英文:
How to remove duplicate rows in R based on condition?
问题
以下是翻译好的内容：
我有以下数据：
df <- data.frame(id = c("001", "001", "001", "002", "002", "003", "003"),
x = c(0, 0, 0, 0, 1, 0, 1))
数据的性质是，某些 id 只可能有 x = 0 的行。在给定的 id 中，如果 x = 1，那么只会在该 id 的最后一行出现。我想要删除每个 id 的重复行，但是如果某个 id 的情况是 x = 1，则只保留该行。
期望的输出：
id x
001 0
002 1
003 1
最好使用 tidyverse 来解决。谢谢！
英文:
I have the following data:
df <- data.frame(id = c("001", "001", "001", "002", "002", "003", "003"),
x = c(0, 0, 0, 0, 1, 0, 1))
id x
001 0
001 0
001 0
002 0
002 1
003 0
003 1
The nature of the data is such that it is possible for some id to only have x = 0 rows. In the case where x = 1 for a given id, it only occurs once, and that too in the last row for that id. I want to remove duplicate rows for each id, but in case x = 1 for an id, I want to keep only that row.
The desired output:
id x
001 0
002 1
003 1
A tidyverse solution is preferable. Thanks!
答案1
得分: 5
在基本的 R 中，你可以使用 aggregate 函数：
aggregate(x ~ id, df, max)
id x
1 001 0
2 002 1
3 003 1
英文:
in base R you could use aggregate function:
aggregate(x ~ id, df, max)
id x
1 001 0
2 002 1
3 003 1
答案2
得分: 4
可能是slice_max
df %>%
slice_max(x, by = id) %>%
distinct()
或者（来自@r2evans的评论）
df %>%
slice_max(x, by = id, with_ties = FALSE)
这将得到
id x
1 001 0
2 002 1
3 003 1
英文:
Probably slice_max
df %>%
slice_max(x, by = id) %>%
distinct()
or (as comments from @r2evans)
df %>%
slice_max(x, by = id, with_ties = FALSE)
which gives
id x
1 001 0
2 002 1
3 003 1
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