我有一个简单的数据集。
structure(list(ID = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 4L, 4L,
4L, 5L, 5L), Primrely = c(0L, 2L, 1L, 1L, 1L, 1L, 3L, 4L, 4L,
3L, 1L, 2L, 2L), Primset = c(-4L, -3L, 1L, 2L, -4L, 5L, 3L, 1L,
2L, -4L, -2L, -3L, 3L), Primvalue = c(45L, 5L, 6L, 15L, 53L,
45L, 44L, 65L, 1L, 5L, 1L, 12L, 5L), Secrely = c(5L, 7L, 2L,
1L, 2L, 0L, 4L, 5L, 1L, 1L, 1L, 0L, 2L), Secset = c(-3L, 1L,
2L, -2L, -3L, 2L, 5L, 7L, 7L, 4L, 3L, 2L, 1L), Secvalue = c(38L,
-2L, -1L, 8L, 46L, 38L, 37L, 58L, -6L, -2L, -6L, 5L, -2L), Desired = structure(c(NA,
1L, NA, NA, 2L, 2L, NA, NA, NA, NA, NA, 1L, 1L), .Label = c("Primary",
"Secondary"), class = "factor")), .Names = c("ID", "Primrely",
"Primset", "Primvalue", "Secrely", "Secset", "Secvalue", "Desired"
), class = "data.frame", row.names = c(NA, -13L))
ID Primrely Primset Primvalue Secrely Secset Secvalue Desired
1 1 0 -4 45 5 -3 38 <NA>
2 1 2 -3 5 7 1 -2 Primary
3 1 1 1 6 2 2 -1 <NA>
4 1 1 2 15 1 -2 8 <NA>
5 2 1 -4 53 2 -3 46 Secondary
6 2 1 5 45 0 2 38 Secondary
7 2 3 3 44 4 5 37 <NA>
8 3 4 1 65 5 7 58 <NA>
9 4 4 2 1 1 7 -6 <NA>
10 4 3 -4 5 1 4 -2 <NA>
11 4 1 -2 1 1 3 -6 <NA>
12 5 2 -3 12 0 2 5 Primary
13 5 2 3 5 2 1 -2 Primary
对于每个ID，我想选择符合条件(Prim = primary, Sec = secondary)的行：如果Primrely为0或2且{{1 } {是Primset，为每个-3:3选择所有行。如果给定ID没有符合主要条件的行，请选择符合次要条件ID的行。理想情况下，我想添加一列(Secrely is 0 or 2 and Secset is -3:3)来表明符合哪些条件(Desired)。
我一直在使用(primary/secondary/NA)和ifelse函数而没有太多运气，主要是因为我不知道如果命令if else加入给定的R主要标准已经满足（例如ID符合第二个标准，但不需要它，因为它已经符合第一个标准）。换句话说，如果“主要”出现在给定的ID中，它将胜过满足的所有“次要”标准。我很感激任何建议。
答案 0 :(得分：2)
如果我现在正确理解你：
（在步骤中留下来向您展示我在做什么，您可以删除它们和/或如果您愿意，可以一步完成）
dat <- structure(list(ID = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 4L, 4L,
4L, 5L, 5L), Primrely = c(0L, 2L, 1L, 1L, 1L, 1L, 3L, 4L, 4L,
3L, 1L, 2L, 2L), Primset = c(-4L, -3L, 1L, 2L, -4L, 5L, 3L, 1L,
2L, -4L, -2L, -3L, 3L), Primvalue = c(45L, 5L, 6L, 15L, 53L,
45L, 44L, 65L, 1L, 5L, 1L, 12L, 5L), Secrely = c(5L, 7L, 2L,
1L, 2L, 0L, 4L, 5L, 1L, 1L, 1L, 0L, 2L), Secset = c(-3L, 1L,
2L, -2L, -3L, 2L, 5L, 7L, 7L, 4L, 3L, 2L, 1L), Secvalue = c(38L,
-2L, -1L, 8L, 46L, 38L, 37L, 58L, -6L, -2L, -6L, 5L, -2L), Desired = structure(c(NA,
1L, NA, NA, 2L, 2L, NA, NA, NA, NA, NA, 1L, 1L), .Label = c("Primary",
"Secondary"), class = "factor")), .Names = c("ID", "Primrely",
"Primset", "Primvalue", "Secrely", "Secset", "Secvalue", "Desired"
), class = "data.frame", row.names = c(NA, -13L))
within(dat, {
Desired_step1 <- ifelse(Primrely %in% c(0,2) & Primset %in% -3:3,
1, ifelse(Secrely %in% c(0,2) & Secset %in% -3:3,
2, 3))
Desired_new <- factor(ave(Desired_step1, ID, FUN = function(x)
ifelse(x == min(x), x, NA)),
levels = 1:3, labels = c('Primary', 'Secondary', 'NA'))
Desired_step1 <- c('1'='Primary','2'='Secondary','3'=NA)[Desired_step1]
})
# ID Primrely Primset Primvalue Secrely Secset Secvalue Desired Desired_new Desired_step1
# 1 1 0 -4 45 5 -3 38 <NA> <NA> <NA>
# 2 1 2 -3 5 7 1 -2 Primary Primary Primary
# 3 1 1 1 6 2 2 -1 <NA> <NA> Secondary
# 4 1 1 2 15 1 -2 8 <NA> <NA> <NA>
# 5 2 1 -4 53 2 -3 46 Secondary Secondary Secondary
# 6 2 1 5 45 0 2 38 Secondary Secondary Secondary
# 7 2 3 3 44 4 5 37 <NA> <NA> <NA>
# 8 3 4 1 65 5 7 58 <NA> NA <NA>
# 9 4 4 2 1 1 7 -6 <NA> NA <NA>
# 10 4 3 -4 5 1 4 -2 <NA> NA <NA>
# 11 4 1 -2 1 1 3 -6 <NA> NA <NA>
# 12 5 2 -3 12 0 2 5 Primary Primary Primary
# 13 5 2 3 5 2 1 -2 Primary Primary Primary
答案 1 :(得分：0)
这是我的快速＆amp;假设您的data.frame名为df，则为脏解决方案。您可以自己完善它：
df$Desired <- ifelse((df$Primrely==0 | df$Primrely==2) & (df$Primset >= -3 & df$Primset <= 3),
"Primary",
NA)
idx <- is.na(df$Desired)
df$Desired[idx] <- ifelse((df$Secrely[idx]==0 | df$Secrely[idx]==2) & (df$Secset[idx] >= -3 & df$Secset[idx] <= 3),
"Secondary",
NA)