这是一个可能的解决方案。您还需要为ID
.
样本数据
df <- data.frame("ID" = c(1,1,2,2,2,2), "Year" = c(2012, 2015,2012,2013,2015,2016), "AH" = c(1, NA, 1,1,1,1))
代码ID == 2
current_year <- df[df$ID == 2, "Year"]
n <- length(current_year)
i = 0
df$retention <- 0
while(i<n){
i = i + 1
df_temp <- subset(df, df$Year == (current_year[i]+1) & df$ID == 2 )
n_temp <- nrow(df_temp)
if(n_temp>0)
if(df[df$Year == (current_year[i]+1), "ID" ] == 2 & df[df$Year == (current_year[i]+1), "AH"] == 1)
{
df[df$Year == current_year[i] & df$ID == 2, "retention"] <- 1
}
}
编辑 - 更通用的代码
如果你想对所有人进行概括ID
,你需要创建一个唯一 ID 的列表,计算 ID 的数量并执行一个 while 循环。下面的代码
df <- data.frame("ID" = c(1,1,2,2,2,2), "Year" = c(2012, 2015,2012,2013,2015,2016), "AH" = c(1, NA, 1,1,1,1))
ID_list <- unique(df$ID)
n_ID <- length(ID_list)
j = 0
while(j < n_ID)
{
j = j + 1
current_year <- df[df$ID == ID_list[j], "Year"]
n <- length(current_year)
i = 0
df$retention <- 0
while(i<n){
i = i + 1
df_temp <- subset(df, df$Year == (current_year[i]+1) & df$ID == ID_list[j] )
n_temp <- nrow(df_temp)
if(n_temp>0)
if(df[df$Year == (current_year[i]+1), "ID" ] == ID_list[j] & df[df$Year == (current_year[i]+1), "AH"] == 1)
{
df[df$Year == current_year[i] & df$ID == ID_list[j], "retention"] <- 1
}
}
}