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我已经尝试了几个小时，但我无法弄清楚。我有一个包含主题和条件的数据框df1，我想从中排除具有特定值的观察值（变量“值”中小于 3）df2。我无法使其工作，因为我需要从df1不同级别的组合中删除的两个变量。

这是df1：

df1 <- structure(list(subject = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,2L, 2L, 2L, 2L, 
                                  2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), 
                      condition = c("A", "A", "A", "B", "B", "B", "C", "C","C", "A", "A", 
                                    "A", "B", "B", "B", "C", "C", "C", "A", "A", "A","B", "B", "B", "C", "C", "C")), 
                 row.names = c(NA, -27L), class = c("tbl_df", "tbl", "data.frame"))

这是df2

df2 <- structure(list(subject = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L,4L, 4L, 4L, 5L, 5L, 5L), 
                      condition = c("A", "B", "C", "A", "B","C", "A", "B", "C", "A", "B", "C", "A", "B", "C"), 
                      value = c(10L, 8L, 7L, 3L, 8L, 5L, 3L, 3L, 9L, 8L, 7L, 8L, 10L, 6L, 2L)), 
                 row.names = c(NA,-15L), class = c("tbl_df", "tbl", "data.frame"))

我想删除df1所有主题和条件的组合，其值低于 3，所以这将是最终的 df：

df3 <- structure(list(subject = c(2L, 3L, 3L, 5L), 
                      condition = c("A","A", "B", "C")), 
                 row.names = c(NA, -4L), 
                 class = c("tbl_df","tbl", "data.frame"))

到目前为止，我一直在这样做，但我不能再这样做了，因为我有数百行......

df3 <- df1 %>% filter(!(subject==2 & condition=="A" |
                        subject==3 & (condition=="A" | condition=="B") |
                        subject==5 & condition=="C"))
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1 回答 1

2

您的示例结果df3与您用于派生它的代码冲突，因此这里是dplyr您想要的每种解释的解决方案df3。

注意：只有当您

...排除具有特定值（小于[或等于] 3 在来自 df2.

所以我使用不等式<= 3而不是< 3.

第一次解释df3

获取版本df3

# A tibble: 4 x 2
  subject condition
    <int> <chr>    
1       2 A        
2       3 A        
3       3 B        
4       5 C

您在此处提供的示例结果

我想在 df1 中删除所有低于 3 的主题和条件的组合，所以这将是最终的 df：

df3 <- structure(list(subject = c(2L, 3L, 3L, 5L), 
                      condition = c("A","A", "B", "C")), 
                 row.names = c(NA, -4L), 
                 class = c("tbl_df","tbl", "data.frame"))

只需filter()使用df2：

library(dplyr)


# ...
# Code to generate 'df1' and 'df2'.
# ...

df3 <- df2 %>% filter(value <= 3)

第二次解释df3

但是，我看来您实际上想要以下版本df3

# A tibble: 18 x 2
   subject condition
     <int> <chr>    
 1       1 A        
 2       1 A        
 3       1 A        
 4       1 B        
 5       1 B        
 6       1 B        
 7       1 C        
 8       1 C        
 9       1 C        
10       2 B        
11       2 B        
12       2 B        
13       2 C        
14       2 C        
15       2 C        
16       3 C        
17       3 C        
18       3 C

你在这里得出的：

df3 <- df1 %>% filter(!(subject==2 & condition=="A" |
                        subject==3 & (condition=="A" |condition=="B") |
                        subject==5 & condition=="C"))

在这种情况下，您应该使用anti_join()ed版本：df1filter()df2

library(dplyr)


# ...
# Code to generate 'df1' and 'df2'.
# ...


df3 <- df1 %>%
  anti_join(df2 %>% filter(value <= 3), by = c("subject", "condition"))
于 2021-07-16T15:10:54.830 回答