英文:
cannot pass argument to which with pipeline in R
问题
我想根据它们的属性筛选数据表的列。答案实际上基于Nera在这里的回答 https://stackoverflow.com/questions/7813578/convert-column-classes-in-data-table
# 例子DT <- data.table(x = c("a", "b", "c"),y = c(1L, 2L, 3L),z = c(1.1, 2.1, 3.1))# 使用嵌套结构按类别筛选changeCols <- colnames(DT)[which(as.vector(DT[,lapply(.SD, class)]) == "character")]changeCols# 使用管道按类别筛选DT[,lapply(.SD, class)] %>%as.vector(.) == "character" %>%which(.)# 出现错误：argument to 'which' is not logical# 简单地中断管道可以纠正错误cols <- DT[,lapply(.SD, class)] %>% as.vector(.) == "character"which(cols)
我不明白为什么在使用管道时会出现错误，而在中断管道时错误被解决。谢谢！
英文:
I want to filter columns of a data.table based on their attribute. The answer is actually based on Nera's answer here https://stackoverflow.com/questions/7813578/convert-column-classes-in-data-table
# exampleDT <- data.table(x = c("a", "b", "c"),y = c(1L, 2L, 3L),z = c(1.1, 2.1, 3.1))# filter by class with nested structurechangeCols <- colnames(DT)[which(as.vector(DT[,lapply(.SD, class)]) == "character")]changeCols# filter by class with pipelineDT[,lapply(.SD, class)] %>%as.vector(.) == "character" %>%which(.)# Error in which(.) : argument to 'which' is not logical# simply break the pipeline corrects the errorcols <- DT[,lapply(.SD, class)] %>% as.vector(.) == "character"which(cols)
I don't understand why there's an error when I use pipeline while the error was solved when breaking the pipe. Thanks!
答案1
得分: 2
你需要封装 as.vector(.) == "character" 这一部分，否则我认为被传递的只是字符串 "character"：
DT[, lapply(.SD, class)] %>%{as.vector(.) == "character"} %>%which(.)#x#1
在这种情况下，如果你将 %>% 替换为基本的 |>，将 . 替换为 _，然后使用 quote() 对其进行引用，你可以看到问题出在哪里（尽管基本管道不完全等同于 %>%，但这次用于调试问题非常有用）：
quote(DT[,lapply(.SD, class)] |>as.vector(x=_) == "character" |>which(x=_))##as.vector(x = DT[, lapply(.SD, class)]) == which(x = "character")
你也可以通过在 DT[i, j, by] 的 data.table j 参数内工作来避免这个问题：
DT[, which(lapply(.SD, class) == "character")]#x#1
英文:
You need to encapsulate the as.vector(.) == "character" bit, otherwise I think what is being piped forwards is just the string "character":
DT[,lapply(.SD, class)] %>%{as.vector(.) == "character"} %>%which(.)#x#1
On this occasion, you can see where it went wrong if you switch out the %>% for the base |>, . for _ and then quote() it (though the base pipe isn't identical to %>% it was useful to debug the issue this time).
quote(DT[,lapply(.SD, class)] |>as.vector(x=_) == "character" |>which(x=_))##as.vector(x = DT[, lapply(.SD, class)]) == which(x = "character")
You could also avoid this by just working inside the data.table j argument of DT[i, j, by] like:
DT[, which(lapply(.SD, class) == "character")]#x#1
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