英文:
How to create sequence of integers in R?
问题
对于每个 id，我想将变量 year 转换为从第一个观察开始，每个连续行增加一的序列。期望的输出：
id year
001 1988
001 1989
001 1990
001 1991
001 1992
002 1972
002 1973
002 1974
002 1975
如何实现这个目标？我更喜欢使用 tidyverse 的解决方案。谢谢。
我的数据：
df <- data.frame(id = c("001", "001", "001", "001", "001", "002", "002", "002", "002"),
year = c(1988, 1988, 1988, 1988, 1988, 1972, 1972, 1972, 1972))
英文:
I've got dataframe like so:
id year
001 1988
001 1988
001 1988
001 1988
001 1988
002 1972
002 1972
002 1972
002 1972
For each id, I would like to convert the variable year into a sequence beginning from the first observation, increasing by one in each consecutive row. The desired output:
id year
001 1988
001 1989
001 1990
001 1991
001 1992
002 1972
002 1973
002 1974
002 1975
How do I do this? A tidyverse solution would be preferred. Thanks.
My data:
df <- data.frame(id = c("001", "001", "001", "001", "001", "002", "002", "002", "002"),
year = c(1988, 1988, 1988, 1988, 1988, 1972, 1972, 1972, 1972))
答案1
得分: 3
使用 seq 函数：
library(dplyr)
df %>%
mutate(year = seq(first(year), length.out = n()), .by = id)
# id year
# 1 1 1988
# 2 1 1989
# 3 1 1990
# 4 1 1991
# 5 1 1992
# 6 2 1972
# 7 2 1973
# 8 2 1974
# 9 2 1975
希望这对您有所帮助。
英文:
With seq:
library(dplyr)
df %>%
mutate(year = seq(first(year), length.out = n()), .by = id)
# id year
# 1 1 1988
# 2 1 1989
# 3 1 1990
# 4 1 1991
# 5 1 1992
# 6 2 1972
# 7 2 1973
# 8 2 1974
# 9 2 1975
答案2
得分: 2
一个“基础”解决方案：
transform(df, year = ave(year, id, FUN = \(x) seq_along(x) + x[1] - 1))
# id year
# 1 1 1988
# 2 1 1989
# 3 1 1990
# 4 1 1991
# 5 1 1992
# 6 2 1972
# 7 2 1973
# 8 2 1974
# 9 2 1975
英文:
A base solution:
transform(df, year = ave(year, id, FUN = \(x) seq_along(x) + x[1] - 1))
# id year
# 1 1 1988
# 2 1 1989
# 3 1 1990
# 4 1 1991
# 5 1 1992
# 6 2 1972
# 7 2 1973
# 8 2 1974
# 9 2 1975
答案3
得分: 1
对于完整性，可以使用data.table方法：
library(data.table)
setDT(df)
df[, year := .SD + 1:.N - 1, id]
或者如果最初每个id的年份不都相同，那么您需要使用first(year)：
df[, year := first(year) + 1:.N - 1, id]
英文:
For completeness, a data.table approach
library(data.table)
setDT(df)
df[, year := .SD + 1:.N - 1, id]
or in case not all years are the same initially per id then you need first(year)
df[, year := first(year) + 1:.N - 1, id]
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论