Re: [R] How to rank matrix data by deciles?

[Peter Ehlers]

On 2010-05-13 17:50, Phil Spector wrote:

Vincent -
I'm afraid there's no solution other than artificially modifying
the zeroes:

vec

[1] 26.58950617 5.73074074 5.96222222 5.64777778 20.95728395 0.00000000
0.07000000 12.86888889
[9] 3.64543210 0.05049383 25.60888889 3.53246914 0.00000000 31.39049383
3.77641975 13.19617284
[17] 0.00000000

cut(vec,quantile(vec,(0:10)/10),include.lowest=TRUE,label=FALSE)

Error in cut.default(vec, quantile(vec, (0:10)/10), include.lowest =
TRUE, :
'breaks' are not unique

vec[vec==0] = jitter(vec[vec==0])
cut(vec,quantile(vec,(0:10)/10),include.lowest=TRUE,label=FALSE)

[1] 10 6 7 5 9 1 3 7 4 2 9 4 2 10 5 8 1

It gives an answer, but it may not make sense for all data.

- Phil

The problem is that quantile() produces multiple values
for the breaks used in cut(). Phil's suggestion modifies
the data. It might be preferable to modify the breaks:

  eps <- .Machine$double.eps  #or use something like 1e-10
  brks <- quantile(vec, (0:10)/10) + eps*(0:10)
  cut(vec, brks, include.lowest=TRUE, labels=FALSE)
  #[1] 10  6  7  5  9  1  3  7  4  2  9  4  1 10  5  8  1

 -Peter Ehlers

On Thu, 13 May 2010, vincent.deluard wrote:

Dear Phil,

You helped me with a request to rand matrix columns by deciles two weeks
ago.

This really un-blocked me on this project but I found a little bug.

As in before, my data is in a matrix:

madebt[1:16,1:2]

X4.19.2010 X4.16.2010
[1,] 26.61197531 26.58950617
[2,] 5.72765432 5.73074074
[3,] 5.95839506 5.96222222
[4,] 5.64333333 5.64777778
[5,] 20.93814815 20.95728395
[6,] 0.00000000 0.00000000
[7,] 0.07000000 0.07000000
[8,] 12.87802469 12.86888889
[9,] 3.64407407 3.64543210
[10,] 0.05037037 0.05049383
[11,] 25.59024691 25.60888889
[12,] 3.47987654 3.53246914
[13,] 0.00000000 0.00000000
[14,] 31.39037037 31.39049383
[15,] 3.78296296 3.77641975
[16,] 13.17876543 13.19617284

The apply function will work for this sample of my data:

debtdeciles = apply(madebt[1:16,1:2],2,function(x)
cut(x,quantile(x,(0:10)/10,
na.rm=TRUE),label=FALSE,include.lowest=TRUE))

debtdeciles

X4.19.2010 X4.16.2010
[1,] 10 10
[2,] 6 6
[3,] 6 6
[4,] 5 5
[5,] 8 8
[6,] 1 1
[7,] 2 2
[8,] 7 7
[9,] 4 4
[10,] 2 2
[11,] 9 9
[12,] 3 3
[13,] 1 1
[14,] 10 10
[15,] 4 4
[16,] 8 8

However, it will fail for

madebt[1:17,1:2]

X4.19.2010 X4.16.2010
[1,] 26.61197531 26.58950617
[2,] 5.72765432 5.73074074
[3,] 5.95839506 5.96222222
[4,] 5.64333333 5.64777778
[5,] 20.93814815 20.95728395
[6,] 0.00000000 0.00000000
[7,] 0.07000000 0.07000000
[8,] 12.87802469 12.86888889
[9,] 3.64407407 3.64543210
[10,] 0.05037037 0.05049383
[11,] 25.59024691 25.60888889
[12,] 3.47987654 3.53246914
[13,] 0.00000000 0.00000000
[14,] 31.39037037 31.39049383
[15,] 3.78296296 3.77641975
[16,] 13.17876543 13.19617284
[17,] 0.00000000 0.00000000

debtdeciles = apply(madebt[1:17,1:2],2,function(x)

+ cut(x,quantile(x,(0:10)/10,
na.rm=TRUE),label=FALSE,include.lowest=TRUE))
Error in cut.default(x, quantile(x, (0:10)/10, na.rm = TRUE), label =
FALSE,
:
'breaks' are not unique

My guess is that we now have 3 "zeros" in each column. For each
decile, we
cannot have more than 2 elements (total of 17 numbers in each column)
and I
believe R cannot determine where to put the third "zero". Do you have any
solution for this problem?

Many thanks,

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