similar to: Unique Values per Column

Hi, I would like to replace the last tree characters of the values of a certain column in a dataframe. This replacement should only take place if the last three characters correspond to the value "/:/" and they should be replaced with ""(blank) I cannot perform a simple gsub because the characters /:/ might also be present somewhere else in the string values and then they

Hi, I'm looking for some lines of code that does the following: I have a dataframe with 160 Columns and a number of rows (max 30): Col1 Col2 Col3 ... Col 159 Col 160 Row 1 0 0 LD ... 0 VD Row 2 HD 0 0 0 MD Row 3 0 HD HD 0 LD Row 4 LD HD HD 0 LD ... ... LastRow HD HD LD 0 MD Now I want a dataframe that looks like this. As you see

I believe you need to spend time with an R tutorial or two: a data frame (presumably the "table" data structure you describe) can *not* contain "blanks" -- all columns must be the same length, which means NA's are filled in as needed. Also, 8e^5 * 7e^4 = 5.6e^10, which almost certainly will not fit into any local version of R (maybe it would in some server version --

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Hi, I'm trying to run on Windows 7 a scriptfile with Rscript.exe from within Excel 2010 with the following code: Call Shell(rPath & "\Rscript.exe C:\Work\Latest\_Test.R", vbHide) The good news is: the above code works perfectly, but ... If I add white spaces to my map directory, like: Call Shell(rPath & "\Rscript.exe C:\Work\Latest 1\_Test.R", vbHide) In the

Hi, I have a dataframe SDF1 that looks like this: Char1 Char2 Char 3 W.2007.02 W.2007.09 W.2007.16 W.2008.13 A C1 F1 F2 F3 A C2 F4 B C3 F5 F6 I have another dataframe SDF2 with 163 cols that has the following column names Char1 Char2 Char 3 W.2007.02 W.2007.03 W.2007.04

Hi, I have the following data frame, where col2 is a startdate and col3 an enddate COL1 COL2 COL3 A 40462 40482 B 40462 40478 The above timeframe of 3 weeks I would like to splits it in weeks like this COL1 COL2 COL3 COL4 A 40462 40468 1 A 40469 40475 1 A 40476 40482 1 B

Hi Allaisone, I took a slightly different approach but you might find this either as or more useful than your approach, or at least a start on the path to a solution you need. df1 <- data.frame(CustId=c(1,1,1,2,3,3,4,4,4),DietType=c("a","c","b","f","a","j","c","c","f"),

Hi All I have a datafram which looks like this : CustomerID DietType 1 a 1 c 1 b 2 f 2 a 3 j 4 c 4 c 4 f And I would like to reshape this so I can

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: "/

Hi, I''m having a dataframe ''Subset1'' with a number of factor variables and 160 numerical variables Now I want to make sums for all rows that have the same values for the different factor variables, except for the factor variables: VAR1,VAR2,VAR3 who may have the same values. With the formula given below this works great, but in a situation with 15000 rows and 13

Hi Duncan and Bert; I do appreciate for your replies. I just figured out that after x1= noquotes(x) commend my 733*22 matrix returns into n*1 vector. Is there way to keep this as matrix with the dimension of 733*22? Regards, Greg On Tue, Sep 19, 2017 at 10:04 AM, Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > On 19/09/2017 9:47 AM, greg holly wrote: > >> Hi all;

Well, let's see: 1) You do not appear to understand basic flow control statements in R. Note that (from ?if): if(cond) expr if(cond) cons.expr else alt.expr where "cond A length-one logical vector that is not NA." Your cond is a vector of length nrow(DF), so you don't want if, you want ifelse(). Did you fail to show us your warning messages?? 2. Revising your code and

Works fine for me. What do you object to in the following? Calling the above df "d", > dm <- as.matrix(d) > dm Sub_Pathways BMI_beta SAT_beta VAT_beta 1 "Alanine_and_Aspartate" " 0.23820" "-0.02409" " 0.94180" 2 "Alanine_and_Aspartate" "-0.31300" "-1.97510" "-2.22040" 3

Thank you. The problem was not finding the mode but applying it the R way (I have the tendency to loop into each line of the dataframes, which I believe is NOT the R way). I'll try them. Best regards Luigi On Sat, Oct 31, 2020 at 5:40 PM Bert Gunter <bgunter.4567 at gmail.com> wrote: > > As usual, a web search ("find statistical mode in R") brought up something that is

... and a slightly more efficient non-dplyr 1-liner: > sapply(strsplit(dd$Lower,"[^[:digit:]]"), function(x)sum(as.numeric(x), na.rm=TRUE)) [1] 105 67 60 100 80 Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On

Dear @William<mailto:wdunlap at tibco.com>, thanks for the feedback. I have tested it on the larger dataset and noticed that it created two variables, pf_raw and pf_curated. The output we were looking for, was one that takes the variable pf_mcl in curated dataset and replaces pf_mcl in matching rows within the raw dataset. @Eric<mailto:ericjberger at gmail.com>?s solution was able to

yes, it works, even if I do not really get how and why it's working the combination of logical results (could you provide some insights for that?) moreover, and most of all, I was hoping for a compact solution because I need to deal with MANY columns (more than 40) in data frame with the same basic structure as the simplified example I posted thanks m ----- Messaggio originale ----- Da:

Hello all, I have a set B and a dataframe df. I want to name the columns of the dataframe after the elements of the set B. For example, for set B with elements {{"P1"}, {"P2"}, {"P3", "P4"}} I want to create a new dataframe with 3 columns named {"P1"} and {"P2"} and {"P3","P4"}. I tried colnames(df)<-(B). But it

Do you mean like this: mydf <- within(mydf, { is.na(A)<- !A_flag is.na(B)<- !B_flag } ) > mydf A A_flag B B_flag 1 8 10 5 12 2 NA 0 6 9 3 10 1 NA 0 4 NA 0 1 5 5 5 2 NA 0 Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into