Создайте новый столбец из условий в вложенных циклах

Question

Я пытаюсь создать новый столбец в моем наборе данных, который говорит мне, если доход продукта равен 0 с за все 3 месяца, некоторые 0 за все 3 месяца или не 0 для всех 3 месяца.

Я представил NewColumn как бы хотел, чтобы результат выглядел.

data$ZEROES <- 0 
data$ZEROES2 <- 0 
for (i in unique(data$product_id)){ 
    for (j in unique(data$Revenue)){ 
     n[j] <-ifelse(all(data$Value == 0)," ALL 0", 
     ifelse(any(data$Value == 0),"Some 0", 
     ifelse(all(data$Value != 0), "None 0", "Blank"))) 
    data$ZEROES[j] <-n[j] 
    data$ZEROES2[i] <-long$ZEROES[j] 
    } 
} 

product_ id Date   Revenue   Value NewColumn 
1   January  in    0   Some 0 
1   February  in    1   Some 0 
1   March   in    0   Some 0 
1   January  out    0   All 0 
1   February  out    0   All 0 
1   March   out    0   All 0 
2   January  in    1   No 0 
2   February  in    2   No 0 
2   March   in    3   No 0 
2   January  out    1   Some 0 
2   February  out    1   Some 0 
2   March   out    0   Some 0 

данных

structure(list(product_id = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), 
       Date = c("January", "February", "March", "January", "February", "March", "January", "February", "March", "January", "February", "March"), 
       Revenue = c("in", "in", "in", "out", "out", "out", "in", "in", "in", "out", "out", "out"), 
       Value = c(0L, 1L, 0L, 0L, 0L, 0L, 1L, 2L, 3L, 1L, 1L, 0L), 
       NewColumn = c("Some 0", "Some 0", "Some 0", "All 0", "All 0", "All 0", "No 0", "No 0", "No 0", "Some 0", "Some 0", "Some 0")), 
      .Names = c("product_id", "Date", "Revenue", "Value", "NewColumn"), 
      class = "data.frame", row.names = c(NA, -12L)) 

Answer 1

@ решение thelatemail является кратким. Вот два альтернативные решения:

1. Основание R:

   do.call("rbind", by(dat, list(dat$product_id, dat$Revenue), FUN = function(df) { 
       within(df, NewColumn <- ifelse(all(Value == 0), "All 0", 
              ifelse(all(Value != 0), "No 0", "Some 0"))) 
   })) 
   # product_id  Date Revenue Value NewColumn 
   # 1   1 January  in  0 Some 0 
   # 2   1 February  in  1 Some 0 
   # 3   1 March  in  0 Some 0 
   # 7   2 January  in  1  No 0 
   # 8   2 February  in  2  No 0 
   # 9   2 March  in  3  No 0 
   # 4   1 January  out  0  All 0 
   # 5   1 February  out  0  All 0 
   # 6   1 March  out  0  All 0 
   # 10   2 January  out  1 Some 0 
   # 11   2 February  out  1 Some 0 
   # 12   2 March  out  0 Some 0 
   

2. С dplyr

   library(dplyr) 
   dat %>% 
       group_by(product_id, Revenue) %>% 
       mutate(
       NewColumn = ifelse(all(Value == 0), "All 0", 
            ifelse(all(Value != 0), "No 0", "Some 0")) 
   ) 
   # Source: local data frame [12 x 5] 
   # Groups: product_id, Revenue [4] 
   # product_id  Date Revenue Value NewColumn 
   #   <int> <chr> <chr> <int>  <chr> 
   # 1   1 January  in  0 Some 0 
   # 2   1 February  in  1 Some 0 
   # 3   1 March  in  0 Some 0 
   # 4   1 January  out  0  All 0 
   # 5   1 February  out  0  All 0 
   # 6   1 March  out  0  All 0 
   # 7   2 January  in  1  No 0 
   # 8   2 February  in  2  No 0 
   # 9   2 March  in  3  No 0 
   # 10   2 January  out  1 Some 0 
   # 11   2 February  out  1 Some 0 
   # 12   2 March  out  0 Some 0 
   

Answer 2

В базовой R вы можете сделать собственную функцию, а затем использовать ave делать вычисления в каждой группе:

f <- function(x) if(all(x)) 3 else if(any(x)) 2 else 1 
c("None","Some","All")[with(dat, ave(Value==0, list(product_id,Revenue), FUN=f))] 
# [1] "Some" "Some" "Some" "All" "All" "All" "None" "None" "None" "Some" 
#[11] "Some" "Some" 

Answer 3

Использование data.table

library(data.table) 
setDT(df1)[, NewColumn := c("No 0", "Some 0", "All 0")[(all(!Value) + 
         any(!Value))+1], .(product_id, Revenue)] 
# product_id  Date Revenue Value NewColumn 
# 1:   1 January  in  0 Some 0 
# 2:   1 February  in  1 Some 0 
# 3:   1 March  in  0 Some 0 
# 4:   1 January  out  0  All 0 
# 5:   1 February  out  0  All 0 
# 6:   1 March  out  0  All 0 
# 7:   2 January  in  1  No 0 
# 8:   2 February  in  2  No 0 
# 9:   2 March  in  3  No 0 
#10:   2 January  out  1 Some 0 
#11:   2 February  out  1 Some 0 
#12:   2 March  out  0 Some 0