在R中重新格式化数据帧

我有多个COL，希望重新格式化dataframe以减少COL

这是我的建议：

# Dataframe
df <- data.frame(
    ~Location, ~Product_Name, ~Category, ~Machine1, ~Machine2 ~Machine1_adds, ~Machine2_adds, ~Sales1, ~Saless2, Spoils1, Spoils2
    A,  "Snickers",   Candy,  0, 1,  $2.5, $3, 2, 1
    A,  "Kitcat",   Candy,  0, 1,  $3, $3, 2, 1
    A,  "Pepsi",   Bev,  1, 1,  $5, $4, 3, 0
    B,  "Coke",   Bev,  1, 0, $5,  $6.45, 1, 1
    B,  "Gatoraid",   Bev,  0, 1, $4,  $4.45, 1, 0
    B,  "Sprite",   Bev,  1, 1,  $8, $6, 1, 0
)
df

#数据帧
df这里有一个选项，带有
melt
from
data.table

library(data.table)
melt(setDT(df), measure = patterns("^Machine", "^Sales", "^Spoils"), 
      value.name = c("Machine_adds", "Sales", "Spoils"))[, variable := NULL][]
#    Location Product_Name Category Machine_adds Sales Spoils
# 1:        A     Snickers    Candy       0  $2.5      2
# 2:        A       Kitcat    Candy       0    $3      2
# 3:        A        Pepsi      Bev       1    $5      3
# 4:        B         Coke      Bev       1    $5      1
# 5:        B     Gatoraid      Bev       0    $4      1
# 6:        B       Sprite      Bev       1    $8      1
# 7:        A     Snickers    Candy       1    $3      1
# 8:        A       Kitcat    Candy       1    $3      1
# 9:        A        Pepsi      Bev       1    $4      0
#10:        B         Coke      Bev       0 $6.45      1
#11:        B     Gatoraid      Bev       1 $4.45      0
#12:        B       Sprite      Bev       1    $6      0

更新
根据OP的更新示例，如果有“Machine”和Machine_adds”列，我们可以稍微将
模式更改为

# creating new columns in the dataset
df[c('Machine1', 'Machine2')] <- df[c("Machine1_adds", "Machine2_adds")]
melt(setDT(df), measure = patterns("^Machine\\d+$", 
         "^Machine\\d+_adds$", "^Sales", "^Spoils"), 
   value.name = c("Machine", "Machine_adds", "Sales", "Spoils"))[, 
           variable := NULL][]

更新
数据
df根据您所拥有的，您正在寻找
重塑
函数，其中变量对齐为
列表（c（4,5），c（6,7），c（8,9））
。您可以使用：

reshape(df,t(matrix(4:ncol(df),2)),idvar = 1:3,dir="long")

或

要获得您拥有的名称，我将使用
v.names
参数

reshape(df,list(c(4,5),c(6,7),c(8,9)),idvar = 1:3,dir="long",
         v.names = c("Machine_adds","Sales","Spoils"))[-4]# -4 removes the time variable.
                   Location Product_Name Category Machine_adds Sales Spoils
A.Snickers.Candy.1        A     Snickers    Candy            0  $2.5      2
A.Kitcat.Candy.1          A       Kitcat    Candy            0    $3      2
A.Pepsi.Bev.1             A        Pepsi      Bev            1    $5      3
B.Coke.Bev.1              B         Coke      Bev            1    $5      1
B.Gatoraid.Bev.1          B     Gatoraid      Bev            0    $4      1
B.Sprite.Bev.1            B       Sprite      Bev            1    $8      1
A.Snickers.Candy.2        A     Snickers    Candy            1    $3      1
A.Kitcat.Candy.2          A       Kitcat    Candy            1    $3      1
A.Pepsi.Bev.2             A        Pepsi      Bev            1    $4      0
B.Coke.Bev.2              B         Coke      Bev            0 $6.45      1
B.Gatoraid.Bev.2          B     Gatoraid      Bev            1 $4.45      0
B.Sprite.Bev.2            B       Sprite      Bev            1    $6      0

您的示例是给出错误扫描，您将
dput（df）
粘贴到此处。这是否回答了您的问题？如果您的示例数据采用可以直接运行的格式，则会更容易提供帮助。您将此设置为对
tribble
的调用，但使用实际不起作用的语法，例如，不带引号的字符串，
$2.5
``除了获取机器标签（1、2或3）之外，其他一切都在起作用……它只是显示变量。@Dinho它是
variable
而不是
variable
@Dinho得到了它，我想你既有
机器
又有
机器
。我使用了一个仅包含
Machine\u adds
column@Dinho您可以将代码更改为
melt（setDT（df），measure=patterns（“^Machine\\d+$”、“^Machine\\d+\u adds$”、“^Sales”、“^battles”）、value.name=c（“Machine”、“Machine\u adds”、“Sales”、“battles”）[，变量：=NULL][
df %>%
   rename_at(vars(matches('^Machine.*adds$')), ~ 
         str_replace(., '(\\d+)_(\\w+)$', '_\\2\\1')) %>% 
   rename_at(3:ncol(.), ~ str_replace(., "(\\d+)_?.*", ":\\1")) %>%
   pivot_longer(cols =  matches("^(Machine|Sales|Spoils)"),
        names_to = c(".value", "group"), names_sep = ":") %>%
   select(-group)

df <- structure(list(Location = c("A", "A", "A", "B", "B", "B"), 
     Product_Name = c("Snickers", 
"Kitcat", "Pepsi", "Coke", "Gatoraid", "Sprite"), Category = c("Candy", 
"Candy", "Bev", "Bev", "Bev", "Bev"), Machine1_adds = c(0, 0, 
1, 1, 0, 1), Machine2_adds = c(1, 1, 1, 0, 1, 1), Sales1 = c("$2.5", 
"$3", "$5", "$5", "$4", "$8"), Sales2 = c("$3", "$3", "$4", "$6.45", 
"$4.45", "$6"), Spoils1 = c(2, 2, 3, 1, 1, 1), Spoils2 = c(1, 
1, 0, 1, 0, 0)), row.names = c(NA, -6L), class = c("tbl_df", 
"tbl", "data.frame"))

reshape(df,t(matrix(4:ncol(df),2)),idvar = 1:3,dir="long")

reshape(df,list(c(4,5),c(6,7),c(8,9)),idvar = 1:3,dir="long")

reshape(df,list(c(4,5),c(6,7),c(8,9)),idvar = 1:3,dir="long",
         v.names = c("Machine_adds","Sales","Spoils"))[-4]# -4 removes the time variable.
                   Location Product_Name Category Machine_adds Sales Spoils
A.Snickers.Candy.1        A     Snickers    Candy            0  $2.5      2
A.Kitcat.Candy.1          A       Kitcat    Candy            0    $3      2
A.Pepsi.Bev.1             A        Pepsi      Bev            1    $5      3
B.Coke.Bev.1              B         Coke      Bev            1    $5      1
B.Gatoraid.Bev.1          B     Gatoraid      Bev            0    $4      1
B.Sprite.Bev.1            B       Sprite      Bev            1    $8      1
A.Snickers.Candy.2        A     Snickers    Candy            1    $3      1
A.Kitcat.Candy.2          A       Kitcat    Candy            1    $3      1
A.Pepsi.Bev.2             A        Pepsi      Bev            1    $4      0
B.Coke.Bev.2              B         Coke      Bev            0 $6.45      1
B.Gatoraid.Bev.2          B     Gatoraid      Bev            1 $4.45      0
B.Sprite.Bev.2            B       Sprite      Bev            1    $6      0