-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of eliza botto
Sent: Friday, May 24, 2013 2:10 PM
To: ruipbarra...@sapo.pt; r-help@r-project.org
Subject: Re: [R] Continuous columns of matrix
Dear Rui,
Regarding your last reply there is a small additional question which i want to
ask. For the
following column
dput(mat[,1])c(0.563879907485297, 0.749077642331436, 0.681023650957497,
1.30140773002346, 1.46377246795771, 1.20312609775816, 0.651886452442823,
0.853749099839423, 1.041608733313, 0.690719733451964, 1.49415144965002,
1.30559703478921)
Your loop gives following results
[,1] [,2] [,3] [,4]
index 4.000000 5.000000 6.000000 11.000000
values 1.301408 1.463772 1.203126 1.494151
which is very accurate. Index 11 had the maximum value and 4,5,6 were the
neibhours. i
want to add an other condition which is that if a column has more
than 1 neibhours, than just like the maximum value, those neibhour can not be
next to
each other as well.
So what i should have is the following
[,1] [,2]
index 5.000000 11.000000
values 1.463772 1.494151
Is there anyway of doing it??
Thanks in advance
Elisa
Date: Fri, 24 May 2013 17:02:56 +0100
From: ruipbarra...@sapo.pt
To: eliza_bo...@hotmail.com
CC: b...@xs4all.nl; r-help@r-project.org
Subject: Re: [R] Continuous columns of matrix
Hello,
No problem. Just change <0 to >= and Inf to -Inf:
fun2 <- function(x){
n <- length(x)
imx <- which.max(x)
if(imx == 1){
x[2] <- x[n] <- -Inf
}else if(imx == n){
x[1] <- x[n - 1] <- -Inf
}else{
x[imx - 1] <- -Inf
x[imx + 1] <- -Inf
}
index <- which(x >= 0.8*x[imx])
values <- x[index]
list(index = index, values = values)
}
apply(mat, 2, fun2)
Rui Barradas
Em 24-05-2013 16:23, eliza botto escreveu:
Dear Rui,
I infact wanted to have something like the following..
suppose the columns are
structure(c(0.706461987893674, 0.998391468394261, 0.72402995269242,
1.70874688194537, 1.93906363083693, 0.89540353128442, 0.328327645695443,
0.427434603701202, 0.591932250254601, 0.444627635494183, 1.44407704434405,
1.79150336746345, 2.06195904001605, 1.41493262330451, 1.35748791897328,
1.19490680241894, 0.702488756183322, 0.338258418490199, 0.123398398622741,
0.138548982660226, 0.16170889185798, 0.414543218677095, 1.84629295875002,
2.24547399004563), .Dim = c(12L, 2L))
For col 1
[[1]]
$Index
5 12
$value
1.939 1.79
Although value 1.708 of index 4 also has value which is above 80% of the maximum
value but as it is in the neighbor of maxmimum value so we wont consider it.
similarly for the column 2
[[1]]
$Index
12
$value
2.245
Although values 1.846 of index 11 and 2.0619 of index 1 also have values which
are
above 80% of the maximum value but as they are in the neighbor of maxmimum
value so
we wont consider them.
i am sorry if the manner in which i asked my question was not conclusive.
i hope you wont mind...
Elisa
Date: Fri, 24 May 2013 15:59:50 +0100
From: ruipbarra...@sapo.pt
To: eliza_bo...@hotmail.com
CC: b...@xs4all.nl; r-help@r-project.org
Subject: Re: [R] Continuous columns of matrix
Hello,
Something like this?
