Início / coding / Perguntas / 77909260
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Salvador
Asked: 2024-01-31 04:32:54 +0800 CST

Como usar floor_date com instrução if condicional

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Estou tentando adicionar uma coluna ao meu conjunto de dados com floor_date. Se o mês tiver 31 dias quero selecionar 16 dias, caso contrário, divido um mês de 30 dias em dois de 15 dias, mas estou recebendo um erro. Considere o pequeno conjunto de dados:

library(tidyverse)

a <- structure(list(StartDate = structure(c(17774, 17773, 17772, 17771, 
17770, 17769, 17768, 17767, 17766, 17765, 17764, 17763, 17762, 
17761, 17760, 17759, 17758, 17757, 17756, 17755, 17754, 17753, 
17752, 17751, 17750, 17749, 17748, 17747, 17746, 17745, 17744, 
17651, 17650, 17649, 17648, 17647, 17646, 17645, 17644, 17643, 
17642, 17641, 17640, 17639, 17638, 17637, 17636, 17635, 17634, 
17633, 17632, 17631, 17630, 17629, 17628, 17627, 17626, 17625, 
17624, 17623, 17622, 17562, 17561, 17560, 17559, 17558, 17557, 
17556, 17555, 17554, 17553, 17552, 17551, 17550, 17549, 17548, 
17547, 17546, 17545, 17544, 17543, 17542, 17541, 17540, 17539, 
17538, 17537, 17536, 17535, 17534, 17533, 17532), class = "Date"), 
    Minimum = c(6.306, 6.636, 5.745, 6.102, 7.895, 7.895, 5.847, 
    6.484, 5.847, 6.788, 6.458, 7.87, 7.795, 7.87, 8.767, 8.618, 
    6.585, 7.293, 7.645, 8.717, 9.188, 8.841, 8.643, 8.369, 8.02, 
    7.645, 7.77, 7.142, 8.02, 8.891, 9.114, 0.577, 0.439, 0.329, 
    -0.032, -0.004, 0.024, 0.079, 0.687, 0.19, 0.797, 1.071, 
    1.208, -0.004, 0.163, 1.344, 1.044, 1.017, 0.051, 0.55, 1.317, 
    0.522, 0.163, 0.797, 1.507, 2.128, 2.262, 1.208, -0.004, 
    0.384, 0.687, -0.032, 0.66, 0.989, 1.044, 0.439, -0.032, 
    -0.032, 0.024, -0.06, 0.163, -0.032, -0.032, 0.384, 0.163, 
    0.024, -0.032, -0.06, 0.301, 0.742, -0.032, -0.032, 0.495, 
    1.48, 0.495, 0.163, 0.962, -0.004, -0.004, -0.032, -0.032, 
    -0.032), month = c("August", "August", "August", "August", 
    "August", "August", "August", "August", "August", "August", 
    "August", "August", "August", "August", "August", "August", 
    "August", "August", "August", "August", "August", "August", 
    "August", "August", "August", "August", "August", "August", 
    "August", "August", "August", "April", "April", "April", 
    "April", "April", "April", "April", "April", "April", "April", 
    "April", "April", "April", "April", "April", "April", "April", 
    "April", "April", "April", "April", "April", "April", "April", 
    "April", "April", "April", "April", "April", "April", "January", 
    "January", "January", "January", "January", "January", "January", 
    "January", "January", "January", "January", "January", "January", 
    "January", "January", "January", "January", "January", "January", 
    "January", "January", "January", "January", "January", "January", 
    "January", "January", "January", "January", "January", "January"
    )), class = "data.frame", row.names = c(NA, -92L))

#This works but floor_date is applied to every month, however, it fails with the conditional statement below:
a <- a %>% mutate(new = lubridate::floor_date(as.Date(paste(StartDate)), '16 days'))
a  

#Fails
m <- c('January','August')
if(lubridate::month %in% m) {   #January and August
  a %>% mutate(new = lubridate::floor_date(as.Date(paste(StartDate)), '16 days'))
   
   } else {   #April
   
      a %>% mutate(new = lubridate::floor_date(as.Date(paste(StartDate)), '15 days'))
   }
 
Error in match(x, table, nomatch = 0L) : 
  'match' requires vector arguments

1 respostas

  1. Best Answer

    if/ elseé uma construção de fluxo de contol, mas aqui está uma variante vetorizada como if_else(), case_when(), ... que pode ser usada com vetores / em mutate()chamadas:

    library(tidyverse)

a %>% 
  # to tibble for more descriptive print-out
  as_tibble() %>% 
  mutate(new = case_when(
    month %in% m ~ floor_date(StartDate,'16 days'),
    .default     = floor_date(StartDate,'15 days')
    )
  )
#> # A tibble: 92 × 4
#>    StartDate  Minimum month  new       
#>    <date>       <dbl> <chr>  <date>    
#>  1 2018-08-31    6.31 August 2018-08-17
#>  2 2018-08-30    6.64 August 2018-08-17
#>  3 2018-08-29    5.74 August 2018-08-17
#>  4 2018-08-28    6.10 August 2018-08-17
#>  5 2018-08-27    7.90 August 2018-08-17
#>  6 2018-08-26    7.90 August 2018-08-17
#>  7 2018-08-25    5.85 August 2018-08-17
#>  8 2018-08-24    6.48 August 2018-08-17
#>  9 2018-08-23    5.85 August 2018-08-17
#> 10 2018-08-22    6.79 August 2018-08-17
#> # ℹ 82 more rows
    • 1

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