h <NA> 2017-08-13 8: i <NA> 2017-04-29 9: j <NA> 2018-01-28 10: k <NA> 2017-10-08 11: l <NA> 2017-01-10 12: m <NA> 2017-11-01

基准

创建基准数据：

library(dplyr) library(data.table) n_row <- 1e5L n_key <- 500L share_na <- 0.5 set.seed(123L) DT0 <- data.table( key1 = sprintf("%04i", sample.int(n_key, n_row, TRUE)), date1 = as.Date("2017-01-01") + sample.int(n_key, n_row, TRUE) ) # set NA values DT0[sample.int(n_row, share_na * n_row), date1 := NA] # coerce to tibble df0 <- as_tibble(DT0)

运行基准：

library(microbenchmark) bm <- microbenchmark( dplyr = { copy(df0) %>% group_by(key1) %>% mutate(date1 = case_when( is.na(date1) ~ max(date1, na.rm = TRUE), TRUE ~ date1) ) }, dt = { DT <- copy(DT0) tmp <- DT[, .(date1 = as.Date(NA), max(date1, na.rm = TRUE)), by = key1] DT[tmp, on = .(key1, date1), date1 := V2][] }, times = 21L ) print(bm)

Unit: milliseconds expr min lq mean median uq max neval cld dplyr 131.02040 136.81967 142.63845 137.78741 141.36084 191.37755 21 b dt 18.14997 18.68349 19.65384 19.32424 19.54815 26.87965 21 a

对于给定问题大小为100 k行，500个组和50％NA值，data.table方法比dplyr版本快7倍左右。

请注意，DT0和df0的新副本用于每次重复，因为DT已就地更新。对copy()的调用包含在两个案例的时间安排中。已修改dplyr版本以更新date1，而不是在输出中创建第三列。

R - 使用重复键将NA替换为日期

2 个答案:

基准