其他解决方案：

Question

我有要分组的数据，执行计算然后是最终结果，将其用于下一组中的计算。

我们使用条件语句并按组执行计算，例如：

# Example Data 
condition <- c(0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,1,1,0)
col_a <- c(0,0,0,2,3,4,0,0,0,2,4,5,6,0,0,0,0,0,0,0,0,1,2,0)
col_b <- c(0,0,0,10,131,14,0,0,0,22,64,75,96,0,0,0,0,0,0,0,0,41,52,0)
df <- data.frame(condition,col_a,col_b)

这是进行计算的代码，按RunID分组

# Group by RunID
# Perform calculations 
# Last value, brought forward to next group
require(dplyr) 
output <- df %>%
  dplyr::mutate(RunID = data.table::rleid(condition)) %>%
  group_by(RunID) %>%
  dplyr::mutate(calculation = ifelse(condition == 0,0, ifelse(row_number() == n(),first(col_a) * last(col_b),0))) %>%

dplyr :: mutate（last = tag = ifelse（condition == 0,0，ifelse（row_number（）== n（），2,0）））％＆gt;％＃添加助手ID号。在下面的回答中帮助循环取消组合（）％＆gt;％选择（-RunID）输出＆lt; - data.frame（输出）头（输出15）

输出：

      condition col_a col_b calculation
1          0     0     0           0
2          0     0     0           0
3          0     0     0           0
4          1     2    10           0
5          1     3   131           0
6          1     4    14          28
7          0     0     0           0
8          0     0     0           0
9          0     0     0           0
10         1     2    22           0
11         1     4    64           0
12         1     5    75           0
13         1     6    96         192
14         0     0     0           0
15         0     0     0           0

我想做的是。在第一个结果上，计算列中的结果是28.我想将该值传递给下一个组并插入col_a，第10行（28个替换，2）。然后更新该值。第二组计算结果为96 * 28 = 2688对比（96 * 2 = 192）

结转将始终插入每组的第一行，如上例所示。

预期产出：

      condition col_a col_b calculation
1          0     0     0           0
2          0     0     0           0
3          0     0     0           0
4          1     2    10           0
5          1     3   131           0
6          1     4    14          28
7          0     0     0           0
8          0     0     0           0
9          0     0     0           0
10         1     28    22           0
11         1     4    64           0
12         1     5    75           0
13         1     6    96         2688
14         0     0     0           0
15         0     0     0           0

其他解决方案：

我将子集删除所有0，s。添加了识别到每个组底部的2个数字以进行连续运行，然后使用for循环进行抓取和替换。可能不是最优雅，但似乎有效：

# Subset to remove all 0 
subset.no.zero <- subset(output,condition >0)
# Loop to move values
for (i in 1:nrow(subset.no.zero)) {
  temp <- ifelse(subset.no.zero$last.tag[i-1] == 2, subset.no.zero$calculation[i-1],subset.no.zero$col_a[i])
  subset.no.zero$new_col_a[i] <- data.frame(temp)
}

# Re join by index no.
final_out <- full_join(output,subset.no.zero, by="index")

Answer 1

我只能提供data.table解决方案，但也许您可以将逻辑转换为dplyr：

library(data.table)
setDT(df)

#first group multiply 2 and 14
df[rleid(condition) %in% 1:2 & condition != 0, 
   calculation := {
     res <- rep(NA_real_, .N)
     res[.N] <- col_b[.N] * col_a[1]
     res
   }
   ]

#all groups other than first copy col_b
df[, calculation := if (condition[.N] != 0) {
  if(is.na(calculation[.N])) {
    res <- rep(NA_real_, .N)
    res[.N] <- col_b[.N]
    res
  } else calculation
} else NA_real_,
by = rleid(condition)
]    

#cumulative product
df[!is.na(calculation), 
   calculation := cumprod(calculation)] 

#copy values into col_a
df[i = df[, .(condition = condition[1], i = .I[1]), 
          by = rleid(condition)][condition == 1L][-1, i], #finds rows to replace values
   col_a := head(df[!is.na(calculation), calculation], -1) 
   ]

#    condition col_a col_b calculation
# 1:         0     0     0          NA
# 2:         0     0     0          NA
# 3:         0     0     0          NA
# 4:         1     2    10          NA
# 5:         1     3   131          NA
# 6:         1     4    14          28
# 7:         0     0     0          NA
# 8:         0     0     0          NA
# 9:         0     0     0          NA
#10:         1    28    22          NA
#11:         1     4    64          NA
#12:         1     5    75          NA
#13:         1     6    96        2688
#14:         0     0     0          NA
#15:         0     0     0          NA
#16:         0     0     0          NA
#17:         0     0     0          NA
#18:         0     0     0          NA
#19:         0     0     0          NA
#20:         0     0     0          NA
#21:         0     0     0          NA
#22:         1  2688    41          NA
#23:         1     2    52      139776
#24:         0     0     0          NA
#    condition col_a col_b calculation

RunID的dplyr组将值传递给下一组

其他解决方案：

1 个答案: