similar to: dividing values of each column in a dataframe

How can I divide a unit by an number or average a vector of units, e.g.: u1 <- unit( 3, 'npc' ) u2 <- unit( 6, 'npc' ) u1 / 2 ( u1 + u2 ) / 2 mean( unit.c(u1,u2) ) I would use that e.g. to to calculate the coordinates of the midpoint of a line. Wolfram

Hello everyone, I've a problem and dont know how to solve. This is my first posting and it would be fantastic if you could help me. I want to divide n objects in k classes and need an output with all (n+1)(n+2)/2 possibilities. For example n=4, k=3: That would be: 4 0 0 3 1 0 3 0 1 2 2 0 2 1 1 2 0 2 1 3 0 1 2 1 1 1 2 1 0 3 0 3 1 0 2 2 0 1 3

I have this problem... I am working on implimenting a school network. I have two kinds of users on my network: students and facutly. I have also two types of operating system on my network: win 98 and win XPP. Here is the rights and privilages I am thinking about giving my users Students (usage only in computer labs): - limited roaming profile (basically, deny desktop roaming profile) - give

Hai All, Can any one please help me to divide theora video into individual frames. what is the procedure for it?. I have referred theora documentation for the same issue, but didn''t get any answer...!!?? *Thanks & Regards,JIBIN THOMAS* _______________________________________________ theora mailing list theora@xiph.org http://lists.xiph.org/mailman/listinfo/theora

Hai All, Can any one please help me to divide theora video into individual frames. what is the procedure for it?. I have referred theora documentation for the same issue, but didn''t get any answer...!!?? *Thanks & Regards,JIBIN THOMAS* _______________________________________________ theora mailing list theora@xiph.org http://lists.xiph.org/mailman/listinfo/theora

Hello, I have two time series objects, 1 is yearly (population) and the other is quarterly (bankruptcy statistics). I would like to produce a quarterly time series object that consists of bankruptcy/population. Is there a pre-built function to intelligently divide these time series: br.ts = ts(data=br.df[,-1], frequency = 4, start=c(2001,1), end=c(2008,2)) distPop.ts = ts(data=distPop.df[,-1],

to whomever that was. i deleted your email but I think below does what you want. as always, there's probably some improvement that could be done. whether it's ? minor or major, i'm not sure ? also, 3 things:? ? A)there are some Inf's in the output because of division by zero but I wasn't sure how you wanted to handle that. B) it also assumes that an n=1

hey I'd like to divide my data into four seasons. for this I made a function: Jahreszeit <- function(x) { if (x<=02 || x==12) {return("Winter") }else{ if (x>=03 && x<=05) {return("Fruehling") }else{ if (x>=06 && x<=08) {return("Sommer") }else{ if (x>=09 && x<=11) {return("Herbst") }}}}} Now, I have some

Hello all. I am not 100% familiar with the Linux advanced routing capbalieties yet, so I thought , after reading source code, documentation and more that I might be better off asking the experts on this list. I am not using this ina production environment, but at home, as a test case. First of all, here is the network topology. Internal LAN <--> Switch <--> [NAT/FIREWALL/INTERNAL LAN

Hello R community, I have two questions about using R. The first is about dividing each element of a list with another similar sized list. So, if the first list has two elements and so does the second, then the result should also be a list with two elements. For example, the inputs are: list(matrix(1:6,ncol=2),matrix(1:6,ncol=2))->l1 l2<-list(1:3,2) I want to get a list, l3 with the

This will handle all the columns; it assumes the ones you want to start with are in column 2 through the end: > library(dplyr) > df1 <- read.table(text = "ID A B + 1 1 2 + 1 0 3 + 2 5 NA + 2 1 3 + 3 1 4 + 4 NA NA + 4 0 1 + 4

Hi I have two data frames as shown below (second one is obtained by aggregating rows of similar IDs in df1.). They both have similar number of columns but rows of df2 are lesser than rows of df1. df1: ID A B 1 1 2 1 0 3 2 5 NA 2 1 3 3 1 4 4 NA NA 4

Hi, I have two dataframes one with 144 rows and 160 columns (SDF1) and one with 12 rows and 160 columns (SDF2). Now I'm trying to divide rows 1:12 with SDF2, rows 13:24 with SDF2, rows 25:36 with SDF 2, . In S-Plus the following code works fine: DFS = SDF1[1:144,1:60] / as.vector(SDF2[1:12,1:160]) but in R when I try to implement the formula I get the following error: "/

hello i have a very long column of numbers. i want R to make a new column every time the value changes from zero. example for the column: 90.1194354 87.94788274 80.34744843 64.06080347 30.40173724 0 0 0 0 0 16.28664495 23.88707926 29.31596091 48.85993485 13.02931596 0 0 0 7.600434311 20.62975027 29.31596091 32.5732899 for this example i want to get 3 columns. thanks! [[alternative HTML version

[I originally sent this mail to Martin and Richard, but then realised it's of wider interest, so I'm resending it to the list with Martin's reply attached (with his consent)] I was talking to Richard about trying to use shortened dividing keys in the B-trees. Shorter keys are better because you can then fit more in each block, so you need fewer B-tree levels for the same data.

Hi, I have a numeric pixel image which I would like to divide into factors for analysis in Spatstat. I have found that I can use cut.im() function to divide the range of pixel values into a series of equal length intervals (e.g. if my pixels values range from 0 to 60, cut.im(X.im,breaks=2) will produce two factors one containing pixel values 0-30 and one containing pixel values of 30 - 60, or

Hi R-users, I am trying to divide a vector (say X) into equal frequency bins. If one uses the hist() function, then a histogram is plotted, but with bins of equal widths, and not with bins having the same number of data points. I have then tried the histogram() function as follows: histogram(X, nint=10, breaks=NULL, equal.widths=F) This works as I want. However, I can't extract which

R 2.1.1 Win 2k Would someone suggest a method (or methods) that can be used to determine ntile cutpoints of a vector, i.e. to determine values that can be used to divide a vector into thirds such as 0-33 centile, 34-66 centile, 67-100 centile. If for example I had a vector: 1,2,3,4,5,6,7,8,9 and wanted to divide the vector into thirds I would have cut-points of 3, and 6. Thanks, John John

I'm trying to calculate the percent change for a time-series variable. Basically the first several observations often look like this, x <- c(100, 0, 0, 150, 130, 0, 0, 200, 0) and then later in the life of the variable they're are generally no more 0's. So when I try to calculate the percent change from one observation to the next, I end up with a lot of NA/Nan/INF, and

Hi everybody, I am sorry that I am kind of spamming this forum, but I have searched for some input everywhere and cant really find a nice solution for my problem. Data looks like: price 2011-11-01 08:00:00 0.000000000 2011-11-01 08:00:00 0.000000000 2011-11-01 08:02:00 0.000000000 2011-11-01 08:03:00 -0.017033339 2011-11-01 08:13:00 0.000690001 2011-11-01