R 将宽变长，使用对象作为输入

Question

我有一个宽格式的数据框列表，第 1 列中有一个因子变量，第 2 列中有年化数据。我想绘制这些数据。这样做需要将它们重塑为长格式。这是一个数据框的示例：

# SAMPLE DATA
x <- structure(list(State = structure(1:3, .Label = c("Alabama", "Alaska", "Arizona", "Arkansas"), class = "factor"), Green.And.Blue.Score.2001 = c(0L, 40L, 65L), Green.And.Blue.Score.2002 = c(20L, 5L, 60L), Green.And.Blue.Score.2003 = c(35L, 15L, 30L)), .Names = c("State", "Green.And.Blue.Score.2001", "Green.And.Blue.Score.2002", "Green.And.Blue.Score.2003"), row.names = c(NA, 3L), class = "data.frame")

x
#    State Green.And.Blue.Score.2001 Green.And.Blue.Score.2002 Green.And.Blue.Score.2003
#1 Alabama          0                          20                         35
#2  Alaska         40                           5                         15
#3 Arizona         65                          60                         30

我通常使用 reshape() 执行此操作。例如，这很好用：

# RESHAPE WIDE TO LONG (MANUALLY)
y <- reshape(x, 
    idvar = 'State',
  varying = c('Green.And.Blue.Score.2001', 'Green.And.Blue.Score.2002', 'Green.And.Blue.Score.2003'),
  v.names = 'Green.And.Blue.Score.',
    times = c('2001', '2002', '2003'),
direction = 'long')

y
# State    time   Green.And.Blue.Score.
# Alabama  2001      0
# Alaska   2001     40
# Arizona  2001     65
# Alabama  2002     20
# Alaska   2002      5
# Arizona  2002     60
# Alabama  2003     35
# Alaska   2003     15
# Arizona  2003     30

但是，我不想为我的几十个数据框手动输入idvar、varying、v.name和times变量有，并且因为一些列名非常长和复杂，并且在数据帧之间有很大差异，所以简单的 reshape() 命令不能自动解析它们。我的想法是创建一个函数来从数据框中获取这些输入，其前身如下：

# RESHAPE WIDE TO LONG (FUNCTIONALIZED)
id <- noquote(paste("'", names(x[1]), "'", sep = ""))
va <- noquote(paste("c('", paste(names(x)[2:length(x)], collapse = "', '"), "')", sep = ""))
vn <- noquote(paste("'", sub("(\..*)$", ".", names(x)[2]) , "'", sep = ""))
ti <- noquote(paste("c('", paste(sub(".*(\d{4})$", "\1", names(x[2:length(x)])), collapse = "', '"), "')", sep = ""))

其中每一个的输出都匹配上面 #RESHAPE WIDE TO LONG (MANUALLY) 的 idvar、varying、v.name 和 times 输入：

id
# 'State'
va
# c(''Green.And.Blue.Score.2001', ''Green.And.Blue.Score.2002', ''Green.And.Blue.Score.2003')
vn
# ''Green.And.Blue.Score.'
ti
# c('2001', '2002', '2003')

但是，当我尝试在 reshape() 函数中使用这些对象时，我收到一条错误消息：

y <- reshape(x, 
    idvar = id,
  varying = va,
  v.names = vn,
    times = ti,
direction = 'long')

Error in [.data.frame(data, , varying[[i]][1L]) : undefined columns selected

我确定我对 'functionalize' reshape() 的解决方案并不理想。我应该怎么做？

Answer 1

在从名称中提取的 material 周围加上引号的努力导致了错误。这是该代码的简化。请注意，我删除了 v.names 和时间，因为当列名由“.”正确分隔时，这些是自动计算的。

y <- reshape(x, 
     idvar = names(x)[1],
   varying = names(x)[-1],

 direction = 'long')
 y
#-----    
           State time Score
Alabama.2001 Alabama 2001     0
Alaska.2001   Alaska 2001    40
Arizona.2001 Arizona 2001    65
Alabama.2002 Alabama 2002    20
Alaska.2002   Alaska 2002     5
Arizona.2002 Arizona 2002    60
Alabama.2003 Alabama 2003    35
Alaska.2003   Alaska 2003    15
Arizona.2003 Arizona 2003    30

如果我们在您的新示例中使用它，我们可以在“.S”处得到一个 "split"，从而给出一个合理的结果。第一个句点和 "split" 模式之间的列名文本被移动到列名，而领先的州名称和年份作为行名一起附加：

 y <- reshape(x, 
      idvar = names(x)[1],
    varying = names(x)[-1],
    split = list(regexp = "\.S", include = TRUE),
  direction = 'long')
  y

                     State       time Green.And.Blue.
Alabama.Score.2001 Alabama Score.2001               0
Alaska.Score.2001   Alaska Score.2001              40
Arizona.Score.2001 Arizona Score.2001              65
Alabama.Score.2002 Alabama Score.2002              20
Alaska.Score.2002   Alaska Score.2002               5
Arizona.Score.2002 Arizona Score.2002              60
Alabama.Score.2003 Alabama Score.2003              35
Alaska.Score.2003   Alaska Score.2003              15
Arizona.Score.2003 Arizona Score.2003              30

R 将宽变长，使用对象作为输入

R reshape wide to long, using objects for the inputs

r

reshape