similar to: sorting data whilst ignoring NA's

Hello I have the following problem : The dataframe TEST has multiple lines for a same person because : there are differents values of Nom or differents values of Prenom but the values of Matricule or Sexe or Date.de.naissance are the same. TEST <- structure(list(Matricule = c(66L, 67L, 67L, 68L, 89L, 90L, 90L, 91L, 108L, 108L, 108L), Nom = structure(c(1L, 2L, 2L, 4L, 8L, 5L, 6L, 9L, 3L, 3L,

Hi, again i am stuck in my presentation, and i have never learn R before in my life but need this to be done, so please help me out for a favour: http://www.nabble.com/file/p20333155/kew.dat kew.dat run this in R and these comes up: Month Year Rain 1 Jan 1900 74.400000 2 Feb 1900 80.500000 3 Mar 1900 23.600000 4 Apr 1900 23.600000 5 May 1900 25.100000 6

Hello, I have a weird problem here. What I want to do is that I need to draw 1000 samples from a matrix, and use glm on them. when I used this command, it runs without the problem > qdata.glm = glm(X258 ~ ., family = binomial, data = q2data[sample(dim(q2data)[1], 1000), ]) but if I drew the sample first and run glm() on that sample, it gave a warning. > qdata.sample =

On 09/11/2017 11:14 AM, Aki Tuomi wrote: > > On 11.09.2017 11:59, Nagy, Attila wrote: >> On 09/11/2017 10:42 AM, Sami Ketola wrote: >>>> On 11 Sep 2017, at 11.24, Nagy, Attila <bra at fsn.hu> wrote: >>>> I use dovecot with a broken IMAP server (which doesn't properly >>>> implement command pipelining amongst others) as an imapc backend.

Hola!! tengo un data.frame donde cada fila corresponde a un año y cada columna a un mes (De enero a diciembre) > head(valT) V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 1941 18.0 16.3 15.2 10.1 8.1 8.3 8.8 9.2 7.9 12.2 11.9 14.6 1942 17.2 15.9 13.6 11.6 8.7 6.2 6.4 7.2 9.7 12.0 14.1 16.7 1943 17.6 17.3 13.5 12.5 10.5 7.0 8.2 7.9 -999.9 -999.9

Hello all! I have a problem to extract values greater that for example 1820. I try this code: x[x[,1]>1820,]->x1 Please help me! Thank you! The data structure is: structure(c(2.576, 1.728, 3.434, 2.187, 1.928, 1.886, 1.2425, 1.23, 1.075, 1.1785, 1.186, 1.165, 1.732, 1.517, 1.4095, 1.074, 1.618, 1.677, 1.845, 1.594, 1.6655, 1.1605, 1.425, 1.099, 1.007, 1.1795, 1.3855, 1.4065, 1.138, 1.514,

Hello all! I have a problem with R. I try to merge data like this: structure(c(2.1785, 1.868, 2.1855, 2.5175, 2.025, 2.435, 1.809, 1.628, 1.327, 1.3485, 1.4335, 2.052, 2.2465, 2.151, 1.7945, 1.79, 1.6055, 1.616, 1.633, 1.665, 2.002, 2.152, 1.736, 1.7985, 1.9155, 1.7135, 1.548, 1.568, 1.713, 2.079, 1.875, 2.12, 2.072, 1.906, 1.4645, 1.3025, 1.407, 1.5445, 1.437, 1.463, 1.5235, 1.609, 1.738, 1.478,

Hello all! I have a problem to group my data (years) in 10 years classes. For example for year year decade 1598 1590-1600 1599 1590-1600 1600 1590-1600 1601 1600-1610 --- my is like this> [1] 1598 1599 1600 1601 1602 1603 1604 1605 1606 1607 1608 1609 1610 1611 1612 [16] 1613 1614 1615 1616 1617 1618 1619 1620 1621 1622 1623 1624 1625 1626 1627 [31] 1628 1629 1630 1631 1632 1633

