如何在R中创建整数序列？

huangapple
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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：

  1. id    year
  2. 001   1988
  3. 001   1989
  4. 001   1990
  5. 001   1991
  6. 001   1992
  7. 002   1972
  8. 002   1973
  9. 002   1974
  10. 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：

  1. df <- data.frame(id = c("001", "001", "001", "001", "001", "002", "002", "002", "002"),
  2.                  year = c(1988, 1988, 1988, 1988, 1988, 1972, 1972, 1972, 1972))
英文:

I've got dataframe like so:

  1. id    year
  2. 001   1988
  3. 001   1988
  4. 001   1988
  5. 001   1988
  6. 001   1988
  7. 002   1972
  8. 002   1972
  9. 002   1972
  10. 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:

  1. id    year
  2. 001   1988
  3. 001   1989
  4. 001   1990
  5. 001   1991
  6. 001   1992
  7. 002   1972
  8. 002   1973
  9. 002   1974
  10. 002   1975

How do I do this? A tidyverse solution would be preferred. Thanks.

My data:

  1. df &lt;- data.frame(id = c(&quot;001&quot;, &quot;001&quot;, &quot;001&quot;, &quot;001&quot;, &quot;001&quot;, &quot;002&quot;, &quot;002&quot;, &quot;002&quot;, &quot;002&quot;),
  2.                  year = c(1988, 1988, 1988, 1988, 1988, 1972, 1972, 1972, 1972))

答案1

得分: 3

使用 seq 函数：

  1. library(dplyr)
  2. df %>%
  3.   mutate(year = seq(first(year), length.out = n()), .by = id)
  5. #   id year
  6. # 1  1 1988
  7. # 2  1 1989
  8. # 3  1 1990
  9. # 4  1 1991
  10. # 5  1 1992
  11. # 6  2 1972
  12. # 7  2 1973
  13. # 8  2 1974
  14. # 9  2 1975

希望这对您有所帮助。

英文:

With seq:

  1. library(dplyr)
  2. df %&gt;% 
  3.   mutate(year = seq(first(year), length.out = n()), .by = id)
  5. #   id year
  6. # 1  1 1988
  7. # 2  1 1989
  8. # 3  1 1990
  9. # 4  1 1991
  10. # 5  1 1992
  11. # 6  2 1972
  12. # 7  2 1973
  13. # 8  2 1974
  14. # 9  2 1975

答案2

得分: 2

一个“基础”解决方案：

  1. transform(df, year = ave(year, id, FUN = \(x) seq_along(x) + x[1] - 1))
  3. #   id year
  4. # 1  1 1988
  5. # 2  1 1989
  6. # 3  1 1990
  7. # 4  1 1991
  8. # 5  1 1992
  9. # 6  2 1972
  10. # 7  2 1973
  11. # 8  2 1974
  12. # 9  2 1975
英文:

A base solution:

  1. transform(df, year = ave(year, id, FUN = \(x) seq_along(x) + x[1] - 1))
  3. #   id year
  4. # 1  1 1988
  5. # 2  1 1989
  6. # 3  1 1990
  7. # 4  1 1991
  8. # 5  1 1992
  9. # 6  2 1972
  10. # 7  2 1973
  11. # 8  2 1974
  12. # 9  2 1975

答案3

得分: 1

对于完整性，可以使用data.table方法：

  1. library(data.table)
  3. setDT(df)
  5. 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)：

  1. df[, year := first(year) + 1:.N - 1, id]
英文:

For completeness, a data.table approach

  1. library(data.table)
  3. setDT(df)
  5. df[, year := .SD + 1:.N - 1, id]

or in case not all years are the same initially per id then you need first(year)

  1. df[, year := first(year) + 1:.N - 1, id]

huangapple
  • 本文由 发表于 2023年7月17日 21:43:48
  • 转载请务必保留本文链接：https://go.coder-hub.com/76705072.html
  • dataframe
  • dplyr
  • r
  • sequential
  • tidyverse
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