您可以使用cumsum()：

library(dplyr)

df %>% 
  mutate(stock_over_time = lag(stock + cumsum(purchase - sales)))

# A tibble: 5 x 6
  article      week sales purchase stock stock_over_time
  <chr>       <dbl> <dbl>    <dbl> <dbl>           <dbl>
1 article one     1    10        5    50              NA
2 article one     2    10        0    50              45
3 article one     3    10        5    50              35
4 article one     4    10        5    50              30
5 article one     5    10        0    50              25

我们可以使用递归的方法来做这件事，它也应该工作的复杂情况

df$stock_over_time <- df$stock
for(i in 2:nrow(df)) {
    df$stock_over_time[i] <- df$stock_over_time[i-1] - 
           df$sales[i-1] + df$purchase[i-1]
 }
 
df
# A tibble: 5 x 6
#  article      week sales purchase stock stock_over_time
#  <chr>       <dbl> <dbl>    <dbl> <dbl>           <dbl>
#1 article one     1    10        5    50              50
#2 article one     2    10        0    50              45
#3 article one     3    10        5    50              35
#4 article one     4    10        5    50              30
#5 article one     5    10        0    50              25

或者另一个选项是purrr中的accumulate

library(purrr)
library(dplyr)
df %>% 
    mutate(stock_over_time = accumulate((purchase- sales)[-1], 
            ~ .x + .y, .init = first(stock)))
# A tibble: 5 x 6
#  article      week sales purchase stock stock_over_time
#  <chr>       <dbl> <dbl>    <dbl> <dbl>           <dbl>
#1 article one     1    10        5    50              50
#2 article one     2    10        0    50              40
#3 article one     3    10        5    50              35
#4 article one     4    10        5    50              30
#5 article one     5    10        0    50              20

也可以写成

df %>% 
    mutate(stock_over_time = accumulate(c(first(stock), 
         (purchase- sales)[-1]), ~ .x + .y))

下面的方法怎么样？

df$stock_over_time <- df$stock - df$sales + df$purchase

如果在dfs列中存在计算所需的任何NA，我将在以下之前执行此操作：

df[is.na(df)] <- 0