具有先前行依赖性的R data.table计算

本文介绍了具有先前行依赖性的R data.table计算的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

下面是一些当前我在Excel中计算的数据。

Below is some data which I currently calculate in Excel.

col_A col _B col_C col_D col_E col_F col_G -1.5% 0.010 1.00 1 1.00 - - -5.4% 0.024 1.00 1 1.00 0.01 -0.00 -7.9% 0.036 1.00 1 1.00 0.02 -0.00 -12.7% 0.052 0.99 1 0.99 0.06 -0.01 -4.6% 0.049 0.98 1 0.98 0.19 -0.01 -8.3% 0.051 0.95 1 0.95 0.39 -0.03 -7.3% 0.052 0.88 1 0.88 1.00 -0.07 -9.2% 0.055 0.69 1 0.69 2.31 -0.21 -7.9% 0.055 0.38 1 0.38 5.63 -0.44 -2.2% 0.051 0.29 1 0.29 11.13 -0.24

我一直在尝试使用data.table在R中执行计算。我的问题是data.table按列执行计算。由于依赖于先前行值的结果，我需要按行执行计算。下面给出了计算列的Excel公式，其中 T表示当前行， T-1表示上一行

I have been trying to perform the calculations in R using data.table. The problem I have is that data.table performs calculation column-wise. I need the calculations to be performed row-wise, because of dependencies on the results of previous row values. The Excel-formulas for the calculated columns are given below, with "T" indicating "current row" and "T-1" indication "previous row"

col_C：（ col_C.T-1）*（1 + col_G.T）

col_C: (col_C.T-1) * (1 + col_G.T)

col_D：最大值（Col_C.T，col_D.T-1）

col_D: max (Col_C.T, col_D.T-1)

col_E：（col_C.T / col_D.T）

col_E: (col_C.T / col_D.T)

col_F：max（（1-（col_C.T-1 / col_D。 T-1））/ col BT-1），0.01）

col_F: max ((1 - (col_C.T-1 / col_D.T-1)) / col B.T-1), 0.01)

col_G：col_A * col_F

col_G: col_A * col_F

任何

推荐答案

如果没有其他条件需要使用 data.table 我建议使用矩阵实现按行计算：

If there are no other conditions which require to use data.table I suggest to implement the rowwise calculations using a matrix:

m <- data.matrix(dt) m[, 3:7] <- NA for (i in seq.int(nrow(m))) { if (i == 1L) { m[i, "col_F"] <- 0 m[i, "col_G"] <- 0 m[i, "col_C"] <- 1 m[i, "col_D"] <- 1 } else { m[i, "col_F"] <- max((1 - (m[i-1, "col_C"] / m[i-1, "col_D"])) / m[i-1, "col_B"], 0.01) m[i, "col_G"] <- m[i, "col_A"] * m[i, "col_F"] m[i, "col_C"] <- m[i-1, "col_C"] * (1 + m[i, "col_G"]) m[i, "col_D"] <- max(m[i, "col_C"], m[i-1, "col_D"]) } m[i, "col_E"] <- m[i, "col_C"] / m[i, "col_D"] } m

col_A col_B col_C col_D col_E col_F col_G [1,] -0.015 0.010 1.0000000 1 1.0000000 0.00000000 0.000000000 [2,] -0.054 0.024 0.9994600 1 0.9994600 0.01000000 -0.000540000 [3,] -0.079 0.036 0.9976835 1 0.9976835 0.02250000 -0.001777500 [4,] -0.127 0.052 0.9895302 1 0.9895302 0.06434834 -0.008172239 [5,] -0.046 0.049 0.9803653 1 0.9803653 0.20134322 -0.009261788 [6,] -0.083 0.051 0.9477596 1 0.9477596 0.40070748 -0.033258721 [7,] -0.073 0.052 0.8768905 1 0.8768905 1.02432085 -0.074775422 [8,] -0.092 0.055 0.6858958 1 0.6858958 2.36749020 -0.217809099 [9,] -0.079 0.055 0.3764416 1 0.3764416 5.71098585 -0.451167882 [10,] -0.022 0.051 0.2825483 1 0.2825483 11.33742486 -0.249423347

col_F 的最后4行与OP的预期结果之间的偏差可能是由于过帐值的精度有限 col_A 和 col_B 。

The deviations in the last 4 rows of col_F from OP's expected result might be due to the limited precision of the posted values of col_A and col_B.

library(data.table) dt <- fread("col_A col_B col_C col_D col_E col_F col_G -1.5% 0.010 1.00 1 1.00 - - -5.4% 0.024 1.00 1 1.00 0.01 -0.00 -7.9% 0.036 1.00 1 1.00 0.02 -0.00 -12.7% 0.052 0.99 1 0.99 0.06 -0.01 -4.6% 0.049 0.98 1 0.98 0.19 -0.01 -8.3% 0.051 0.95 1 0.95 0.39 -0.03 -7.3% 0.052 0.88 1 0.88 1.00 -0.07 -9.2% 0.055 0.69 1 0.69 2.31 -0.21 -7.9% 0.055 0.38 1 0.38 5.63 -0.44 -2.2% 0.051 0.29 1 0.29 11.13 -0.24 ", na.strings = "-") # convert percent string to numeric dt[, col_A := readr::parse_number(col_A) / 100]