代码之家 › 专栏 › 技术社区 › iod

在dplyr::mutate中从ifelse中的另一列传输值

mutate dplyr if-statement r

1

iod · 技术社区 · 7 年前

dd (问题底部的dput):

# A tibble: 6 x 2
# Groups:   Date [5]
  Date     keeper
  <chr>    <lgl> 
1 1/1/2018 TRUE  
2 2/1/2018 TRUE  
3 3/1/2018 FALSE 
4 4/1/2018 FALSE 
5 3/1/2018 TRUE  
6 5/1/2018 TRUE

注意,它已按日期分组。我正在尝试创建另一列,如果组中只有一行,则该列将“keeper”变为TRUE,否则保留keeper的值。这看起来相当简单,但当我尝试这一点时,我得到了以下结果:

dd %>% mutate(moose=ifelse(n()==1,TRUE,keeper))
# A tibble: 6 x 3
# Groups:   Date [5]
  Date     keeper moose
  <chr>    <lgl>  <lgl>
1 1/1/2018 TRUE   TRUE 
2 2/1/2018 TRUE   TRUE 
3 3/1/2018 FALSE  FALSE
4 4/1/2018 FALSE  TRUE 
5 3/1/2018 TRUE   FALSE
6 5/1/2018 TRUE   TRUE

预期产出:

  Date     keeper moose
  <chr>    <lgl>  <lgl>
1 1/1/2018 TRUE   TRUE 
2 2/1/2018 TRUE   TRUE 
3 3/1/2018 FALSE  FALSE
4 4/1/2018 FALSE  TRUE 
5 3/1/2018 TRUE   TRUE
6 5/1/2018 TRUE   TRUE

(注第5行)

以下是数据帧的dput:

dd<-structure(list(Date = c("1/1/2018", "2/1/2018", "3/1/2018", "4/1/2018", 
"3/1/2018", "5/1/2018"), keeper = c(TRUE, TRUE, FALSE, FALSE, 
TRUE, TRUE)), class = c("grouped_df", "tbl_df", "tbl", "data.frame"
), row.names = c(NA, -6L), vars = "Date", drop = TRUE, indices = list(
    0L, 1L, c(2L, 4L), 3L, 5L), group_sizes = c(1L, 1L, 2L, 1L, 
1L), biggest_group_size = 2L, labels = structure(list(Date = c("1/1/2018", 
"2/1/2018", "3/1/2018", "4/1/2018", "5/1/2018")), class = "data.frame", row.names = c(NA, 
-5L), vars = "Date", drop = TRUE, indices = list(0L, 1L, 2L, 
    4L, 3L, 5L), group_sizes = c(1L, 1L, 1L, 1L, 1L, 1L), biggest_group_size = 1L, labels = structure(list(
    Date = c("1/1/2018", "2/1/2018", "3/1/2018", "3/1/2018", 
    "4/1/2018", "5/1/2018"), keeper = c(TRUE, TRUE, FALSE, TRUE, 
    FALSE, TRUE)), class = "data.frame", row.names = c(NA, -6L
), vars = c("Date", "keeper"), drop = TRUE, .Names = c("Date", 
"keeper")), .Names = "Date"), .Names = c("Date", "keeper"))

当我继续使用这个数据帧时,我发现如果我首先创建一个列 n add_count ,并参考我的 ifelse n() n() 给我同样的结果?

1 回复 | 直到 7 年前

1

akrun 7 年前

有一种循环效应。对于 ifelse length 属于 n() 是1。第二个论点 TRUE 长度为1。因此,这方面存在不匹配 长度 第三个参数“keeper”是 长度 n() . 这造成了循环利用的不平衡。OP在附录中提到,如果创建了一列,那么问题就不存在了。原因是,一旦创建了列 长度 “n”列的值不是1,而是 n()

dd %>% 
   mutate(moose = ifelse(rep(n(), n()) == 1, TRUE, keeper))
# A tibble: 6 x 3
# Groups:   Date [5]
#  Date     keeper moose
#  <chr>    <lgl>  <lgl>
#1 1/1/2018 TRUE   TRUE 
#2 2/1/2018 TRUE   TRUE 
#3 3/1/2018 FALSE  FALSE
#4 4/1/2018 FALSE  TRUE 
#5 3/1/2018 TRUE   TRUE 
#6 5/1/2018 TRUE   TRUE

而且, 属于 n() if/else

dd %>% 
    mutate(moose = if(n()==1) TRUE else keeper)
# A tibble: 6 x 3
# Groups:   Date [5]
#  Date     keeper moose
#  <chr>    <lgl>  <lgl>
#1 1/1/2018 TRUE   TRUE 
#2 2/1/2018 TRUE   TRUE 
#3 3/1/2018 FALSE  FALSE
#4 4/1/2018 FALSE  TRUE 
#5 3/1/2018 TRUE   TRUE 
#6 5/1/2018 TRUE   TRUE