how do you subset a data frame based on column names? - r

Question

I have this data frame:
dput(df)
structure(list(Server = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "servera", class = "factor"),
Date = structure(1:6, .Label = c("7/13/2017 15:01", "7/13/2017 15:02",
"7/13/2017 15:03", "7/13/2017 15:04", "7/13/2017 15:05",
"7/13/2017 15:06"), class = "factor"), Host_CPU = c(1.812950134,
2.288070679, 1.563278198, 1.925239563, 5.350669861, 2.612503052
), UsedMemPercent = c(38.19, 38.19, 38.19, 38.19, 38.19,
38.22), jvm1 = c(10.91, 11.13, 11.34, 11.56, 11.77, 11.99
), jvm2 = c(11.47, 11.7, 11.91, 12.13, 12.35, 12.57), jvm3 = c(75.65,
76.88, 56.93, 58.99, 65.29, 67.97), jvm4 = c(39.43, 40.86,
42.27, 43.71, 45.09, 45.33), jvm5 = c(27.42, 29.63, 31.02,
32.37, 33.72, 37.71)), .Names = c("Server", "Date", "Host_CPU",
"UsedMemPercent", "jvm1", "jvm2", "jvm3", "jvm4", "jvm5"), class = "data.frame", row.names = c(NA,
-6L))
I only want to be able to subset this data frame based on the vectors names in this variable:
select<-c("jvm3", "jvm4", "jvm5")
so, my final df should look like this:
structure(list(Server = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "servera", class = "factor"),
Date = structure(1:6, .Label = c("7/13/2017 15:01", "7/13/2017 15:02",
"7/13/2017 15:03", "7/13/2017 15:04", "7/13/2017 15:05",
"7/13/2017 15:06"), class = "factor"), Host_CPU = c(1.812950134,
2.288070679, 1.563278198, 1.925239563, 5.350669861, 2.612503052
), UsedMemPercent = c(38.19, 38.19, 38.19, 38.19, 38.19,
38.22), jvm3 = c(75.65, 76.88, 56.93, 58.99, 65.29, 67.97
), jvm4 = c(39.43, 40.86, 42.27, 43.71, 45.09, 45.33), jvm5 = c(27.42,
29.63, 31.02, 32.37, 33.72, 37.71)), .Names = c("Server",
"Date", "Host_CPU", "UsedMemPercent", "jvm3", "jvm4", "jvm5"), class = "data.frame", row.names = c(NA,
-6L))
any ideas?

Answer 1

Please revisit indices. If you use the index mechanism [ in R, you can use mainly three types of indices:
logical vectors: same length as the number of columns, TRUE means select the column
numeric vectors: selects columns based on position
character vectors: select columns based on name
If you use the index mechanism for data frames, you can treat these objects in two ways:
as a list, because they are internally lists
as a matrix, because they mimick matrix behaviour in many cases
Take the iris data frame as example to compare the multiple ways you can select columns from a data frame. If you treat it as a list, you have the following two options:
Use [[ if you want a single column in the form of a vector:
iris[["Species"]]
# [1] setosa setosa setosa ... : is a vector
Use [ if you want one or more columns, but you need a data frame back :
iris["Species"]
iris[c("Sepal.Width", "Species")]
If you treat it as a matrix, you just do the exact same as you would do with a matrix. If you don't specify any row indices, these commands are actually equivalent to the ones used above:
iris[ , "Species"] # is the same as iris[["Species"]]
iris[ , "Species", drop = FALSE] # is the same as iris["Species"]
iris[ , c("Sepal.Width", "Species")] # is the same as iris[c("Sepal.Width", "Species")]
So in your case, you simply need:
select <- c("Server","Date","Host_CPU","UsedMemPercent",
"jvm3","jvm4","jvm5")
df[select]
Note on subset: subset works, but should ONLY be used interactively. There's a warning on the help page stating :
This is a convenience function intended for use interactively. For
programming it is better to use the standard subsetting functions like
[, and in particular the non-standard evaluation of argument subset
can have unanticipated consequences.

Answer 2

Saving your dataframe to a variable df:
df <-
structure(
list(
Server = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "servera", class = "factor"),
Date = structure(
1:6,
.Label = c(
"7/13/2017 15:01",
"7/13/2017 15:02",
"7/13/2017 15:03",
"7/13/2017 15:04",
"7/13/2017 15:05",
"7/13/2017 15:06"
),
class = "factor"
),
Host_CPU = c(
1.812950134,
2.288070679,
1.563278198,
1.925239563,
5.350669861,
2.612503052
),
UsedMemPercent = c(38.19, 38.19, 38.19, 38.19, 38.19,
38.22),
jvm1 = c(10.91, 11.13, 11.34, 11.56, 11.77, 11.99),
jvm2 = c(11.47, 11.7, 11.91, 12.13, 12.35, 12.57),
jvm3 = c(75.65,
76.88, 56.93, 58.99, 65.29, 67.97),
jvm4 = c(39.43, 40.86,
42.27, 43.71, 45.09, 45.33),
jvm5 = c(27.42, 29.63, 31.02,
32.37, 33.72, 37.71)
),
.Names = c(
"Server",
"Date",
"Host_CPU",
"UsedMemPercent",
"jvm1",
"jvm2",
"jvm3",
"jvm4",
"jvm5"
),
class = "data.frame",
row.names = c(NA,-6L)
)
df[,select] should be what youre looking for

Answer 3

Here's one way:
df[,c(1:4,7:9)]
You can also use dplyr to select columns:
select(df, Server,Date,Host_CPU,UsedMemPercent,jvm3,jvm4,jvm5)