英文:
Filter rows with positive and appropriate negative values
问题
在R中，我想要使用"first"列的group_by筛选出在"second"列中存在正值和相应的负值的行。
所以在下面的情况下，前两行和最后两行。
# 创建第一列first <- c(1,2,3,4,5,6,7,1,1,2)# 创建第二列second <- c(88,52,-28,-19,39,86,-31,10,-88,-52)df <- data.frame(first, second)
结果应该是
data.frame(first=c(1,2,1,2), second=c(88,52,-88,-52))
英文:
In R I want to filter with a group_by of column "first" the rows where there is a positive value in "second" and the appropriated negative value in "second".
So in the case below the first two rows and the last two rows.
# creating first columnfirst <- c(1,2,3,4,5,6,7,1,1,2)# creating second columnsecond <- c(88,52,-28,-19,39,86,-31,10,-88,-52)df<-data.frame(first, second)
Result should be
data.frame(first=c(1,2,1,2), second=c(88,52,-88,-52))
答案1
得分: 2
你可以在 ave 中使用 %in% 与 intersect。
df[ave(df$second, df$first, FUN=\(x) x %in% intersect(x, -x)) == 1,]# first second#1 1 88#2 2 52#9 1 -88#10 2 -52
英文:
You can use %in% with intersect in ave.
df[ave(df$second, df$first, FUN=\(x) x %in% intersect(x, -x)) == 1,]# first second#1 1 88#2 2 52#9 1 -88#10 2 -52
答案2
得分: 1
library(dplyr)df %>%mutate(abs = abs(second)) %>%group_by(first, abs) %>%filter(n() == 2 & (sum(second > 0) == 1 & sum(second < 0) == 1)) %>%ungroup()
英文:
First create a column representing the absolute values of second, then group_by both first and the abs column. filter for groups that have exactly two entries while one of them is positive and the other one is negative.
library(dplyr)df %>%mutate(abs = abs(second)) %>%group_by(first, abs) %>%filter(n() == 2 & (sum(second > 0) == 1 & sum(second < 0) == 1)) %>%ungroup()# A tibble: 4 × 3first second abs<dbl> <dbl> <dbl>1 1 88 882 2 52 523 1 -88 884 2 -52 52
答案3
得分: 1
这可能会有所帮助：
df %>%按first分组, abs_second = abs(second)) %>%筛选(n() == 2, sum(second) == 0) %>%解除分组() %>%选择(-abs_second)
这将产生以下结果：
# 一个数据框: 4 × 2first second1 1 882 2 523 1 -884 2 -52
英文:
Probably this could help
df %>%group_by(first, abs_second = abs(second)) %>%filter(n() == 2, sum(second) == 0) %>%ungroup() %>%select(-abs_second)
which gives
# A tibble: 4 × 2first second<dbl> <dbl>1 1 882 2 523 1 -884 2 -52
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