通过与上面的单元进行比较来更改单元格值

发布于 2025-02-13 15:33:02 字数 568 浏览 3 评论 0原文

我有以下数据框架：

ID year level
1  2000  NA
1  2001  3
1  2002  3
1  2003  2
1  2004  1
2  2000  1
2  2001  3
2  2002  3
2  2003  3
2  2004  3

如果上一个较小，我想根据上一个在“级别”列中按ID更新每个值。

数据帧应该看起来像

ID year level
1  2000  NA
1  2001  3
1  2002  3
1  2003  2
1  2004  1
2  2000  1
2  2001  1
2  2002  1
2  2003  1
2  2004  1

我尝试使用从数据表的移动来尝试的，但它只会更改一个单元格。我得到了这个结果

ID year level
1  2000  NA
1  2001  3
1  2002  3
1  2003  2
1  2004  1
2  2000  1
2  2001  1
2  2002  3
2  2003  3
2  2004  3

原文

I have the following dataframe:

ID year level
1  2000  NA
1  2001  3
1  2002  3
1  2003  2
1  2004  1
2  2000  1
2  2001  3
2  2002  3
2  2003  3
2  2004  3

I want to update each value in "level" column by ID based on the previous one if the previous one is smaller.

the dataframe should look like this

ID year level
1  2000  NA
1  2001  3
1  2002  3
1  2003  2
1  2004  1
2  2000  1
2  2001  1
2  2002  1
2  2003  1
2  2004  1

I tried using shift from data table but it only changes one cell.
I got this result

ID year level
1  2000  NA
1  2001  3
1  2002  3
1  2003  2
1  2004  1
2  2000  1
2  2001  1
2  2002  3
2  2003  3
2  2004  3

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

青柠芒果 2025-02-20 15:33:02

library(data.table)
df <- data.frame(
          ID = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L),
          year = c(2000L,2001L,2002L,2003L,2004L,
                   2000L,2001L,2002L,2003L,2004L),
       level = c(NA, 3L, 3L, 2L, 1L, 1L, 1L, 1L, 1L, 1L)
  )

setDT(df)[!is.na(level), level := cummin(level), by = ID][]
#>     ID year level
#>  1:  1 2000    NA
#>  2:  1 2001     3
#>  3:  1 2002     3
#>  4:  1 2003     2
#>  5:  1 2004     1
#>  6:  2 2000     1
#>  7:  2 2001     1
#>  8:  2 2002     1
#>  9:  2 2003     1
#> 10:  2 2004     1

^由

library(data.table)
df <- data.frame(
          ID = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L),
          year = c(2000L,2001L,2002L,2003L,2004L,
                   2000L,2001L,2002L,2003L,2004L),
       level = c(NA, 3L, 3L, 2L, 1L, 1L, 1L, 1L, 1L, 1L)
  )

setDT(df)[!is.na(level), level := cummin(level), by = ID][]
#>     ID year level
#>  1:  1 2000    NA
#>  2:  1 2001     3
#>  3:  1 2002     3
#>  4:  1 2003     2
#>  5:  1 2004     1
#>  6:  2 2000     1
#>  7:  2 2001     1
#>  8:  2 2002     1
#>  9:  2 2003     1
#> 10:  2 2004     1

^{Created on 2022-07-06 by the reprex package (v2.0.1)}

回复收藏 0 原文

昨迟人 2025-02-20 15:33:02

tidyverse带有accomeculate（）的解决方案来自purrr：

library(tidyverse)

df %>%
  group_by(ID) %>%
  mutate(level2 = accumulate(level, min, na.rm = TRUE)) %>%
  ungroup()

# # A tibble: 10 × 4
#       ID  year level level2
#    <int> <int> <int>  <int>
#  1     1  2000    NA     NA
#  2     1  2001     3      3
#  3     1  2002     3      3
#  4     1  2003     2      2
#  5     1  2004     1      1
#  6     2  2000     1      1
#  7     2  2001     3      1
#  8     2  2002     3      1
#  9     2  2003     3      1
# 10     2  2004     3      1

A tidyverse solution with accumulate() from purrr:

library(tidyverse)

df %>%
  group_by(ID) %>%
  mutate(level2 = accumulate(level, min, na.rm = TRUE)) %>%
  ungroup()

# # A tibble: 10 × 4
#       ID  year level level2
#    <int> <int> <int>  <int>
#  1     1  2000    NA     NA
#  2     1  2001     3      3
#  3     1  2002     3      3
#  4     1  2003     2      2
#  5     1  2004     1      1
#  6     2  2000     1      1
#  7     2  2001     3      1
#  8     2  2002     3      1
#  9     2  2003     3      1
# 10     2  2004     3      1

回复收藏 0 原文

~没有更多了~