similar to: subsetting comparison problem

> On Mar 11, 2018, at 3:32 PM, Neha Aggarwal <aggarwalneha2000 at gmail.com> wrote: > > Hello All, > I am facing a unique problem and am unable to find any help in R help pages > or online. I will appreciate your help for the following problem: > I have 2 data-frames, samples below and there is an expected output > > R Dataframe1: > C1 C2

Hello All, I am facing a unique problem and am unable to find any help in R help pages or online. I will appreciate your help for the following problem: I have 2 data-frames, samples below and there is an expected output R Dataframe1: C1 C2 C3 C4...... CN R1 0 1 0 1 R2 1 0 1 1 R3

Hello All, I am facing a unique issue and am unable to find any help in R help pages or online. I will appreciate your help for the following problem: I have 2 data-frames, samples below and there is an expected output R Dataframe1: C1 C2 C3 C4...... CN R1 0 1 0 1 R2 1 0 1 1 R3

Dataframe1 contains a list of specific dates. Dataframe2 contains a large dataset, one element of which is Date. How do I create a subset of Dataframe2 that excludes the dates from Dataframe1? I know how to do it with a left outer join vs null in SQL, but I can't figure out how to do it more directly via the subcripts that already exist? Dateframe1 Date 1/1/2010 1/18/2010 Dataframe2 Date

hello R users, I didn't find a solution for a special problem. I have two dataframes. dataframe1: X value row col ID 1 8.973498062 5512625 3460000 1 2 11.656658570 5501625 3464000 2 3 11.121777570 5495625 3473000 3 4 9.310465964 5508625 3477000 4 5 8.883483845 5515625 3496000 5 dataframe2: X value

dear all, I have two dataframes dataframe1 ID a b c dataframe2 ID value a;W 100 X;c 200 Y;Z 300 I wanted to match the IDs from the two dataframes and record the values into a new column of dataframe1 at the corresponding rows. This is what I expect: dataframe1 ID value a 100 b c 200 I tried doing it like this: for (i in seq(1:nrow(dataframe1))) {

Hi Thanks Gabor for your suggestion. I am posting the code that worked for me. dataframe1 = data.frame(cbind(Src = c(1,1,1,2,3), Target1 = c('aaa','bbb','ccc','aaa','ddd'))); #must be data frame dataframe2 = data.frame(cbind(Src = c(2,3,4,4,4), Target2 = c('aaaa','dddd','bbbb','eeee','ffff'))); dataframe3 =

hello, the example below does not work. (i know it's not supposed, but it makes it clear what i'm trying to achieve) par(mfrow=c(2,1)) xyplot(y~x2|x1,data=dataframe1,pch=20) xyplot(y~x2|x1,data=dataframe2,pch=20) i know i could probably merge the two datasets and do something like xyplot(y~x2|x1+dataset,data=merged) any other suggestion? thanks. [[alternative HTML version deleted]]

HI, I have the array list: X<-vector("list", 2) X[[1]] : data frame 1 X[[2]]: dataframe2 now i want to change index 1 and 2 into: "0-10" , "11-20" ,. finally I want to have X[["0-10"]]:dataframe1 X[["11-20"]]:dataframe2 how do I get them? Thanks a lot. Kind regards, Tammy [[alternative HTML version deleted]]

If this is posted elsewhere I cannot find it. I need to perform multiple integration where some of the variables are in the bounds of the other variables. I was trying to use R2Cuba function but cannot set the upper and lower bounds. My code so far is : int <- function(y){ u2 = y[1] z2 = y[2] u1 =y[3] z1 = y[4] ff <- u1*(z1-u1)*u2*(z2-u2)*exp(-0.027*(12-z2)) return(ff) }

Dear all, Suppose I have a very long expression e. Lets assume, for simplicity, that it is e = expression(u1+u2+u3) Now I wish to replace u2 with x and u3 with 1. I.e. the 'new' expression, after replacement, should be: > e expression(u1+x+1) My question is how to do the replacement? I have tried using: > e = parse(text=gsub("u2","x",e)) > e =

I have a dataframe named ?temp?, and another dataframe named ?descriptions?. I wish to ?rename? temp, and to ?call? it the names of a certain column in the dataframe ?descriptions?. Is there a good way to do this? A similar question: I am using a ?for loop? to create several new dataframes. e.g. for(j in 1:9){ .. I?d like each dataframe to be named d1, d2, d3, with the number being tied to

Dear all, I am using the function 'D' in the 'stats' package to perform symbolic derivation. This works very well and it is much faster than e.g. Mathematica (at least for my purposes). First, I would like to thank the development team for this excellent function. However, I run into trouble in some cases, particularly when I am to do some operations on long expressions

Dear all, I would like to solve the following problem, which can be done with optimal control theory or dynamic programming: max(x,y) a*u1+b*u2+c*f1(u2) s.t. 0<u1<x, 0<u2<f2(x,u2), x'=f3(u1,u2,x) which can be rewritten if optimal control theory should be applied as H=a*u1+b*u2+c*f1(u2)+lambda*(x') s.t. 0<u1<x, 0<u2<f2(x,u2) The maximum principle

Hi, in the code below, I am drawing 1000 samples from two beta distributions, each time using the same random number seed. Using set.seed(80) produces results I expect, in that the differences between the distributions are very small. Using set.seed(20) produces results I can't make sense of. Around half of the time, it behaves as with set.seed(80), but around half of the time, it behaves

OS: CentosOS 7 I have installed samba + openldap + smbldap-tools + pam by: yum --enablerepo=extras install -y epel-release yum install -y smbldap-tools yum install -y samba openldap openldap-clients openldap-servers migrationtools yum install -y nss-pam* I know that smbldap-tools is a dead project, but I'm interested in it and would like research on it. I create users and

Dear all, Can anyone take a look at my program below? There are two functions: f1 (lambda,z,p1) and f2(p1,cl, cu). I fixed p1=0.15 for both functions. For any fixed value of lambda (between 0.01 and 0.99), I solve f1(p1=0.15, lambda=lambda, z)=0 for the corresponding cl and cu values. Then I plug the calculated cl and cu back into the function f2. Eventually, I want to find the lambda value

Hi, If I have an R object UUU, where the second element is U2, based on "g" column of my.table my.table of UUU is: mmm ggg gindex map Info aaa123 U1 1 1 1 aaa124 U1 1 2 1 bbb1378 U2 2 1 1 bbb8888 U2 2 2 0 bbb1389 U2 2 3

On Tuesday, March 07, 2000 5:40 PM, Richard Bilonick wrote: >I need to merge several data.frames into one data.frame. In S-Plus I would >use >"merge" but I don't see a merge command in R. What is the best way to >accomplish >this? The easiest way to to this, I think, is as follows: if you have several data frames, dataframe1, dataframe2, . . . , dataframen, you

Hi, In grid 4.0, adding two complex units by `sum()` seems to give wrong results. In the following example, `u1 + u2` gives the correct result, but `sum(u1, u2)` also `sum(unit.c(u1, u2))` give the wrong results. ``` library(grid) u1 = 0.4*sum(unit(1, "inch"), unit(1, "mm")) u2 = 0.1*sum(unit(1, "inch"), unit(1, "mm")) u1 # [1] 0.4*sum(1inches, 1mm)