我在下面粘贴了一个解决方案。希望我的评论能很好地解释这个解决方案。

#Packages used
library(dplyr)
#Some reproducible data
dta <- data.frame(
  date = c(1, 1, 1, 2, 2, 2, 3, 3, 3),
  a = c(NA, NA, NA, NA, NA, NA, NA, NA, NA),
  x = c(123, NA, NA, 3456, NA, NA, 2345, NA, NA),
  y = c(NA, 123, NA, NA, 3456, NA, NA, 2345, NA),
  z = c(NA, NA, 123, NA, NA, 3456, NA, NA, 2345)
)
dta <- dta |> 
  group_by(date) |> #To group by the dates
  dplyr::summarise(a = sum(a, na.rm = TRUE), #just summarise the single value (min(), mean(), etc. work just as well)
            x = sum(x, na.rm = TRUE),
            y = sum(y, na.rm = TRUE),
            z = sum(z, na.rm = TRUE)) |> 
  select_if(~sum(.) > 0) #Remove columns with sum of 0 (columns with all NA)

如果每列中每个日期只有一个非缺失值，则以下代码有效：

library(tidyverse)
df <- tibble::tribble(
     ~date, ~col1, ~col2, ~col3, ~col4, ~col5,
   "2019-01-01",  NA,  1966439,    NA,  NA,    NA,
   "2019-01-01",  NA,  NA,          NA,  133.6, NA,
  "2019-01-01",  NA,  NA,          NA,  NA,    6.2,
   "2019-02-01",  NA,  1962946,     NA,  NA,    NA,
   "2019-02-01",  NA,  NA,          NA,  134.5, NA,
  "2019-02-01",  NA,  NA,          NA,  NA,    6.1,
   "2019-03-01",  NA,  1974072,     NA,  NA,    NA,
   "2019-03-01",  NA,  NA,          NA,  135.4, NA,
  "2019-03-01",  NA,  NA,          NA,  NA,    6.3,
   "2019-04-01",  NA,  1984086,     NA,  NA,    NA
  )
remove_na <- function(x) {
  if (all(is.na(x))) return(NA)
  discard(x, is.na)
}
df |> 
  group_by(date) |> 
  summarize(across(starts_with("col"), remove_na))
#> # A tibble: 4 × 6
#>   date       col1     col2 col3   col4  col5
#>   <chr>      <lgl>   <dbl> <lgl> <dbl> <dbl>
#> 1 2019-01-01 NA    1966439 NA     134.   6.2
#> 2 2019-02-01 NA    1962946 NA     134.   6.1
#> 3 2019-03-01 NA    1974072 NA     135.   6.3
#> 4 2019-04-01 NA    1984086 NA      NA   NA

创建于2023年3月3日，使用reprex v2.0.2
请包括一些生成数据集的代码（就像我在这里做的），下次你张贴一个问题！