如何将dput（）的输出加载到对象中？

时间：2015-03-20 07:14:47

标签： r variable-assignment

例如，我有一个类似的代码：

structure(list(mpg = c(21, 21, 22.8, 21.4, 18.7, 18.1, 14.3, 
24.4, 22.8, 19.2, 17.8, 16.4, 17.3, 15.2, 10.4, 10.4, 14.7, 32.4, 
30.4, 33.9, 21.5, 15.5, 15.2, 13.3, 19.2, 27.3, 26, 30.4, 15.8, 
19.7, 15, 21.4), cyl = c(6, 6, 4, 6, 8, 6, 8, 4, 4, 6, 6, 8, 
8, 8, 8, 8, 8, 4, 4, 4, 4, 8, 8, 8, 8, 4, 4, 4, 8, 6, 8, 4), 
    disp = c(160, 160, 108, 258, 360, 225, 360, 146.7, 140.8, 
    167.6, 167.6, 275.8, 275.8, 275.8, 472, 460, 440, 78.7, 75.7, 
    71.1, 120.1, 318, 304, 350, 400, 79, 120.3, 95.1, 351, 145, 
    301, 121), hp = c(110, 110, 93, 110, 175, 105, 245, 62, 95, 
    123, 123, 180, 180, 180, 205, 215, 230, 66, 52, 65, 97, 150, 
    150, 245, 175, 66, 91, 113, 264, 175, 335, 109), drat = c(3.9, 
    3.9, 3.85, 3.08, 3.15, 2.76, 3.21, 3.69, 3.92, 3.92, 3.92, 
    3.07, 3.07, 3.07, 2.93, 3, 3.23, 4.08, 4.93, 4.22, 3.7, 2.76, 
    3.15, 3.73, 3.08, 4.08, 4.43, 3.77, 4.22, 3.62, 3.54, 4.11
    ), wt = c(2.62, 2.875, 2.32, 3.215, 3.44, 3.46, 3.57, 3.19, 
    3.15, 3.44, 3.44, 4.07, 3.73, 3.78, 5.25, 5.424, 5.345, 2.2, 
    1.615, 1.835, 2.465, 3.52, 3.435, 3.84, 3.845, 1.935, 2.14, 
    1.513, 3.17, 2.77, 3.57, 2.78), qsec = c(16.46, 17.02, 18.61, 
    19.44, 17.02, 20.22, 15.84, 20, 22.9, 18.3, 18.9, 17.4, 17.6, 
    18, 17.98, 17.82, 17.42, 19.47, 18.52, 19.9, 20.01, 16.87, 
    17.3, 15.41, 17.05, 18.9, 16.7, 16.9, 14.5, 15.5, 14.6, 18.6
    ), vs = c(0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 
    0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1), am = c(1, 
    1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 
    0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1), gear = c(4, 4, 4, 3, 
    3, 3, 3, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 4, 4, 4, 3, 3, 3, 
    3, 3, 4, 5, 5, 5, 5, 5, 4), carb = c(4, 4, 1, 1, 2, 1, 4, 
    2, 2, 4, 4, 3, 3, 3, 4, 4, 4, 1, 2, 1, 1, 2, 2, 4, 2, 1, 
    2, 2, 4, 6, 8, 2)), .Names = c("mpg", "cyl", "disp", "hp", 
"drat", "wt", "qsec", "vs", "am", "gear", "carb"), row.names = c("Mazda RX4", 
"Mazda RX4 Wag", "Datsun 710", "Hornet 4 Drive", "Hornet Sportabout", 
"Valiant", "Duster 360", "Merc 240D", "Merc 230", "Merc 280", 
"Merc 280C", "Merc 450SE", "Merc 450SL", "Merc 450SLC", "Cadillac Fleetwood", 
"Lincoln Continental", "Chrysler Imperial", "Fiat 128", "Honda Civic", 
"Toyota Corolla", "Toyota Corona", "Dodge Challenger", "AMC Javelin", 
"Camaro Z28", "Pontiac Firebird", "Fiat X1-9", "Porsche 914-2", 
"Lotus Europa", "Ford Pantera L", "Ferrari Dino", "Maserati Bora", 
"Volvo 142E"), class = "data.frame")

我认为只需要函数dget就很容易，但看起来我无法像这样加载data.frame。你能告诉我如何轻松加载上面的代码吗？

1 个答案:

答案 0 :(得分：10)

