Question

这是另一个问题 Extracting from Nested list to data frame

使用更新的答案，我将获得我的数据框。

然后我使用df <- data.frame(start = df3[5,])

所以我离开了：

dput(df)
structure(list(start.X1_1 = structure(4L, .Names = "experience.start", .Label = c("", 
" ", "1", "2015"), class = "factor"), start.X2_2 = structure(3L, .Names = "experience.start", .Label = c(" ", 
"1", "2011"), class = "factor"), start.X3_2 = structure(3L, .Names = "experience.start", .Label = c(" ", 
"1", "2007"), class = "factor"), start.X4_2 = structure(NA_integer_, .Names = "experience.start", .Label = c(" ", 
"1"), class = "factor"), start.X5_2 = structure(NA_integer_, .Names = "experience.start", .Label = c(" ", 
"1"), class = "factor"), start.X6_2 = structure(NA_integer_, .Names = "experience.start", .Label = c(" ", 
"1"), class = "factor"), start.X7_2 = structure(NA_integer_, .Names = "experience.start", .Label = c(" ", 
"1"), class = "factor"), start.X8_2 = structure(NA_integer_, .Names = "experience.start", .Label = c(" ", 
"1"), class = "factor"), start.X9_2 = structure(NA_integer_, .Names = "experience.start", .Label = c(" ", 
"1"), class = "factor"), start.X10_3 = structure(3L, .Names = "experience.start", .Label = c(" ", 
"1", "2016", "3000"), class = "factor"), start.X11_3 = structure(3L, .Names = "experience.start", .Label = c(" ", 
"1", "2015", "3000"), class = "factor"), start.X12_3 = structure(4L, .Names = "experience.start", .Label = c("", 
" ", "1", "2015", "2016", "EE"), class = "factor"), start.X13_3 = structure(4L, .Names = "experience.start", .Label = c("", 
" ", "1", "2014", "2015"), class = "factor"), start.X14_3 = structure(3L, .Names = "experience.start", .Label = c(" ", 
"1", "2013", "2014"), class = "factor"), start.X15_3 = structure(3L, .Names = "experience.start", .Label = c(" ", 
"1", "2010", "2011", "Virtusa"), class = "factor")), .Names = c("start.X1_1", 
"start.X2_2", "start.X3_2", "start.X4_2", "start.X5_2", "start.X6_2", 
"start.X7_2", "start.X8_2", "start.X9_2", "start.X10_3", "start.X11_3", 
"start.X12_3", "start.X13_3", "start.X14_3", "start.X15_3"), row.names = "experience.start", class = "data.frame")

现在我想要采用以下格式：

  v1    v2  v3   v4   v5   v6   v7   v8
1 2015
2 2011 2007 null null null null null null
3 2016 2015 2015 2015 2013 2010

我可以使用以下内容查找匹配

的列

sR <- function(x, n){
    substr(x, nchar(x)-n+1, nchar(x))}

 sR(names(df),2)
 [1] "_1" "_2" "_2" "_2" "_2" "_2" "_2" "_2" "_2" "_3" "_3" "_3" "_3" "_3" "_3"

所以我想从这里必须有一种方法可以达到我想要的输出。

或者我确定有人会告诉我更好的方式

Answer 1

主要思想是根据下划线后面的后缀split数据框。通过这种方式，您可以获得包含3个元素的列表，每个后缀为1个（在您的情况下为1，2，3）

df[] <- lapply(df[], as.character)
l1 <- lapply(split(stack(df), as.numeric(sub('.*_', '', stack(df)[,2]))), '[', 1)
lapply(l1, head, 2)

#$`1`
#  values
#1   2015

#$`2`
#  values
#2   2011
#3   2007

#$`3`
#   values
#10   2016
#11   2015

现在我们需要做的就是将cbind这3个元素放在一起，这有点棘手，因为它们的长度不同。幸运的是，我们可以使用SO（我们可以使用下面的免责声明）来解决这个问题。

t(do.call(cbindPad, l1))

#       1      2      3      4      5      6      7  8 
#values "2015" NA     NA     NA     NA     NA     NA NA
#values "2011" "2007" NA     NA     NA     NA     NA NA
#values "2016" "2015" "2015" "2014" "2013" "2010" NA NA

<强>声明

函数cbindPad取自@ Joran在this post

中的回答

或者，rbind.fill包中的函数plyr可以在转置后使用，以提供cbind.fill种结果。

plyr::rbind.fill(lapply(l1, function(i) as.data.frame(t(i))))

#     1    2    3    4    5    6    7    8    9   10   11   12   13   14   15
#1 2015 <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA>
#2 <NA> 2011 2007 <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA>
#3 <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> 2016 2015 2015 2014 2013 2010

使用列名转换数据框

1 个答案: