Question

我有一个列表aa，它引用另一个列表bb的索引名称以及包含另一个元素（称为cm）。列表bb项包含字符串。我有一个遍历bb的循环，并且对于与我指定的字符串匹配的每个项目，将其添加到数据框中的新行。我还需要将cm值添加到该行。

示例：

library("tidyverse")

aa <- list(c(123, 1), c(234, 1), c(345, 2), c(456, 3))
bb <- list("123" = c("a", "b", "c"), "234" = c("b", "c", "d"), "345" = c("c", "d", "e"), "456" = c("f", "g", "h"))
cc <- c("a", "b", "c")

tbl <- NULL
for (a in aa){
  for (b in bb) {
      if (any(cc %in% b)) {
        tb <- tibble(cm=a[2],n1=b[1],n2=b[2],n3=b[3])
        tbl <- bind_rows(tbl,tb)
      }
    }
  }

这会针对bb的每个可能组合进行迭代，并将其添加到每个cm对，这是不好的。我的输出应该是这样的：

output <- tibble(cm = c(1, 1, 2), n1 = c("a", "b", "c"),
             n2 = c("b", "c", "d"), n3 = c("c", "d", "e"))

> output
# A tibble: 3 x 4
         cm    n1    n2    n3
      <dbl> <chr> <chr> <chr>
1         1     a     b     c
2         1     b     c     d
3         2     c     d     e

我认为这样的事情可能会起作用，至少我可以稍后循环tbl并使用nm将其替换为适当的cm值：

tbl <- NULL
for (a in aa){
  for (b in bb) {
    if (any(cc %in% b)) {
      tb <- tibble(nm = names(bb)[b], n1=b[1],n2=b[2],n3=b[3])
      tbl <- bind_rows(tbl,tb)
      }
     }
    }

我真的不明白为什么这不起作用，因为names(bb)[1]会返回123所以我认为它在names(bb)[b]的循环中会起作用。

Answer 1

如果您对没有显式循环的基本R解决方案感到满意，这会起作用吗？

# generate data
aa <- list(c(123, 1), c(234, 1), c(345, 2), c(456, 3))
# cm is an element of bb
bb <- list("123" = c("a", "b", "c"), "234" = c("b", "c", "d"),
           "345" = c("c", "d", "e"), "456" = c("f", "g", "h"),
           cm = c(1, 1, 2))
cc <- c("a", "b", "c")

tbl <- data.frame(
  bb[["cm"]],
  # apply to each element of aa
  do.call(rbind, lapply(aa, function(x, y, c) { # function takes 3 args
    # only elements of bb whose names are in aa[[x]]
    names_y <- as.character(intersect(x, names(y)))
    # turn subset of bb into data.frame
    out <- as.data.frame(do.call(rbind, y[names_y]))
    # subset rows for which any row element %in% cc
    out <- out[apply(out, 1, function(x, c) any(x %in% c), c)] 
    return(out)
  }, bb, cc))) # pass bb and cc as args to the function in lapply()
names(tbl) <- c("cm", paste0("n", 1:(ncol(tbl) - 1)))

给出

 > tbl
    cm n1 n2 n3
123  1  a  b  c
234  1  b  c  d
345  2  c  d  e

使用names（）

1 个答案: