为 R 中的回归创建超前年和滞后年假人 [英] Create lead and lag year dummies for regression in R
问题描述
这是一个示例数据框，其中 PRE5_id1、POST5_id1、PRE5_id2、POST5_id2 是我想要获取的变量.我正在寻找一个超前和滞后值，在自然死亡之前的年份 (PRE5) 和自然死亡年份 (POST5) 之后的 5 年中将有五个值为 1.我不确定在创建这些 PRE 和 POST 变量时如何留在国家组内，在这种情况下，PRE 和 POST 变量仅在同一国家/地区内变为 +5 和 -5.
This is an example data frame, where PRE5_id1,POST5_id1, PRE5_id2, POST5_id2 are the variables that I would like to get. I am looking for a lead and lag value which will have five values of 1 in the years before natural death (PRE5) and 5 years after the year of natural death (POST5). I am not sure how to stay within the group of country when creating these PRE and POST variables, in which case the PRE and POST variables go to +5 and -5 only within the same country.
我计划对每个 ID 进行单独的回归(我的数据集中总共有 69 次自然死亡，因此高达 ID69)并为每个回归包括 PRE5 和 POST5，如下所示:lm(gdp.growth.rate~country+year+PRE5_id1+POST5_id1) 所以如果无论如何在回归中创建这些 PRE 和 POST 假人也可以工作.
I am planning to do a separate regressions for each ID (there are overall 69 natural deaths in my dataset and hence up to ID69) and to include PRE5 and POST5 for every regression, something like this: lm(gdp.growth.rate~country+year+PRE5_id1+POST5_id1) so if there is anyway to create these PRE and POST dummies in the regression that could also work.
> df <- data.frame(country = rep("Angola",length(20)), year=c(1940:1959), leader = c("David", "NA", "NA", "NA","Henry","NA","Tom","NA","Chris","NA","NA","NA","NA","Alia","NA","NA","NA","NA","NA","NA"), natural.death = c(0,NA,NA,NA,0,NA,1,NA,0,NA,NA,NA,NA,1,NA,NA,NA,NA,NA,NA),gdp.growth.rate=c(1:20),
+ id1=c(0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0),
+ id2=c(0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0),
+ PRE5_id1=c(0,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0),
+ PRE5_id2=c(0,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0),
+ POST5_id1=c(0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,0),
+ POST5_id2=c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,0))
> df
country year leader natural.death gdp.growth.rate id1 id2 PRE5_id1 PRE5_id2 POST5_id1 POST5_id2
1 Angola 1940 David 0 1 0 0 0 0 0 0
2 Angola 1941 NA NA 2 0 0 1 0 0 0
3 Angola 1942 NA NA 3 0 0 1 0 0 0
4 Angola 1943 NA NA 4 0 0 1 0 0 0
5 Angola 1944 Henry 0 5 0 0 1 0 0 0
6 Angola 1945 NA NA 6 0 0 1 0 0 0
7 Angola 1946 Tom 1 7 1 0 0 0 0 0
8 Angola 1947 NA NA 8 0 0 0 0 1 0
9 Angola 1948 Chris 0 9 0 0 0 1 1 0
10 Angola 1949 NA NA 10 0 0 0 1 1 0
11 Angola 1950 NA NA 11 0 0 0 1 1 0
12 Angola 1951 NA NA 12 0 0 0 1 1 0
13 Angola 1952 NA NA 13 0 0 0 1 0 0
14 Angola 1953 Alia 1 14 0 1 0 0 0 0
15 Angola 1954 NA NA 15 0 0 0 0 0 1
16 Angola 1955 NA NA 16 0 0 0 0 0 1
17 Angola 1956 NA NA 17 0 0 0 0 0 1
18 Angola 1957 NA NA 18 0 0 0 0 0 1
19 Angola 1958 NA NA 19 0 0 0 0 0 1
20 Angola 1959 NA NA 20 0 0 0 0 0 0
任何帮助将不胜感激.谢谢！
Any help will be appreciated. Thanks!
在尝试下面的答案之一并将原始 df 修改为以下内容(见下文)后，我得到以下 output.df(见下文):
After trying one of the answers from below and modified the original df to the following (see below), I get the following output.df (see below):
> df <- data.frame(country=c("Angola","Angola","Angola","Angola",
+ "Angola","Angola","Angola","Angola",
+ "Angola","Angola","US","US","US","US",
+ "US","US","US","US","US","US"),
+ year=c(1940:1949,1940:1949),
+ leader = c("David", "NA", "NA", "NA","Henry","NA",
+ "Tom","NA","Chris","NA","NA","NA","NA",
+ "Alia","NA","NA","NA","NA","NA","NA"),
+ natural.death = c(0,NA,NA,NA,0,NA,1,NA,0,NA,NA,NA,NA,1,NA,NA,NA,NA,NA,NA),gdp.growth.rate=c(1:20),
+ id1=c(0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0),
+ id2=c(0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0))
> output.df
country year leader natural.death gdp.growth.rate id1 id2 id1.PRE
Angola.1 Angola 1940 David 0 1 0 0 0
Angola.2 Angola 1941 NA NA 2 0 0 1
Angola.3 Angola 1942 NA NA 3 0 0 1
Angola.4 Angola 1943 NA NA 4 0 0 1
Angola.5 Angola 1944 Henry 0 5 0 0 1
Angola.6 Angola 1945 NA NA 6 0 0 1
Angola.7 Angola 1946 Tom 1 7 1 0 0
Angola.8 Angola 1947 NA NA 8 0 0 0
Angola.9 Angola 1948 Chris 0 9 0 0 0
Angola.10 Angola 1949 NA NA 10 0 0 0
US.1 US 1940 NA NA 11 0 0 0
US.2 US 1941 NA NA 12 0 0 0
US.3 US 1942 NA NA 13 0 0 0
US.4 US 1943 Alia 1 14 0 1 0
US.5 US 1944 NA NA 15 0 0 0
US.6 US 1945 NA NA 16 0 0 0
US.7 US 1946 NA NA 17 0 0 0
US.8 US 1947 NA NA 18 0 0 0
US.9 US 1948 NA NA 19 0 0 0
US.10 US 1949 NA NA 20 0 0 0
id1.POST id2.PRE id2.POST
Angola.1 0 0 0
Angola.2 0 0 1
Angola.3 0 0 1
Angola.4 0 0 1
Angola.5 0 0 1
Angola.6 0 0 1
Angola.7 0 0 0
Angola.8 1 0 0
Angola.9 1 0 0
Angola.10 1 0 0
US.1 0 1 0
US.2 1 1 0
US.3 1 1 0
US.4 1 0 0
US.5 1 0 1
US.6 1 0 1
US.7 0 0 1
US.8 0 0 1
US.9 0 0 1
US.10 0 0 0
推荐答案
一种使用基础 R 的方法.我们创建了一个函数 generate_dummy，它为每个 "id" 返回两列> 带有 PRE 和 POST 数据的列.
One approach using base R. We create a function generate_dummy which returns two columns for each "id" column with PRE and POST data.
generate_dummy <- function(x) {
inds <- which(x == 1)
if(length(inds) == 1) {
vec <- seq_along(x)
data.frame(PRE = +(vec > (inds - 6) & vec < (inds)),
POST = +(vec > (inds) & vec < (inds + 6)))
}
else data.frame(PRE = rep(0, length(x)),POST = rep(0, length(x)))
}
#Columns which start with id
cols <- grep("^id", names(df), value = TRUE)
为了将其应用于每个国家，我们按国家/地区拆分数据并将 generate_dummy 函数应用于每个国家并合并结果.
To apply it for each country we split the data by country and apply generate_dummy function to each one of them and combine the results.
output <- cbind(df, do.call(rbind, lapply(split(df, df$country), function(x)
do.call(cbind, lapply(x[cols], generate_dummy)))))
row.names(output) <- NULL
output
# country year leader natural.death gdp.growth.rate id1 id2 id1.PRE id1.POST id2.PRE id2.POST
#1 Angola 1940 David 0 1 0 0 0 0 0 0
#2 Angola 1941 NA NA 2 0 0 1 0 0 0
#3 Angola 1942 NA NA 3 0 0 1 0 0 0
#4 Angola 1943 NA NA 4 0 0 1 0 0 0
#5 Angola 1944 Henry 0 5 0 0 1 0 0 0
#6 Angola 1945 NA NA 6 0 0 1 0 0 0
#7 Angola 1946 Tom 1 7 1 0 0 0 0 0
#8 Angola 1947 NA NA 8 0 0 0 1 0 0
#9 Angola 1948 Chris 0 9 0 0 0 1 1 0
#10 Angola 1949 NA NA 10 0 0 0 1 1 0
#11 Angola 1950 NA NA 11 0 0 0 1 1 0
#12 Angola 1951 NA NA 12 0 0 0 1 1 0
#13 Angola 1952 NA NA 13 0 0 0 0 1 0
#14 Angola 1953 Alia 1 14 0 1 0 0 0 0
#15 Angola 1954 NA NA 15 0 0 0 0 0 1
#16 Angola 1955 NA NA 16 0 0 0 0 0 1
#17 Angola 1956 NA NA 17 0 0 0 0 0 1
#18 Angola 1957 NA NA 18 0 0 0 0 0 1
#19 Angola 1958 NA NA 19 0 0 0 0 0 1
#20 Angola 1959 NA NA 20 0 0 0 0 0 0
数据
df <- data.frame(country = rep("Angola",length(20)), year=c(1940:1959),
leader = c("David", "NA", "NA", "NA","Henry","NA","Tom","NA","Chris","NA",
"NA","NA","NA","Alia","NA","NA","NA","NA","NA","NA"),
natural.death = c(0,NA,NA,NA,0,NA,1,NA,0,NA,NA,NA,NA,1,NA,NA,NA,NA,NA,NA),
gdp.growth.rate=c(1:20),
id1=c(0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0),
id2=c(0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0))
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