我有一个名为“confidence_table”的tibble。有谁知道为什么如果我尝试使用mutate动词添加新列，这不起作用？
# A tibble: 12 x 3
# Groups: Age [2]
Age Condition Prop
<fctr> <fctr> <dbl>
0 old 0.73993056
1 old 0.75590278
0 old 0.15069444
1 old 0.13090278
0 new 0.06388889
1 new 0.04965278
0 new 0.05902778
1 new 0.05416667
0 lure 0.23055556
1 lure 0.23645833
0 lure 0.13819444
1 lure 0.12013889
我在base r中使用了这个函数，它确实有效
confidence_table$Confidence <- as.factor(rep(c("HC", "LC"), times = 3, each = 2))
# A tibble: 12 x 4
# Groups: Age [2]
Age Condition Prop Confidence
<fctr> <fctr> <dbl> <fctr>
0 old 0.73993056 HC
1 old 0.75590278 HC
0 old 0.15069444 LC
1 old 0.13090278 LC
0 new 0.06388889 HC
1 new 0.04965278 HC
0 new 0.05902778 LC
1 new 0.05416667 LC
0 lure 0.23055556 HC
1 lure 0.23645833 HC
0 lure 0.13819444 LC
1 lure 0.12013889 LC
这是与基本r代码一起使用的预期输出。但是，如果我使用：
confidence_table <- confidence_table %>%
mutate(Confidence = rep(c("HC", "LC"), times = 3, each = 2))
它说：mutate_impl（.data，dots）中的错误：列置信度必须是长度6（组大小）或一个，而不是12
这有什么问题？
在这种情况下，错误消息实际上应该可以帮助您找出问题所在。请注意2 x 3 x 2 = 12。
confidence_table %>%
mutate(Confidence = rep(c("HC", "LC"), times = 3, each = 2))
# Error in mutate_impl(.data, dots) :
# Column `Confidence` must be length 6 (the group size) or one, not 12
正如评论中指出的那样，解决这个问题的一种方法是首先对ungroup。
confidence_table %>%
ungroup() %>%
mutate(Confidence = rep(c("HC", "LC"), times = 3, each = 2))
# # A tibble: 12 x 4
# Age Condition Prop Confidence
# <int> <chr> <dbl> <chr>
# 1 0 old 0.73993056 HC
# 2 1 old 0.75590278 HC
# 3 0 old 0.15069444 LC
# 4 1 old 0.13090278 LC
# 5 0 new 0.06388889 HC
# 6 1 new 0.04965278 HC
# 7 0 new 0.05902778 LC
# 8 1 new 0.05416667 LC
# 9 0 lure 0.23055556 HC
# 10 1 lure 0.23645833 HC
# 11 0 lure 0.13819444 LC
# 12 1 lure 0.12013889 LC
你也可以在没有ungrouping的情况下做到：
confidence_table %>%
mutate(Confidence = rep(c("HC", "LC"), times = 3)) # 2x3 = 6
# # A tibble: 12 x 4
# # Groups: Age [2]
# Age Condition Prop Confidence
# <int> <chr> <dbl> <chr>
# 1 0 old 0.73993056 HC
# 2 1 old 0.75590278 HC
# 3 0 old 0.15069444 LC
# 4 1 old 0.13090278 LC
# 5 0 new 0.06388889 HC
# 6 1 new 0.04965278 HC
# 7 0 new 0.05902778 LC
# 8 1 new 0.05416667 LC
# 9 0 lure 0.23055556 HC
# 10 1 lure 0.23645833 HC
# 11 0 lure 0.13819444 LC
# 12 1 lure 0.12013889 LC
另一种选择是将“条件”分组 - 可能是这样的：
confidence_table %>%
group_by(Condition) %>%
mutate(Confidence = c("HC", "LC")[cumsum(Age == 0)])
样本数据：
confidence_table <- structure(list(Age = c(0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L,
0L, 1L), Condition = c("old", "old", "old", "old", "new", "new",
"new", "new", "lure", "lure", "lure", "lure"), Prop = c(0.73993056,
0.75590278, 0.15069444, 0.13090278, 0.06388889, 0.04965278, 0.05902778,
0.05416667, 0.23055556, 0.23645833, 0.13819444, 0.12013889)), .Names = c("Age",
"Condition", "Prop"), row.names = c(NA, -12L), class = c("grouped_df",
"tbl_df", "tbl", "data.frame"), vars = "Age", drop = TRUE, indices = list(
c(0L, 2L, 4L, 6L, 8L, 10L), c(1L, 3L, 5L, 7L, 9L, 11L)), group_sizes = c(6L,
6L), biggest_group_size = 6L, labels = structure(list(Age = 0:1), row.names = c(NA,
-2L), class = "data.frame", vars = "Age", drop = TRUE, .Names = "Age"))