Pull specific column to beginning after columns have been ordered alphabetically

Question

I have a dataframe in R. I'm ordering the columns alphabetically like this:

df_ordered <- df[,order(names(df))]

Now I want to select a specific column by name (column name is "city") and "pull" it to the front while leaving the rest unchanged, i.e. if the columns are:

alpha, beta, city, delta

I want them to be

city, alpha, beta, delta

Thanks for your help.

If you use [my `moveme` function](http://stackoverflow.com/questions/3369959/moving-columns-within-a-data-frame-without-retyping/18540144#18540144), the code would be: `moveme(df, "city first")`. — A5C1D2H2I1M1N2O1R2T1, Sep 08 '14 at 16:50

score 3 · Accepted Answer · answered Sep 08 '14 at 15:31

3

Using @nrussell sample Df data.frame, you can also do a trick with relevel.

Df[, levels(relevel(factor(names(Df)),"G"))]

We use the implicit sorting of factor() and use relevel to bring a particular column forward.

answered Sep 08 '14 at 15:31

MrFlick

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nrussell · Answer 2 · 2014-09-08T14:40:59.380

How about this:

Df <- data.frame(
  matrix(rnorm(52),
         ncol=26))
names(Df) <- LETTERS
##
Df <- Df[,c("C",sort(names(Df)[names(Df)!="C"]))]
##
> head(Df)
           C          A         B          D         E          F         G          H         I          J           K
1 -1.1685829 -1.0733238 0.1991507 -0.5315889  1.417376  0.3473343 0.8068851 -1.3548045 1.5587990  0.1617466 -0.02262477
2 -0.9241126 -0.3910728 1.6685727 -0.7506446 -1.275201 -0.1603301 0.9326244 -0.8349584 0.4534975 -0.3170741  0.63663999
          L           M          N         O          P          Q         R         S          T          U          V
1 0.9524257 -0.04871767 -0.8686760 -1.971867  2.0080797 -0.6866727 0.4532914 -0.328414 0.05844098  0.8595681  1.1198372
2 1.7291060  1.52756996  0.4328425  1.560806 -0.4236301 -1.1098159 0.3315950  1.328299 2.34595615 -0.3624254 -0.2300019
           W          X          Y          Z
1 -0.2515527 -0.2934354 -0.9494282 -0.9744378
2  1.0197277 -0.6822827 -1.0165435 -0.3204356

Or if you prefer to sort the columns before hand, like in your example, the sort is not needed:

Df <- data.frame(
  matrix(rnorm(52),
         ncol=26))
names(Df) <- sample(LETTERS,26)
Df <- Df[,order(names(Df))]
Df <- Df[,c("C",names(Df)[names(Df)!="C"])]
##
> head(Df)
           C        A          B         D           E          F          G          H          I          J         K
1 -0.4508683 0.605652  1.0056819 0.3410927 -0.08246848  0.4173749  1.7259750  1.0860495 0.06465027 -1.5329574 0.1268806
2  0.5212941 1.264273 -0.3907112 0.4043759 -0.76360427 -0.7712454 -0.2604928 -0.8549233 0.53874782  0.1891443 1.9644381
           L          M          N         O          P         Q         R           S           T          U          V
1  0.8127268 -0.9294953 -0.9134001 2.0891866 -0.1560481  1.480088 0.8328550 -0.47946694 -0.55311500 -0.1393134  0.7400875
2 -0.9649869 -0.6869047 -0.8529926 0.1224387 -0.4958187 -1.273009 0.2728192  0.07974988 -0.01028435  0.9705935 -0.8299274
           W         X         Y         Z
1 -1.9356416 0.9440026 0.9074300 -1.082567
2 -0.2265877 0.4264449 0.9681043 -1.819203

score 0 · Answer 3 · answered Sep 08 '14 at 14:48

We can define a function which returns your desired column subsettor.

pullToFront<-function(entry, vect){
    c(entry, vect[vect!=entry])
}

Where vect is a vector of your column names, entry is the specific column you want to move to the front.

Say you have the following column names.

vect<-letters[1:5]
vect
#output
#[1]"a" "b" "c" "d" "e"

We want to pull "c" to the front, so we say

 subsettor<-pullToFront("c", vect)
 subsettor
 #output
 #[1]"c" "a" "b" "d" "e"

Then given a data frame with those column names, you would simply say

 df[,subsettor]

You would use this function by saying

 vect<-colnames(df)

Then isolating the one you want to move to the front in whatever way you choose

Pull specific column to beginning after columns have been ordered alphabetically

3 Answers3