英文:
How to remove rows if column 1 value is lower than column 2 and column 3?
问题
我想知道如何在列1的值低于列2和列3时删除行。
以下是虚拟数据集的更新：
c <- c(10, 20, 30, 40, 50)b <- c(40, 2, 40, 10, 50)a <- c(10, 50, 0.0000891, 60, 100)id <- c("a", "b", "c", "d", "e")data <- data.frame(id, a, b, c)head(data)# id a b c#1 a 10 40 10#2 b 50 2 20#3 c 8.91e-05 1 0#4 d 60 10 40#5 e 100 50 50
在上述虚拟数据集中，我们可以看到例如：data$id "c"，列a的值低于列b和列c。
在条件删除后，以下是预期输出：
# id a b c#1 b 50 2 20#2 d 60 10 40#3 e 100 50 50
如果可能的话，更倾向于使用tidyverse方法。谢谢。
英文:
I would like to know how to remove rows if column 1 value is lower than column 2 and column 3 ?
Here is dummy datasets Updated:
c <- c(10, 20, 30, 40, 50)b <- c(40, 2, 40, 10, 50)a <- c(10, 50, 0.0000891, 60, 100)id <- c("a", "b", "c", "d", "e")data <- data.frame(id, a, b, c)head(data)# id a b c#1 a 10 40 10#2 b 50 2 20#3 c 8.91e-05 1 0#4 d 60 10 40#5 e 100 50 50
In the above dummy dataset we can see that for example: data$id "c", a column value is lower than "b" and "c" column.
After conditional removalhere is the expected output:
# id a b c#1 b 50 2 20#2 d 60 10 40#3 e 100 50 50
If its possible tidyverse approach more preferred. Thank you.
答案1
得分: 3
另一种允许使用 if_any 进行动态选择的方式：
library(dplyr)data |>filter(if_any(b:c, ~ a >= .x))# id a b c#1 a 10 40 10#2 b 50 2 20#3 d 60 10 40#4 e 100 50 50
（请注意，这里的代码部分没有进行翻译。）
英文:
Another way that allows dynamic selection with if_any:
library(dplyr)data |>filter(if_any(b:c, ~ a >= .x))# id a b c#1 a 10 40 10#2 b 50 2 20#3 d 60 10 40#4 e 100 50 50
答案2
得分: 2
以下是代码部分的翻译：
library(dplyr)# 原始数据集c <- c(10, 20, 30, 40, 50)b <- c(40, 2, 40, 10, 50)a <- c(10, 50, 1, 60, 100)id <- c("a", "b", "c", "d", "e")data <- data.frame(id, a, b, c)data %>%filter(a > b | a > c)#> # A tibble: 4 × 4#> id a b c#> <chr> <dbl> <dbl> <dbl>#> 1 a 10 40 10#> 2 b 50 2 20#> 3 d 60 10 40#> 4 e 100 50 50
更新后的数据集：
c <- c(10, 20, 30, 40, 50)b <- c(40, 2, 40, 10, 50)a <- c(10, 50, 0.0000891, 60, 100)id <- c("a", "b", "c", "d", "e")data <- data.frame(id, a, b, c)data %>%filter(a > b | a > c)#> id a b c#> 1 b 50 2 20#> 2 d 60 10 40#> 3 e 100 50 50
创建于2023年06月19日，使用reprex v2.0.2
英文:
Does this work for you?
Updated to reflect @Mael's comment.
library(dplyr)#original datasetc <- c(10, 20, 30, 40, 50)b <- c(40, 2, 40, 10, 50)a <- c(10, 50, 1, 60, 100)id <- c("a", "b", "c", "d", "e")data <- data.frame(id, a, b, c)data |>filter(a > b | a > c)#> # A tibble: 4 × 4#> id a b c#> <chr> <dbl> <dbl> <dbl>#> 1 a 10 40 10#> 2 b 50 2 20#> 3 d 60 10 40#> 4 e 100 50 50
With updated dataset:
c <- c(10, 20, 30, 40, 50)b <- c(40, 2, 40, 10, 50)a <- c(10, 50, 0.0000891, 60, 100)id <- c("a", "b", "c", "d", "e")data <- data.frame(id, a, b, c)data |>filter(a > b | a > c)#> id a b c#> 1 b 50 2 20#> 2 d 60 10 40#> 3 e 100 50 50
<sup>Created on 2023-06-19 with reprex v2.0.2</sup>
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