search for: 82216

Displaying 7 results from an estimated 7 matches for "82216".

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2009 Jan 12
3
merge table rows (\multirow)
Hi: I need help merging rows. I am trying to merge the 'Month' column using \multirow. For example for the column 'Week' I want July to be merged into one row(weeks 27,28,29,30) and so on for the following weeks. Below, I am creating a PDF using Sweave, MikTex,R-2.8.1 and windows XP to show an example. \documentclass[11pt]{article} \usepackage{longtable,verbatim} \title{How to
2008 Jul 12
2
Excel Trend Function
...2006 21 5 0 0 0 0 2006 22 6 0 0 0 0 2006 23 7 -65 285 635 34112 2006 24 6 0 0 0 0 2006 25 7 0 0 0 0 2006 26 0 2006 27 4 228 931 1634 137726 2006 28 4 801 2231 3662 569977 2006 29 4 4544 9242 13939 6147522 2006 30 5 15798 28465 41131 44697915 2006 31 5 25398 41049 56701 68245523 2006 32 5 48197 82216 116235 322416917 2006 33 5 142980 230411 317841 2129630128 2006 34 5 227141 360468 493794 4952314336 2006 35 5 467244 756325 1045405 23281569629 2006 36 5 281049 463331 645614 9256900449 2006 37 2 227636 620330 1013023 42961663047 2006 38 3 478990 983472 1487954 70903343603 2006 39 7 539690 846522...
2009 Mar 03
1
Seperate auth per protocol (pop3/imap)
Hi, During discussions with the boss we've speculated about having users having only specific services available. One of the ideas was to have users being able to specify if they wanted POP3 or IMAP and whether we could limit their access to one or the other server side. We use MySQL for the userdb and passdb lookups at the moment. Is it possible to specify separate auth files with different
2009 Feb 07
2
Time Series Graphics - "cannot plot more than 10 series"
Hi Experts, I would like to present time series data in meaningful way in building some graphics. I've tried: (1) plot(ts(mbaye3)) and (2) plot(ts(mbaye3), start=1990) But I always get this error-message: Fehler [error] in plotts(x = x, y = y, plot.type = plot.type, xy.labels = xy.labels, : cannot plot more than 10 series as "multiple" my data: mbaye3
2009 Feb 27
4
Optimize for loop / find last record for each person
I want to find the last record for each person_id in a data frame (from a SQL database) ordered by date. Is there a better way than this for loop? for (i in 2:length(history[,1])) { if (history[i, "person_id"] == history[i - 1, "person_id"]) history[i, "order"] = history[i - 1, "order"] + 1 # same person else history[i,
2009 Feb 25
4
Have a function like the "_n_" in R ? (Automatic count function )
Have the counter function in R ? if we use the software SAS /*** SAS Code **************************/ data tmp(drop= i); retain seed x 0; do i = 1 to 5; call ranuni(seed,x); output; end; run; data new; counter=_n_; ***** this keyword _n_ ****; set tmp; run; /* _n_ (Automatic variables) are created automatically by the DATA step or by DATA step statements. */ /*** Output
2009 Feb 09
2
Dataframes: conditional calculations per row .
Dear Sirs: I've been working with several variables in a dataframe that serve as part of a calculation that I need to perform in a different way depending on its value. Let me explain: The main dataframe is called llmcc llmcc : 'data.frame': 283 obs. of 11 variables: $ Area : num 308.8 105.6 51.4 51.4 52.9 ... $ mFondo : num 30.1 10 10.2 10.2 40.4 ... $ mFachada :