我正在尝试将所有行保留在数据框中,但是请删除2行,但至少要保留2年的8行。
library(tidyverse)
forms <- data_frame( CASEID = rep(01012,5), VISIT = c(450, 450, 365, 365, 450), FORM = c(18, 8, 7, 2, 2), DTYvisit = c(2006, 2006, 2003, 2003, 2006) )
> forms # A tibble: 5 x 4
CASEID VISIT FORM YEAR
<dbl> <dbl> <dbl> <dbl>
1 1012 450 18 2006
2 1012 450 8 2006
3 1012 365 7 2003
4 1012 365 2 2003
5 1012 450 2 2004
6 1013 450 8 2003
7 1013 450 18 2003
8 1013 450 2 2003
9 1012 450 2 2009
关于如何丢弃不在FORM 8 DTyvisit的<2年范围内的FORM 2行的任何建议?
效果很好:
form2.matchedOnForm8 <- forms %>% group_by(CASEID) %>% filter(FORM == 8) %>% select(CASEID, VISIT, DTYvisit) %>% left_join(filter(forms, FORM == 2), by = c("CASEID", "VISIT", "DTYvisit")) %>% bind_rows(filter(forms, FORM != 2))
但是现在我正在失去观察力。
我需要以下内容:
library(tidyverse)
forms <- data_frame( CASEID = rep(01012,5), VISIT = c(450, 450, 365, 365, 450), FORM = c(18, 8, 7, 2, 2), DTYvisit = c(2006, 2006, 2003, 2003, 2006) )
> forms # A tibble: 5 x 4
CASEID VISIT FORM YEAR
<dbl> <dbl> <dbl> <dbl>
1 1012 450 18 2006
2 1012 450 8 2006
3 1012 365 7 2003
4 1012 450 2 2004
5 1013 450 8 2003
6 1013 450 18 2003
7 1013 450 2 2003
答案 0 :(得分:1)
这是一个使用outer
来计算给定的YEAR和FROM 8可能具有的所有YEAR值之间的差的解决方案。
min(abs(as.numeric(outer(df[df$FORM==8,'YEAR'],df[1,'YEAR'],'-'))))
[1] 0
df$diff <- apply(df, 1, function(x) min(as.numeric(outer(df[df$FORM==8,'YEAR',drop=TRUE],as.numeric(x['YEAR']),'-'))))
library(dplyr)
df %>% group_by(CASEID) %>%
filter(!(FORM==2 & abs(diff)>2))
df <- read.table(text="
CASEID VISIT FORM YEAR
1 1012 450 18 2006
2 1012 450 8 2006
3 1012 365 7 2003
4 1012 365 2 2003
5 1012 450 2 2004
6 1013 450 8 2003
7 1013 450 18 2003
8 1013 450 2 2003
9 1012 450 2 2009
",header=T, stringsAsFactors = F)