如何计算 r 的滞后投资

Question

我创建了df1

year gvkey  capex   ppent
2004  1004 13.033 139.137
2005  1004 16.296 213.380
2006  1004 29.891 260.167
2007  1004 30.334 310.393
2008  1004 27.535 245.586
2009  1004 28.855 334.430
...

我创建了df2

year gvkey    ROA
2005  1004 0.02796478
2006  1004 0.04665171
2007  1004 0.05976127
2008  1004 0.06255035
2009  1004 0.03549220
2005  1013 0.06882688
...

我想创建df3

year gvkey    ROA               lag_investment
2005  1004 0.02796478  capex from 2004 / ppent from 2004
2006  1004 0.04665171  capex from 2005 / ppent from 2005
2007  1004 0.05976127  capex from 2006 / ppent from 2006
2008  1004 0.06255035  capex from 2007 / ppent from 2007
2009  1004 0.03549220  capex from 2008 / ppent from 2008
2005  1013 0.06882688  capex from 2004 / ppent from 2004
...

我有超过 2,000 家公司年。 gvkey = firm id

我基本上想做的是以下几点：

1) 根据df1计算上一年的投资

2) 在df2 中创建一个名为“lag_investment”的列

2) 在df2的当前年份行中插入步骤 1) 中的值

其他问题：

如果我想执行以下操作，代码会是什么样子？

我创建了df1

  year gvkey        ROA   ppent  capex
1 2004  1004 0.01320911 139.137 13.033
2 2005  1004 0.03005708 213.380 16.296
3 2006  1004 0.05014214 260.167 29.891
4 2007  1004 0.06423255 310.393 30.334
5 2008  1004 0.06723031 245.586 27.535
6 2009  1004 0.03814769 334.430 28.855
...

我想给df1添加一个变量

  year gvkey        ROA   ppent  capex         lag_investment
1 2004  1004 0.01320911 139.137 13.033
2 2005  1004 0.03005708 213.380 16.296  capex from 2004 / ppent from 2004
3 2006  1004 0.05014214 260.167 29.891  capex from 2005 / ppent from 2005
4 2007  1004 0.06423255 310.393 30.334  capex from 2006 / ppent from 2006
5 2008  1004 0.06723031 245.586 27.535  capex from 2007 / ppent from 2007
6 2009  1004 0.03814769 334.430 28.855  capex from 2008 / ppent from 2008
...

我想计算除 2004 年以外所有年份的 lag_investment。

非常感谢！！！

Answer 1

我猜你可以在dplyr 中使用lag

library(dplyr)
df1 %>% mutate(lag_investment = lag(capex)/lag(ppent))

#  year gvkey    ROA ppent capex lag_investment
#1 2004  1004 0.0132   139  13.0             NA
#2 2005  1004 0.0301   213  16.3         0.0937
#3 2006  1004 0.0501   260  29.9         0.0764
#4 2007  1004 0.0642   310  30.3         0.1149
#5 2008  1004 0.0672   246  27.5         0.0977
#6 2009  1004 0.0381   334  28.9         0.1121

如果数据框没有排序，请先使用arrange按年排序。

df1 %>% arrange(year) %>% mutate(lag_investment = lag(capex)/lag(ppent))

---

或shift 中的data.table

library(data.table)
setDT(df1)[, lag_investment := shift(capex)/shift(ppent)]

Answer 2

有了data.table，我们可以做到

library(data.table)
setDT(df1)[, lag_investment :=Reduce(`/`, shift(.SD)), .SDcols = c("capex", "ppent")]
df1
#   year gvkey        ROA   ppent  capex lag_investment
#1: 2004  1004 0.01320911 139.137 13.033             NA
#2: 2005  1004 0.03005708 213.380 16.296     0.09367027
#3: 2006  1004 0.05014214 260.167 29.891     0.07637079
#4: 2007  1004 0.06423255 310.393 30.334     0.11489159
#5: 2008  1004 0.06723031 245.586 27.535     0.09772772
#6: 2009  1004 0.03814769 334.430 28.855     0.11211958

---

或在base R

df1$lag_investment <- with(df1, c(NA, head(capex, -1)/head(ppent, -1)))

也可以写成

df1$lag_investment <- with(df1, c(NA, capex[-nrow(df1)]/ppent[-nrow(df1)]))

数据

df1 <-  structure(list(year = 2004:2009, gvkey = c(1004L, 1004L, 1004L, 
1004L, 1004L, 1004L), ROA = c(0.01320911, 0.03005708, 0.05014214, 
0.06423255, 0.06723031, 0.03814769), ppent = c(139.137, 213.38, 
260.167, 310.393, 245.586, 334.43), capex = c(13.033, 16.296, 
29.891, 30.334, 27.535, 28.855)), class = "data.frame", 
row.names = c("1", 
"2", "3", "4", "5", "6"))