list 如何从 R 中的统一列表中提取值？
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How do I extract values from uniform list in R?
提问by c00kiemonster
For example, how do I get a vector of each and every person's age in the list peoplebelow:
例如，我如何获得以下列表people中每个人年龄的向量：
> people = vector("list", 5)
> people[[1]] = c(name="Paul", age=23)
> people[[2]] = c(name="Peter", age=35)
> people[[3]] = c(name="Sam", age=20)
> people[[4]] = c(name="Lyle", age=31)
> people[[5]] = c(name="Fred", age=26)
> ages = ???
> ages
[1] 23 35 20 31 26
Is there an equivalent of a Python list comprehension or something to the same effect?
是否有等效于 Python 列表理解或具有相同效果的东西？
回答by tflutre
回答by Chase
Given the data structure you provided, I would use sapply:
鉴于您提供的数据结构，我会使用sapply：
sapply(people, function(x) x[2])
> sapply(people, function(x) x[2])
age age age age age
"23" "35" "20" "31" "26"
However, you'll notice that the results of this are character data.
但是，您会注意到此结果是字符数据。
> class(people[[1]])
[1] "character"
One approach would be to coerce to as.numeric()or as.integer()in the call to sapply.
一种方法是强制as.numeric()或as.integer()调用 sapply。
Alternatively - if you have flexibility over how you store the data in the first place, it may make sense to store it as a list of data.frames:
或者 - 如果您首先对如何存储数据具有灵活性，则将其存储为data.frames列表可能是有意义的：
people = vector("list", 5)
people[[1]] = data.frame(name="Paul", age=23)
people[[2]] = data.frame(name="Peter", age=35)
...
If you are going to go that far, you may also want to consider a single data.frame for all of your data:
如果你打算走那么远，你可能还想为你的所有数据考虑一个单一的 data.frame：
people2 <- data.frame(name = c("Paul", "Peter", "Sam", "Lyle", "Fred")
, age = c(23,35,20,31, 26))
There may be some other reason why you didn't consider this approach the first time around though...
不过，您第一次没有考虑这种方法可能还有其他一些原因......
回答by PeterVermont
ages <- sapply(1:length(people), function(i) as.numeric(people[[i]][[2]]))
ages
Output:
输出：
[1] 23 35 20 31 26
[1] 23 35 20 31 26
回答by symbolrush
Alternatively to the apply-family there's @Hadley's purrrpackagewhich offers the map_-functions for this kind of job.
除了apply-family 之外，还有@Hadley's purrrpackage，它map_为这种工作提供了-functions。
(There's a few differences to the apply-family discussed for example here.)
（apply例如这里讨论的-family有一些差异。）
OPs example:
OP示例：
people = vector("list", 5)
people[[1]] = c(name="Paul", age=23)
people[[2]] = c(name="Peter", age=35)
people[[3]] = c(name="Sam", age=20)
people[[4]] = c(name="Lyle", age=31)
people[[5]] = c(name="Fred", age=26)
The sapplyapproach:
该sapply方法：
ages_sapply <- sapply(people, function(x){as.numeric(x[2])})
print(ages_sapply)
[1] 23 35 20 31 26
And the mapapproach:
和map方法：
ages_map <- purrr::map_dbl(people, function(x){as.numeric(x[2])})
print(ages_map)
[1] 23 35 20 31 26
Of course they are identical:
当然，它们是相同的：
identical(ages_sapply, ages_map)
[1] TRUE
回答by CermakM
Though this question is pretty old, I'd like to share my approach to this. It is certainly possible to do with the sapplyas
tflutresuggested. But I find it more intuitive by using the unlistfunction:
虽然这个问题已经很老了，但我想分享我对此的方法。当然可以sapply按照
tflutre 的建议进行处理。但我发现使用该unlist函数更直观：
> ages <- unlist(people, use.names = F)[seq(2, 2 * length(people), 2)]
> ages
[1] "23" "35" "20" "31" "26"
NOTE the multiplication by twoin
2 * length(people), there are two elements stored in thepoeplelist. This can be made more generic by writinglength(people[[1]]) * length(people)
注意把乘2中
2 * length(people)，还有存储在这两个元素poeple列表。这可以通过编写更通用length(people[[1]]) * length(people)
Here unlist(people, use.names = F)yields
这里unlist(people, use.names = F)产量
[1] "Paul" "23" "Peter" "35" "Sam" "20" "Lyle" "31" "Fred"
[10] "26"
and we slice that by every other element using seqcommand.
我们使用seq命令将其按每个其他元素进行切片。