使用赋值运算符<-：

    my.object.df <- structure(list(mpg = c(21, 21, 22.8, 21.4, 18.7, 18.1, 14.3, 
24.4, 22.8, 19.2, 17.8, 16.4, 17.3, 15.2, 10.4, 10.4, 14.7, 32.4, 
30.4, 33.9, 21.5, 15.5, 15.2, 13.3, 19.2, 27.3, 26, 30.4, 15.8, 
19.7, 15, 21.4), cyl = c(6, 6, 4, 6, 8, 6, 8, 4, 4, 6, 6, 8, 
8, 8, 8, 8, 8, 4, 4, 4, 4, 8, 8, 8, 8, 4, 4, 4, 8, 6, 8, 4), 
    disp = c(160, 160, 108, 258, 360, 225, 360, 146.7, 140.8, 
    167.6, 167.6, 275.8, 275.8, 275.8, 472, 460, 440, 78.7, 75.7, 
    71.1, 120.1, 318, 304, 350, 400, 79, 120.3, 95.1, 351, 145, 
    301, 121), hp = c(110, 110, 93, 110, 175, 105, 245, 62, 95, 
    123, 123, 180, 180, 180, 205, 215, 230, 66, 52, 65, 97, 150, 
    150, 245, 175, 66, 91, 113, 264, 175, 335, 109), drat = c(3.9, 
    3.9, 3.85, 3.08, 3.15, 2.76, 3.21, 3.69, 3.92, 3.92, 3.92, 
    3.07, 3.07, 3.07, 2.93, 3, 3.23, 4.08, 4.93, 4.22, 3.7, 2.76, 
    3.15, 3.73, 3.08, 4.08, 4.43, 3.77, 4.22, 3.62, 3.54, 4.11
    ), wt = c(2.62, 2.875, 2.32, 3.215, 3.44, 3.46, 3.57, 3.19, 
    3.15, 3.44, 3.44, 4.07, 3.73, 3.78, 5.25, 5.424, 5.345, 2.2, 
    1.615, 1.835, 2.465, 3.52, 3.435, 3.84, 3.845, 1.935, 2.14, 
    1.513, 3.17, 2.77, 3.57, 2.78), qsec = c(16.46, 17.02, 18.61, 
    19.44, 17.02, 20.22, 15.84, 20, 22.9, 18.3, 18.9, 17.4, 17.6, 
    18, 17.98, 17.82, 17.42, 19.47, 18.52, 19.9, 20.01, 16.87, 
    17.3, 15.41, 17.05, 18.9, 16.7, 16.9, 14.5, 15.5, 14.6, 18.6
    ), vs = c(0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 
    0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1), am = c(1, 
    1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 
    0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1), gear = c(4, 4, 4, 3, 
    3, 3, 3, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 4, 4, 4, 3, 3, 3, 
    3, 3, 4, 5, 5, 5, 5, 5, 4), carb = c(4, 4, 1, 1, 2, 1, 4, 
    2, 2, 4, 4, 3, 3, 3, 4, 4, 4, 1, 2, 1, 1, 2, 2, 4, 2, 1, 
    2, 2, 4, 6, 8, 2)), .Names = c("mpg", "cyl", "disp", "hp", 
"drat", "wt", "qsec", "vs", "am", "gear", "carb"), row.names = c("Mazda RX4", 
"Mazda RX4 Wag", "Datsun 710", "Hornet 4 Drive", "Hornet Sportabout", 
"Valiant", "Duster 360", "Merc 240D", "Merc 230", "Merc 280", 
"Merc 280C", "Merc 450SE", "Merc 450SL", "Merc 450SLC", "Cadillac Fleetwood", 
"Lincoln Continental", "Chrysler Imperial", "Fiat 128", "Honda Civic", 
"Toyota Corolla", "Toyota Corona", "Dodge Challenger", "AMC Javelin", 
"Camaro Z28", "Pontiac Firebird", "Fiat X1-9", "Porsche 914-2", 
"Lotus Europa", "Ford Pantera L", "Ferrari Dino", "Maserati Bora", 
"Volvo 142E"), class = "data.frame")

您也可以创建一个文本文件，然后使用source来读取和执行它。 （当然，使用load创建的'.Rdata'文件save会更快。）

这是一个小例子。 

> test <- c(1.1, 2,3, 4)
> dput(test)
c(1.1, 2, 3, 4)
> dput( test, "mytest.txt")
> dget("mytest.txt")
[1] 1.1 2.0 3.0 4.0
> newtest <- dget("mytest.txt")
> newtest
[1] 1.1 2.0 3.0 4.0

请注意，在使用dget输入时需要分配对象。另一方面，如果您将上面示例的第一行（对文本中的名称赋值）保存为文本文件，则可以使用eval（parse（...））将其带回来

> dget("secondtest.txt")
> test
Error: object 'test' not found
> ?load     # Nope, that only works for binary files
> eval( parse(file= "secondtest.txt"))
> test
[1] 1.1 2.0 3.0 4.0
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