2

我现有的数据框看起来像下面给出的线

NAV_Date    NAV    Year  Day       Units   Amount  Balance_Units
2013-06-01  282.5  2013  Saturday  3.540   1000    3.540
2013-06-08  279.3  2013  Saturday  3.581   1000    3.581
2013-06-15  276.0  2013  Saturday  3.623   1000    3.623
2013-06-22  261.6  2013  Saturday  3.822   1000    3.822
2013-06-29  273.3  2013  Saturday  3.659   1000    3.659

我希望我的新数据框有 Balance_Units 列有如下使用 R 语言给出的条目，即平衡单位列应该是先前值和现值的总和这需要在数据框列表上完成

NAV_Date    NAV    Year  Day       Units   Amount  Balance_Units
2013-06-01  282.5  2013  Saturday  3.540   1000    3.540
2013-06-08  279.3  2013  Saturday  3.581   1000    7.121
2013-06-15  276.0  2013  Saturday  3.623   1000    10.744
2013-06-22  261.6  2013  Saturday  3.822   1000    14.566
2013-06-29  273.3  2013  Saturday  3.659   1000    18.225

我试过这个但是，这不起作用

 for( i in 1:length(W)) {
  W[[i]]$Units  = 1000/W[[i]]$NAV
  W[[i]]$Amount = 1000
  W[[i]]$Balance_Units = 0
  W[[i]]$Balance_Units = W[[i]]$Units + W[[i]]$Balance_Units
   }
4

2 回答 2

1

这可以通过一个方便的函数来cumsum完成base R

df1$Balance_Units <- cumsum(df1$Balance_Units)

dplyr，它可以在内部创建mutate

library(dplyr)
df1 %>%
    mutate(Balance_Units = cumsum(Balance_Units))

如果 'W' 是lists中的一个data.frame，我们可以使用lapply

W <- lapply(W, transform, Balance_Units = cumsum(Balance_Units))

数据

df1 <- structure(list(NAV_Date = c("2013-06-01", "2013-06-08", "2013-06-15", 
"2013-06-22", "2013-06-29"), NAV = c(282.5, 279.3, 276, 261.6, 
273.3), Year = c(2013L, 2013L, 2013L, 2013L, 2013L), Day = c("Saturday", 
"Saturday", "Saturday", "Saturday", "Saturday"), Units = c(3.54, 
3.581, 3.623, 3.822, 3.659), Amount = c(1000L, 1000L, 1000L, 
1000L, 1000L), Balance_Units = c(3.54, 3.581, 3.623, 3.822, 3.659
)), class = "data.frame", row.names = c(NA, -5L))
于 2020-05-31T22:12:06.497 回答
1

这是一个data.table解决方案。我也为顺序总和添加了一些东西..

library(data.table) 
> dat
     NAV_Date   NAV Year      Day Units Amount Balance_Units
1: 2013-06-01 282.5 2013 Saturday 3.540   1000         3.540
2: 2013-06-08 279.3 2013 Saturday 3.581   1000         3.581
3: 2013-06-15 276.0 2013 Saturday 3.623   1000         3.623
4: 2013-06-22 261.6 2013 Saturday 3.822   1000         3.822
5: 2013-06-29 273.3 2013 Saturday 3.659   1000         3.659

# Cumulative sum
> dat[, cumulative_sum := cumsum(Balance_Units)]
> dat
     NAV_Date   NAV Year      Day Units Amount Balance_Units cumulative_sum
1: 2013-06-01 282.5 2013 Saturday 3.540   1000         3.540          3.540
2: 2013-06-08 279.3 2013 Saturday 3.581   1000         3.581          7.121
3: 2013-06-15 276.0 2013 Saturday 3.623   1000         3.623         10.744
4: 2013-06-22 261.6 2013 Saturday 3.822   1000         3.822         14.566
5: 2013-06-29 273.3 2013 Saturday 3.659   1000         3.659         18.225

# Sequential sum
> dat[, sequential_sum := Balance_Units + shift(x = Balance_Units, fill = 0)]
> dat
     NAV_Date   NAV Year      Day Units Amount Balance_Units cumulative_sum sequential_sum
1: 2013-06-01 282.5 2013 Saturday 3.540   1000         3.540          3.540          3.540
2: 2013-06-08 279.3 2013 Saturday 3.581   1000         3.581          7.121          7.121
3: 2013-06-15 276.0 2013 Saturday 3.623   1000         3.623         10.744          7.204
4: 2013-06-22 261.6 2013 Saturday 3.822   1000         3.822         14.566          7.445
5: 2013-06-29 273.3 2013 Saturday 3.659   1000         3.659         18.225          7.481
于 2020-05-31T22:22:17.193 回答