8

Let's say I have a data.frame 

x <- data.frame(a = c('A','A','A','A','A', 'C','C','C','C', 'B','B','B'),
                b = c('a','c','a','a','c', 'd', 'e','e','d', 'b','b','b'),
                c = c( 7,  3,  2,  4,  5,   3,   1,  1,  5,   5,  2,  3),
                stringsAsFactors = FALSE)

> x
   a b c
1  A a 7
2  A c 3
3  A a 2
4  A a 4
5  A c 5
6  C d 3
7  C e 1
8  C e 1
9  C d 5
10 B b 5
11 B b 2
12 B b 3

I would like to sort x by columns b and c but keeping order of a as before. x[order(x$b, x$c),] - breaks order of column a. This is what I want:

   a b c
3  A a 2
4  A a 4
1  A a 7
2  A c 3
5  A c 5
6  C d 3
9  C d 5
7  C e 1
8  C e 1
11 B b 2
12 B b 3
10 B b 5

Is there a quick way of doing it?

Currently I run "for" loop and sort each subset, I'm sure there must be a better way.

Thank you! Ilya

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3 Answers 3

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7

If column "a" is ordered already, then its this simple:

> x[order(x$a,x$b, x$c),]
   a b c
3  A a 2
4  A a 4
1  A a 7
2  A c 3
5  A c 5
6  B d 3
9  B d 5
7  B e 1
8  B e 1
11 C b 2
12 C b 3
10 C b 5

If column a isn't ordered (but is grouped), create a new factor with the levels of x$a and use that.

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2 Comments

column "a" not ordered, but grouped. Order of "a" in data.frame is important.
it means first x$a will be sorted then x$b based on x$a , and then x$c based on x$a and x$b , isn't it?
0

Thank you Spacedman! Your recommendation works well.

x$a <- factor(x$a, levels = unique(x$a), ordered = TRUE)
x[order(x$a,x$b, x$c),]

Following Gavin's comment 

   x$a <- factor(x$a, levels = unique(x$a))
   x[order(x$a,x$b, x$c),]
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4 Comments

obv that fails if x$a is B B B A A C C A A D D D - but if it's completely grouped then you're done.
in my data set there couldn't be separation in groups.
I think you've gotten the wrong idea about factors. The ordering Spacedman referred to was ordering in the levels. ordered = TRUE produces an ordered factor, which is a special type of factor where there is some semi-quantitative ordering. This type of factor is irrelevant to the problem here.
probably. I'm not really familiar with factors. Thank you for clarification.
0 
require(doBy)
orderBy(~ a + b + c, data=x)
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1 Comment

This one will change order of column a

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