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How to calculate ranking of one column based on groups defined by another column?

开发者 https://www.devze.com 2023-02-25 10:15 出处:网络
R Version 2.11.1 32-bit on Windows 7 I get a data set as below: USER_A USER_B SCORE 160.2 170.1 1100.15 260.2

R Version 2.11.1 32-bit on Windows 7

I get a data set as below:

USER_A USER_B SCORE
1        6      0.2
1        7      0.1
1        10     0.15
2        6      0.2
2        9      0.12
3        8      0.15
3        9      0.3

the USER_A is 1:3 and the USER_B is 6:10. Now I need to output the USER_A with the ranking of USER_B by their SCORE:

开发者_运维技巧
USER_A      ranking of USER_B
1  3  1  2  #the ranking of USER_B 6,7,10(which belong to USER_A 1)
2  2  1     #the ranking of USER_B 6,9(which belong to USER_A 2)
3  1  2     #the ranking of USER_B 8,9(which belong to USER_A 3)

in fact, I just need to output the ranking:

3 1 2
2 1
1 2

it is upset because the length of each row is different! I could not store them in a matrix and then output them.

Could anyone help me solve this problem?


df <- read.table(con <- textConnection("USER_A USER_B SCORE
1        6      0.2
1        7      0.1
1        10     0.15
2        6      0.2
2        9      0.12
3        8      0.15
3        9      0.3
"), header = TRUE)
close(con)

One way is to split the data:

sdf <- with(df, split(SCORE, f = USER_A))
lapply(sdf, rank)

The last line gives:

> lapply(sdf, rank)
$`1`
[1] 3 1 2

$`2`
[1] 2 1

$`3`
[1] 1 2

An alternative is to use aggregate() as in:

aggregate(SCORE ~ USER_A, data = df, rank)

Which returns:

> (foo <- aggregate(SCORE ~ USER_A, data = df, rank))
  USER_A   SCORE
1      1 3, 1, 2
2      2    2, 1
3      3    1, 2

But the output is a bit different here, now we have a data frame, with the second component SCORE being a list, just like the lapply() version outputted:

> str(foo)
'data.frame':   3 obs. of  2 variables:
 $ USER_A: int  1 2 3
 $ SCORE :List of 3
  ..$ 0: num  3 1 2
  ..$ 1: num  2 1
  ..$ 2: num  1 2
> foo$SCORE
$`0`
[1] 3 1 2

$`1`
[1] 2 1

$`2`
[1] 1 2
0

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