Displaying 7 results from an estimated 7 matches for "col12".
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col1
2012 Oct 18
4
speeding read.table
...s, each of which is headed by two lines of characters.
The first of these lines is:
TABLE NO. 1
The second is a list of column headers.
For example:
TABLE NO. 1
COL1 COL2 COL3 COL4 COL5 COL6 COL7 COL8 COL9 COL10 COL11 COL12
1.0010E+05 0.0000E+00 1.0000E+00 1.0000E+03 -1.0000E+00 1.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
1.0010E+05 1.0001E+01 1.0000E+00 1.0000E+03 -1.0000E+00 1.0000E+00 2.2737E-14 -2.2737E-14 0.0000E+00 1.9281E-08 0.0000E+00 0.0000E+00
1.0010E...
2013 Jan 03
5
splitting matrices
Dear useRs,
i want to split a matrix having 1116rows and 12 columns. i want to split that matrix into 36 small matrices each having 12 columns and 31 rows. The big matrix should be splitted row wise. which means that the first small matrix should copy values which are in first 31 rows and 12 columns of the big matrix. similarly 2nd small matrix should contain values from 32nd to 63rd row of the
2013 Jan 02
4
list of matrices
dear useRs,
i have a list containing 16 matrices. i want to calculate the column mean of each of them.
i tried
>sr <- lapply(s,function(x) colMeans(x, na.rm=TRUE))
but i am getting the following error
>Error in colMeans(x, na.rm = TRUE) : 'x' must be numeric
can it be done in any other way? and why i am getting this error??
thanks in advance..
elisa
[[alternative
2009 Nov 05
1
Set colors in a PCA plot based on a gradient vector
...col3=rgb(0.250,0.630,1.000)
> col4=rgb(0.450,0.853,1.000)
> col5=rgb(0.670,0.973,1.000)
> col6=rgb(0.880,1.000,1.000)
> col7=rgb(1.000,1.000,0.750)
> col8=rgb(1.000,0.880,0.600)
> col9=rgb(1.000,0.679,0.450)
> col10=rgb(0.970,0.430,0.370)
> col11=rgb(0.850,0.150,0.196)
> col12=rgb(0.650,0.000,0.130)
> colTemperature=c(col1,col2,col3,col4,col5,col6,col7,col8,col9,col10,col11,col12) # Put all colors in a vector
>
> tempLT50_ACC=fieldTrial0809[idx, c(48)]
> tempACC=cut(tempLT50_ACC, breaks=c(-5.4, -5.958333333, -6.516666667, -7.075, -7.633333333, -8.191666667...
2006 Jan 26
0
If you want to disconnect a database properly, there is the code:)!!!
...:database => "db2",
:username => "dbuser2",
:password => "dbuser2")
GenTableAs.set_table_name "table_no_42"
GenTableAs.reset_column_information()
obj1 = GenTableAs.new
@col2 = obj1.class.column_names()
obj1["col12"] = 456
obj1["col34"] = "789"
obj1.save
obj1 = nil
GenTableAs.clear_connection # here we disconnect
again the connection
It should give us the name of the columns of two
different tables of different database in @col1 and
@col2 and save the data to
tables.
So h...
2006 Jan 23
4
Converting from a dataset to a single "column"
I have a dataset of 3 ?columns? and 5 ?rows?.
temp<-data.frame(col1=c(5,10,14,56,7),col2=c(4,2,8,3,34),col3=c(28,4,52,34,67))
I wish to convert this to a single ?column?, with
column 1 on ?top? and column 3 on ?bottom?.
i.e.
5
10
14
56
7
4
2
8
3
34
28
4
52
34
67
Are there any functions that do this, and that will
work well on much larger datasets (e.g. 1000 rows,
6000 columns)?
2005 Apr 14
4
data manipulation
Hello,
my question is about the data handling.
I have a data set that is lined as:
4 1 17 1 1
-5.1536 -0.1668 -2.3412 -0.5062 0.9621 0.3640 0.3678 -0.5081 -0.2227
0.8142 -0.0389 -0.0445 -0.0578 -0.1175 -0.1232 0.8673 -0.1033 -0.0796
-0.0341 -0.1716 -0.1801 -0.7014 0.6578 0.5611
4 1 17 2 1
-5.1536 -0.1668 -2.3412 -0.5062 0.9621 0.3640 0.3678 -0.5081 -0.2227
0.8142 -0.0389 -0.0445