英文:
How to make lapply/sapply output a list in R?
问题
我正在尝试弄清楚如何使用lapply/sapply创建一个列表。根据文档，我了解到可以使用sapply和USE.NAMES = TRUE来实现，但我的示例没有起作用。
我需要将变量名中的空格替换为破折号，然后查找它们在输入列表中。我对这个列表没有直接的控制（它是R Shiny应用程序中的输入反应元素）。
input <- list("Subject-Identifier" = c("1022", "1023", "1024"),"Next-Screening" = c("A", "B", "C"))variables <- c("Subject Identifier", "Next Screening")### 我尝试了以下方法：res <- lapply(variables, function(x){input[[stringr::str_replace_all(x, " ", "-")]]})[[1]][1] "1022" "1023" "1024"[[2]][1] "A" "B" "C"res <- sapply(variables, function(x){input[[stringr::str_replace_all(x, " ", "-")]]}, USE.NAMES = TRUE)Subject Identifier Next Screening[1,] "1022" "A"[2,] "1023" "B"[3,] "1024" "C"### 我想要的是：$`Subject Identifier`[1] "1022" "1023" "1024"$`Next Screening`[1] "A" "B" "C"
如果您有任何其他问题，请告诉我。
英文:
I am trying to figure out how to create a list with lapply/sapply. From the documentation I gather this is possible using sapply with USE.NAMES = TRUE, but my example is not working.
I need to str-replace variable names to look them up in the input list where spaces are replaced with dashes. I have no direct control over this list (it is the input reactive element in an R shiny app)
input <- list("Subject-Identifier" = c("1022", "1023", "1024"),"Next-Screening" = c("A", "B", "C"))variables <- c("Subject Identifier", "Next Screening")### I tried these:res <- lapply(variables, function(x){input[[stringr::str_replace_all(x, " ", "-")]]})[[1]][1] "1022" "1023" "1024"[[2]][1] "A" "B" "C"res <- sapply(variables, function(x){input[[stringr::str_replace_all(x, " ", "-")]]}, USE.NAMES = TRUE)Subject Identifier Next Screening[1,] "1022" "A"[2,] "1023" "B"[3,] "1024" "C"### I want this:$`Subject Identifier`[1] "1022" "1023" "1024"$`Next Screening`[1] "A" "B" "C"
答案1
得分: 2
以下是您要的翻译部分：
We can pass a named vector and lapply returns the named list (which can be done either before or afterwards)
lapply(setNames(variables, variables), function(x){input[[stringr::str_replace_all(x, " ", "-")]})
-output
$`Subject Identifier`[1] "1022" "1023" "1024"$`Next Screening`[1] "A" "B" "C"
Regarding the second case, sapply by default use simplify = TRUE, which can be changed to FALSE and then it will return the expected
sapply(variables, function(x){input[[stringr::str_replace_all(x, " ", "-")]]}, USE.NAMES = TRUE, simplify = FALSE)
-output
$`Subject Identifier`[1] "1022" "1023" "1024"$`Next Screening`[1] "A" "B" "C"
英文:
We can pass a named vector and lapply returns the named list (which can be done either before or afterwards)
lapply(setNames(variables, variables), function(x){input[[stringr::str_replace_all(x, " ", "-")]]})
-output
$`Subject Identifier`[1] "1022" "1023" "1024"$`Next Screening`[1] "A" "B" "C"
Regarding the second case, sapply by default use simplify = TRUE, which can be changed to FALSE and then it will return the expected
sapply(variables, function(x){input[[stringr::str_replace_all(x, " ", "-")]]}, USE.NAMES = TRUE, simplify = FALSE)
-output
$`Subject Identifier`[1] "1022" "1023" "1024"$`Next Screening`[1] "A" "B" "C"
答案2
得分: 1
只需对数据进行子集处理并设置名称：
例如
setNames(input[gsub(" ", "-", variables)], variables)$`Subject Identifier`[1] "1022" "1023" "1024"$`Next Screening`[1] "A" "B" "C"
英文:
You do not need to loop over the data. Just subset and set the names:
ie
setNames(input[gsub(" ", "-", variables)], variables)$`Subject Identifier`[1] "1022" "1023" "1024"$`Next Screening`[1] "A" "B" "C"
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