fun2 <- function(x){
n <- length(x)
imx <- which.max(x)
if(imx == 1){
x[2] <- x[n] <- Inf
}else if(imx == n){
x[1] <- x[n - 1] <- Inf
}else{
x[imx - 1] <- Inf
x[imx + 1] <- Inf
}
index <- which(x <= 0.8*x[imx])
values <- x[index]
list(index = index, values = values)
}
apply(mat, 2, fun2)
Rui Barradas
Em 24-05-2013 13:40, eliza botto escreveu:
Dear Rui,Thankyou very much for your help. just for my own knowledge what if
want the values and index, which are less than or equal to 80% of the maximum
value
other than those in the neighbors?? like if maximum is in row number 5 of any
column
then the second maximum can be in any row other than 4 and 6. similarly if
maximum is
in row number 12 than the second maximum can be in any row other than 1 and
11...thankyou very much for your help
elisa
Date: Fri, 24 May 2013 12:37:37 +0100
From: ruipbarra...@sapo.pt
To: eliza_bo...@hotmail.com
CC: b...@xs4all.nl; r-help@r-project.org
Subject: Re: [R] Continuous columns of matrix
Hello,
Berend is right, it's at least confusing. To get just the index of the
maximum value in each column,
apply(mat, 2, which.max)
To get that index and the two neighbours (before and after, wraping
around) if they are greater than or equal to 80% of the maximum, try
fun <- function(x){
n <- length(x)
imx <- which.max(x)
sec <- numeric(2)
if(imx == 1){
if(x[n] >= 0.8*x[imx]) sec[1] <- n
if(x[2] >= 0.8*x[imx]) sec[2] <- 2
}else if(imx == n){
if(x[n - 1] >= 0.8*x[imx]) sec[1] <- n - 1
if(x[1] >= 0.8*x[imx]) sec[2] <- 1
}else{
if(x[imx - 1] >= 0.8*x[imx]) sec[1] <- imx - 1
if(x[imx + 1] >= 0.8*x[imx]) sec[2] <- imx + 1
}
sec <- sec[sec != 0]
c(imx, sec)
}
apply(mat, 2, fun)
Note that the result comes with the maximum first and the others follow.
Hope this helps,
Rui Barradas
Em 24-05-2013 11:41, eliza botto escreveu:
There you go!!!
structure(c(0.706461987893674, 0.998391468394261, 0.72402995269242,
1.70874688194537, 1.93906363083693, 0.89540353128442, 0.328327645695443,
0.427434603701202, 0.591932250254601, 0.444627635494183, 1.44407704434405,
1.79150336746345, 0.94525563730664, 1.1025988539757, 0.944726401770203,
0.941068515436361, 1.50874009152312, 0.590015480056925, 0.311905493999476,
0.596771673581893, 1.01502499067153, 0.803273181849135, 1.6704085033648,
1.57021117646422, 0.492096635764151, 0.433332688044914, 0.521585941816778,
1.66472272302545, 2.61878329527404, 2.19154489521664, 0.493876245329722,
0.4915787202584, 0.889477365620806, 0.609135860199222, 0.739201878930367,
0.854663750519518, 0.948228727226247, 1.38569091844218, 0.910510759802679,
1.25991218521949, 0.993123416952421, 0.553640392997634, 0.357487763503204,
0.368328033777003, 0.344255688489322, 0.423679560916755, 1.32093576037521,
3.13420679229785, 2.06195904001605, 1.41493262330451, 1.35748791897328,
1.19490680241894, 0.702488!
75618332!
2, 0.338258418490199, 0.123398398622741, 0.138548982660226,
0.16170889185798, 0.414543218677095, 1.84629295875002, 2.24547399004563,
0.0849732189580101, 0.070591276171845, 0.0926010253161898, 0.362209761457517,
1.45769283057202, 3.16165004659667, 2.74903557756267, 1.94633472878995,
1.19319875840883, 0.533232612926756, 0.225531074123974, 0.122949089115578),
.Dim = c(12L, 6L))
Thanks once again..
Elisa
Subject: Re: [R] Continuous columns of matrix
From: b...@xs4all.nl
Date: Fri, 24 May 2013 12:36:47 +0200
CC: r-help@r-project.org
To: eliza_bo...@hotmail.com
On 24-05-2013, at 12:24, eliza botto <eliza_bo...@hotmail.com> wrote:
Dear useRs,If i have a matrix, say, 12 rows and 6 columns. The columns are
continuous. I want to find the index of maximum values and the actual maximum
values.
The maximum values in each column are the highest values and the values greater
than
or equal to 80% of the maximum value. Moreover, if a column has more than one
maximum values than these values should come immediately next to each other.
For
example, if you column 1 has a highest value in 6th row then the second maximum
values
cant be in row 5 or 7. And as the columns are continuous therefore, if maximum
value is
in row 12th, then the second maximum cant be in row 11 and 1.Thankyou very much
indeed in advance
Incomprehensible.
What is a continuous column?
Please give an example input matrix and and the result you want.
Berend
Elisa
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and provide commented, minimal, self-contained, reproducible code.
