Início / coding / Perguntas / 79491785
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Codutie
Asked: 2025-03-07 17:22:27 +0800 CST

Enumerar eventos onde n valores consecutivos não são NA [duplicado]

  • 772

Gostaria de calcular uma nova coluna em um dataframe que encontre n valores consecutivos não sendo nulos e adicione uma enumeração a ela, por grupo de usuários. Seja df nosso dataframe de exemplo:

n <- 4

# Sample DataFrame
df <- tibble(
  user_id = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2),
  date = as.Date('2024-01-01') + 0:19,
  value1 = c(NA, 5, 6, 7, 8, NA, NA, 2, 3, 4, 5, NA, 1, 2, 3, NA, 1, 2, 3, 4)
)

Resultado esperado:

# A tibble: 20 × 4
   user_id date       value1 expected_result
     <dbl> <date>      <dbl>           <dbl>
 1       1 2024-01-01     NA              NA
 2       1 2024-01-02      5               1
 3       1 2024-01-03      6               2
 4       1 2024-01-04      7               3
 5       1 2024-01-05      8               4
 6       1 2024-01-06     NA              NA
 7       1 2024-01-07     NA              NA
 8       1 2024-01-08      2               5
 9       1 2024-01-09      3               6
10       1 2024-01-10      4               7
11       1 2024-01-11      5               8
12       1 2024-01-12     NA              NA
13       2 2024-01-13      1              NA
14       2 2024-01-14      2              NA
15       2 2024-01-15      3              NA
16       2 2024-01-16     NA              NA
17       2 2024-01-17      1               1
18       2 2024-01-18      2               2
19       2 2024-01-19      3               3
20       2 2024-01-20      4               4

3 respostas

  1. Best Answer

    Você pode usarcumsum

    df %>%
  mutate(gp=value1/value1 * cumsum(is.na(value1)), .by=user_id) %>%
  mutate(n=(n()>=n & !is.na(value1)), .by=c(user_id, gp)) %>%
  mutate(expected=na_if(value1/value1 * cumsum(!is.na(value1) & n), 0), .by=user_id) %>%
  select(-c(gp, n))

    Dá

    # A tibble: 20 × 4
   user_id date       value1 expected
     <dbl> <date>      <dbl>    <dbl>
 1       1 2024-01-01     NA       NA
 2       1 2024-01-02      5        1
 3       1 2024-01-03      6        2
 4       1 2024-01-04      7        3
 5       1 2024-01-05      8        4
 6       1 2024-01-06     NA       NA
 7       1 2024-01-07     NA       NA
 8       1 2024-01-08      2        5
 9       1 2024-01-09      3        6
10       1 2024-01-10      4        7
11       1 2024-01-11      5        8
12       1 2024-01-12     NA       NA
13       2 2024-01-13      1       NA 
14       2 2024-01-14      2       NA 
15       2 2024-01-15      3       NA  
16       2 2024-01-16     NA       NA
17       2 2024-01-17      1        1
18       2 2024-01-18      2        2
19       2 2024-01-19      3        3
20       2 2024-01-20      4        4
    • 2

  2. Você pode definir valoresrle de com comprimentos não correspondentes para e usar para criar uma influência.FALSEinvese.rleseq

    > FX <- \(x, n=4) {
+   rl <- rle(!is.na(x))
+   rl$values[rl$lengths < n] <- FALSE
+   irl <- inverse.rle(rl)
+   sirl <- sum(irl)
+   replace(rep_len(NA, sirl), irl, seq_len(sirl))
+ }

    Usar em ave.

    > df |> transform(exp=ave(value1, user_id, FUN=FX))
   user_id       date value1 exp
1        1 2024-01-01     NA  NA
2        1 2024-01-02      5   1
3        1 2024-01-03      6   2
4        1 2024-01-04      7   3
5        1 2024-01-05      8   4
6        1 2024-01-06     NA  NA
7        1 2024-01-07     NA  NA
8        1 2024-01-08      2   5
9        1 2024-01-09      3   6
10       1 2024-01-10      4   7
11       1 2024-01-11      5   8
12       1 2024-01-12     NA  NA
13       2 2024-01-13      1  NA
14       2 2024-01-14      2  NA
15       2 2024-01-15      3  NA
16       2 2024-01-16     NA  NA
17       2 2024-01-17      1   1
18       2 2024-01-18      2   2
19       2 2024-01-19      3   3
20       2 2024-01-20      4   4

    Ou em data.dable.

    > library(data.table)
> DT <- as.data.table(df)
> 
> DT[, exp := FX(value1), by=.(user_id)][]
    user_id       date value1   exp
      <num>     <Date>  <num> <int>
 1:       1 2024-01-01     NA    NA
 2:       1 2024-01-02      5     1
 3:       1 2024-01-03      6     2
 4:       1 2024-01-04      7     3
 5:       1 2024-01-05      8     4
 6:       1 2024-01-06     NA    NA
 7:       1 2024-01-07     NA    NA
 8:       1 2024-01-08      2     5
 9:       1 2024-01-09      3     6
10:       1 2024-01-10      4     7
11:       1 2024-01-11      5     8
12:       1 2024-01-12     NA    NA
13:       2 2024-01-13      1    NA
14:       2 2024-01-14      2    NA
15:       2 2024-01-15      3    NA
16:       2 2024-01-16     NA    NA
17:       2 2024-01-17      1     1
18:       2 2024-01-18      2     2
19:       2 2024-01-19      3     3
20:       2 2024-01-20      4     4
    user_id       date value1   exp
    • 2

  3. Você pode usar uma combinação de rlee collapse::seqid. Basicamente rlecria um vetor lque é 1se xfor NA, e caso contrário é o lcomprimento de não-NAs consecutivos. Então, replacepara NA se o comprimento for abaixo de n, e calcula o seqidnos não-NAs.

    library(collapse)
library(dplyr)

f <- function(x, n){
  l <- with(rle(!is.na(x)), rep(lengths, lengths))
  qF(seqid(replace(l, l < n, NA), na.skip = TRUE))
}

df |>
  mutate(new = f(value1, n), .by = user_id)

   user_id       date value1  new
1        1 2024-01-01     NA <NA>
2        1 2024-01-02      5    1
3        1 2024-01-03      6    2
4        1 2024-01-04      7    3
5        1 2024-01-05      8    4
6        1 2024-01-06     NA <NA>
7        1 2024-01-07     NA <NA>
8        1 2024-01-08      2    5
9        1 2024-01-09      3    6
10       1 2024-01-10      4    7
11       1 2024-01-11      5    8
12       1 2024-01-12     NA <NA>
13       2 2024-01-13      1 <NA>
14       2 2024-01-14      2 <NA>
15       2 2024-01-15      3 <NA>
16       2 2024-01-16     NA <NA>
17       2 2024-01-17      1    1
18       2 2024-01-18      2    2
19       2 2024-01-19      3    3
20       2 2024-01-20      4    4
    • 2

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