从嵌套 for 循环中的条件创建一个新列答案

【问题标题】：Create a new column from conditions in nested for loop从嵌套 for 循环中的条件创建一个新列
【发布时间】：2016-10-27 16:12:12
【问题描述】：

我正在尝试在我的数据集中创建一个新列，用于告诉我产品的收入是否在 3 个月内全部为 0，在所有 3 个月内都为 0，或者在所有 3 个月内都没有 0。

我提供了NewColumn 作为我希望结果的样子。

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

数据

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))

【问题讨论】：

上帝禁止人们将您在答案中使用的数据添加给他人
好电话，我通常会尝试包含它（作为对 OP 的提醒）但这次忘记了，我很高兴@flightless13wings 注意到了:-)
感谢 @rawr 编辑我最初的问题并添加数据。
其中一个答案是否解决了您的问题？如果是这样，通常通过选中答案左侧的复选标记来接受答案。

标签： r for-loop nested subset lapply

【解决方案1】：

在基础 R 中，您可以创建一个自定义函数，然后使用 ave 在每个组内进行计算：

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"

【讨论】：

谢谢！这真的很有帮助。

【解决方案2】：

@thelatemail 的解决方案很简洁。以下是两种替代解决方案：

基础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

与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

【讨论】：

感谢您的帮助。我会用我的真实数据试一试。

【解决方案3】：

使用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

【讨论】：