[英]Is there a more efficient way to calculate the difference in months in R
I have a large data frame in a panel structure (201720 rows; 3 columns) which looks as follows:我在面板结构中有一个大数据框(201720 行;3 列),如下所示:
Name <- c("A", "A", "A", "B", "B", "B")
Inception <- c(as.Date("2007-12-31"), as.Date("2007-12-31"), as.Date("2007-12-31"),
as.Date("1990-12-31"), as.Date("1990-12-31"), as.Date("1990-12-31"))
Months <- c(as.Date("2010-01-01"), as.Date("2010-02-01"), as.Date("2010-03-01"),
as.Date("2010-01-01"), as.Date("2010-02-01"), as.Date("2010-03-01"))
df <- data.frame(Name, Inception, Months)
I want to calculate the difference in months of «Inception» and «Months» for each row and assign it to a new column named «Age».我想为每一行计算 «Inception» 和 «Months» 的月份差异,并将其分配给名为 «Age» 的新列。 If the result is negative, it should fill in with NA.
如果结果为负,则应填写 NA。 I came up with the following solution and it worked.
我想出了以下解决方案并且有效。 However, the computation of it is not very fast.
但是,它的计算速度不是很快。
for (i in 1:nrow(df)){
if(df[i,2]>df[i,3]){
df[i,"Age"] <- NA
} else {
df[i,"Age"] <- interval(df[i,2],
df[i,3]) %/% months(1)
}
}
Is there a more efficient way to calculate this difference?有没有更有效的方法来计算这种差异?
We can use case_when
我们可以使用
case_when
library(dplyr)
library(lubridate)
df <- df %>%
mutate(Age = case_when(Inception <= Months
~ interval(Inception, Months) %/% months(1)))
-output -输出
df
Name Inception Months Age
1 A 2007-12-31 2010-01-01 24
2 A 2007-12-31 2010-02-01 25
3 A 2007-12-31 2010-03-01 26
4 B 1990-12-31 2010-01-01 228
5 B 1990-12-31 2010-02-01 229
6 B 1990-12-31 2010-03-01 230
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