I have following dataset: > res [,1] [,2] [,3] [1,] 1946 4 1.27 [2,] 1946 5 1.27 [3,] 1946 6 1.27 [4,] 1946 7 1.27 [5,] 1946 8 1.52 [6,] 1946 9 1.52 [7,] 1946 10 1.52 [8,] 1946 11 1.52 [9,] 1946 12 1.62 [10,] 1947 1 1.62 [11,] 1947 2 1.62 [12,] 1947 3 1.62 [13,] 1947 4 1.87 [14,] 1947 5 1.87 [15,] 1947 6 1.87 Now I write following code

Taking the Grunfeld data, which is built-in in R, for example, (1)How can I construct a dataset (or dataframe) that consists of the data of all firms in 1951? (2)How can I calculate the average capital in each form over the period 1951-1954? What I can imagine is to categorize the data by firm, and then select the data between 1951 and 1954 for each firm, but how can I do it? Thanks, Miao

On 3/12/16 12:33 AM, peter dalgaard wrote: >> On 12 Mar 2016, at 00:05 , Mick Jordan <mick.jordan at oracle.com> wrote: >> >> This is definitely obscure but we had a unit test that called .Internal(strptime, "1942/01/01", %Y/%m/%d") with timezone (TZ) set to CET. > Umm, that doesn't even parse. And fixing the typo, it doesn't run: > >>

Dear R-users, I have a complex line by xy-values (ordered by z). And I would like to get interpolated y-values on the positions of x = 0:600. How do I get the correct points? x=c(790,790,790,790,790,786,783,778,778,766,763,761,761,761,715,628,521,350,160,134,134,129,108,101,93,111,161,249,288,243,139,45,7)

Hi all, I'm looking at a large data set, and I'm interested in removing rows where only one variable is duplicated. Here's an example: > presidents Qtr1 Qtr2 Qtr3 Qtr4 1945 NA 87 82 75 1946 63 50 43 32 1947 35 60 54 55 1948 36 39 NA NA 1949 69 57 57 51 1950 45 37 46 39 1951 36 24 32 23 1952 25 32 NA 32 1953 59

Hi R-users, I would like to do looping for this process below to estimate alpha beta from gamma distribution: Here are my data: day_data1 <- 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 1943 48.3 18.5 0.0 0.0 18.3 0.0 0.0 0.0 0.0 0.0 0.0 0.0 2.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.8 2.8 1944 0.0 0.0

Hello all! I have a problem with my data in R. When I want to plot the following data, I have a problem with y scale. The maximum value is cc. 10 degrees and in R is about 100. I use this code: fasy<-read.table("gridd1.txt",sep="\t",dec=",",header=T,row.names=1) # here are the years: x <- as.numeric(rownames(fasy)) # extract a series that you want to plot: y

Hi everybody, I've a data.frame "d" like this: 0 2 4 6 8 10 12 14 16 X0 X2 X4 1945 350 NA NA NA NA NA NA NA NA 0.3848451 0.0000000 0.0000000 1946 408 NA NA NA NA NA NA NA NA 1.4202009 0.0000000 0.0000000 1947 511 NA NA NA NA NA NA NA NA 3.2540522 0.0000000 0.0000000 1948 342 215 NA NA NA NA NA NA

On 09/11/2017 10:42 AM, Sami Ketola wrote: >> On 11 Sep 2017, at 11.24, Nagy, Attila <bra at fsn.hu> wrote: >> I use dovecot with a broken IMAP server (which doesn't properly implement command pipelining amongst others) as an imapc backend. >> Dovecot issues the above command sequence (SELECT and UID FETCH pipelined), which doesn't work with this server. >>

Some help with how to re-label the vertical axis in a histogram would be appreciated. qplot(off.sc,weight=rel.freq,binwidth=.29,main="test Figure"+ylab("New from inside"))+ylab("New from outside")+ xlab("off.sc\nAggregated frequency plots for 17 equal intervals.") The code

I wanted to ask how I can make a for loop or a function return an R object with a unique name based on either some XX of the for loop or some input for the function. For example if I have a function: fn<-function(data,year){ which does does some stuff } How do I return an object from the function called X.year, such that if I run fn(data,1989), the output is an object called X.1989? In

Hello all, I changed the subject line of the e-mail, because the question I''m posing now is different than the first one. I hope that this is proper etiquette. However, the original chain is included below. I've incorporated bits of both Ethan and Brian's code into the script below, but there's one aspect I can't get my head around. I'm totally new to programming