数据
我在df中总共有10列，对于3到10列，我需要检查是否有值0或1行。如果任何一行的值为1，则选择id (R language)
test<-df %>%
group_by(id) %>%
filter(all(if_any(sympt3:sympt10, as.logical)))
id date
symptom1
1 2011-08-27 0
2 2012-02-21 1
3 2012-07-12 0
4 2012-07-25 0
5 2010-12-31 0
6 2011-01-02 0
7 2012-08-21 0
8 2013-03-26 0
9 2012-11-07 0
10 2010-11-22 0
symptom 2 symptom 3
1 0 0
2 1 1
3 0 0
4 0 0
5 0 0
6 0 1
7 0 0
8 1 0
9 0 0
10 0 0
symptom 5. symptom 6
1 1 0
2 0 1
3 0 0
4 0 0
5 0 0
6 1 0
7 0 1
8 0 0
9 0 0
10 0 1
symptom 7 symptom 8
1 1 0
2 0 1
3 0 0
4 0 0
5 1 0
6 0 0
7 1 1
8 0 0
9 0 0
10 1 0
我不确定我是否正在选择从我需要的任何列中具有值1的id ?
感谢您的帮助
试试这些选项
(I在& '不等于0& ';和& '等于1& ';之间变化，取决于所有需要。它们都是有效的，为了便于演示，我选择改变它们
基地R
dat[rowSums(subset(dat, select = grep("^symptom", names(dat), value = TRUE)) != 0) > 0,]
# id date symptom1 symptom2 symptom3 symptom5 symptom6 symptom7 symptom8
# 1 1 2011-08-27 0 0 0 1 0 1 0
# 2 2 2012-02-21 1 1 1 0 1 0 1
# 5 5 2010-12-31 0 0 0 0 0 1 0
# 6 6 2011-01-02 0 0 1 1 0 0 0
# 7 7 2012-08-21 0 0 0 0 1 1 1
# 8 8 2013-03-26 0 1 0 0 0 0 0
# 10 10 2010-11-22 0 0 0 0 1 1 0
dplyr
library(dplyr)
dat %>%
filter(if_any(starts_with("symptom"), ~ . == 1))
# id date symptom1 symptom2 symptom3 symptom5 symptom6 symptom7 symptom8
# 1 1 2011-08-27 0 0 0 1 0 1 0
# 2 2 2012-02-21 1 1 1 0 1 0 1
# 3 5 2010-12-31 0 0 0 0 0 1 0
# 4 6 2011-01-02 0 0 1 1 0 0 0
# 5 7 2012-08-21 0 0 0 0 1 1 1
# 6 8 2013-03-26 0 1 0 0 0 0 0
# 7 10 2010-11-22 0 0 0 0 1 1 0
数据
dat <- structure(list(id = 1:10, date = c("2011-08-27", "2012-02-21", "2012-07-12", "2012-07-25", "2010-12-31", "2011-01-02", "2012-08-21", "2013-03-26", "2012-11-07", "2010-11-22"), symptom1 = c(0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), symptom2 = c(0L, 1L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L), symptom3 = c(0L, 1L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L), symptom5 = c(1L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L), symptom6 = c(0L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 1L), symptom7 = c(1L, 0L, 0L, 0L, 1L, 0L, 1L, 0L, 0L, 1L), symptom8 = c(0L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L)), class = "data.frame", row.names = c(NA, -10L))