Question

我有一个包含三列的数据框

SentenceID = c(1,1,1,1,2,2,2,3,3,3,3,3,3,3,3)
Tokens = c("I","went","to","school","nobody","can","find","some","people","know","what","they","are","doing","now")
WordIndex = c(3,4,7,8,9,10,12,54,34,66,33,89,87,23,22)

df = data.frame(SentenceID, Tokens, WordIndex)

期望的结果：

我必须遍历每个SentenceID并创建一个向量X列表，如下所示

X           
[[1]]   3 4 7 8     
[[2]]   9 10 12     
[[3]]   54 34 66 33 89 87 23 22

然后我需要用0填充它们10个斑点

X           
[[1]]   3 4 7 8 0 0 0 0 0 0 0       
[[2]]   9 10 12 0 0 0 0 0 0 0   
[[3]]   54 34 66 33 89 87 23 22 0 0

我该如何做到这一点？

Answer 1

这是一种方法：

> lapply(split(df$WordIndex, df$SentenceID), function(x) c(x, rep(0, pmax(10 - length(x), 0))))
$`1`
 [1] 3 4 7 8 0 0 0 0 0 0

$`2`
 [1]  9 10 12  0  0  0  0  0  0  0

$`3`
 [1] 54 34 66 33 89 87 23 22  0  0

Answer 2

带有aggregate的基础R解决方案：

lapply(aggregate(WordIndex, list(SentenceID), c)$x, 
    function(X) head(c(X, rep(0,10)), 10))
$`1`
 [1] 3 4 7 8 0 0 0 0 0 0
$`2`
 [1]  9 10 12  0  0  0  0  0  0  0
$`3`
 [1] 54 34 66 33 89 87 23 22  0  0

Answer 3

您可以使用tidyverse的{{1}}功能尝试purrr

map

向量列表和序列填充

3 个答案: