使用gre中的grep和子集的多个过滤器

Question

我尝试创建一个过滤器，使用grep和subset一起从数据集中删除行。

示例数据集：

id <- 1:10
problem <- c("a" , "b", "c", "d", "a","b","c","a", "b", "a")
solution1 <- c("eat", "sleep", "drink", "play", "sleep", "play", "play", "drink", "play", "eat")
solution2 <- c("read", "read", "eat", "drink", "eat", "sleep", "eat", "read", "eat", "play")
df <- c(id, problem, solution1, solution2)

我试图删除那些有问题的行＆＃34; a＆＃34;并且吃了＃34;在solution1或solution2中。

结果是它应该删除id 1,5和10。

我尝试过使用：

df <- subset(df, problem=="a" & !(grepl("eat", df)))

和

df <- df[!grepl("eat", df) & grepl("a", df$problem)]

似乎无法在StackOverflow或我用Google搜索的其他网站上找到类似的解决方案。

如果有人能提供帮助，我将不胜感激。谢谢！

Answer 1

首先，如果你想要一个数据帧，你应该使用data.frame，而不是c：

df <- data.frame(id, problem, solution1, solution2)

然后您可以像这样进行子集化（不需要使用子集本身）

df2 <- df[!(grepl("a", df$problem) & 
           (grepl("eat", df$solution1) |
            grepl("eat", solution2))),]

#   id problem solution1 solution2
# 2  2       b     sleep      read
# 3  3       c     drink       eat
# 4  4       d      play     drink
# 6  6       b      play     sleep
# 7  7       c      play       eat
# 8  8       a     drink      read
# 9  9       b      play       eat

Answer 2

我这样做：

df <- df[!(df$problem %in% "a" & (df$solution1 %in% "eat" | df$solution2 %in% "eat")),]

#   id problem solution1 solution2
# 2  2       b     sleep      read
# 3  3       c     drink       eat
# 4  4       d      play     drink
# 6  6       b      play     sleep
# 7  7       c      play       eat
# 8  8       a     drink      read
# 9  9       b      play       eat

如果比较确切的字符串，

正则表达式并不是必需的。使用%in%进行子集化将节省大量时间，因为它会比较向量。例如而不是"a"可能有c("a", "b", "c")等。