I'm trying to convert a column showing the time of road traffic accidents from military time to standard time. The data looks like this:
Col1 Time..24hr.
1 1404
2 322
3 1945
4 1005
5 945
I'd then like to convert to 12hr so for '322' I'd like to make '3', for instance. Thanks a lot
Given your criteria -- that 322 is represented as 3 and 2045 is 20 -- how about dividing by 100 and then rounding towards 0 with trunc().
time_24hr <- c(1404, 322, 1945, 1005, 945)
trunc(time_24hr / 100)
Use GetFitARpMLE(z,4) You will get > GetFitARpMLE(z,4) $loglikelihood [1] -2350.516 $phiHat ar1 ar2 ar3 ar4 0.0000000 0.0000000 0.0000000 -0.9262513 $constantTerm [1] 0.05388392 ...
You can put your records into a data.frame and then split by the cateogies and then run the correlation for each of the categories. sapply( split(data.frame(var1, var2), categories), function(x) cor(x[[1]],x[[2]]) ) This can look prettier with the dplyr library library(dplyr) data.frame(var1=var1, var2=var2, categories=categories) %>% group_by(categories) %>% summarize(cor= cor(var1, var2)) ...
Using dplyr for your first problem: left_join(contacts, listings, by = c("id" = "id")) %>% filter(abs(listing_date - contact_date) < 30) %>% group_by(id) %>% summarise(cnt = n()) %>% right_join(listings) And the output is: id cnt city listing_date 1 6174 2 A 2015-03-01 2 2175 3 B 2015-03-14 3 9176 1 B 2015-03-30...
A boolean would most likely not yield better performance than integers, since the Excel formula engine is dynamically typed. To significantly improve the performance of your spreadsheet, you should probably consider other options. Excel PowerPivot comes to mind, as it can easily handle millions of records with hundreds of calculations,...
You can simply use input$selectRunid like this: content(GET( "http://stats", path="gentrap/alignments", query=list(runIds=input$selectRunid, userId="dev") add_headers("X-SENTINEL-KEY"="dev"), as = "parsed")) It is probably wise to add some kind of action button and trigger download only on click....
Use [[ or [ if you want to subset by string names, not $. From Hadley's Advanced R, "x$y is equivalent to x[["y", exact = FALSE]]." ## Create input input <- `names<-`(lapply(landelist, function(x) sample(0:1, 1)), landelist) filterland <- c() for (landeselect in landelist) if (input[[landeselect]] == TRUE) # use `[[`...
r,optimization,circular,maximization
I would compute all the pairs of rows in df: (pairs <- cbind(1:nrow(df), c(2:nrow(df), 1))) # [,1] [,2] # [1,] 1 2 # [2,] 2 3 # [3,] 3 4 # [4,] 4 5 # [5,] 5 6 # [6,] 6 1 You can find the best pairing with which.max:...
multivariate multiple regression can be done by lm(). This is very well documented, but here follows a little example: rawMat <- matrix(rnorm(200), ncol=2) noise <- matrix(rnorm(200, 0, 0.2), ncol=2) B <- matrix( 1:4, ncol=2) P <- t( B %*% t(rawMat)) + noise fit <- lm(P ~ rawMat) summary( fit )...
Here's a solution for extracting the article lines only. Turned out much more complex and cryptic than I'd been hoping, but I'm pretty sure it works. Also, thanks to akrun for the test data. v1 <- c('ard','b','','','','rr','','fr','','','','','gh','d'); ind <-...
I would create a list of all your matrices using mget and ls (and some regex expression according to the names of your matrices) and then modify them all at once using lapply and colnames<- and rownames<- replacement functions. Something among these lines l <- mget(ls(patter = "m\\d+.m")) lapply(l, function(x)...
excel,vba,excel-vba,excel-2010
You have: Dim RangeNOut as Double Dim RangeNOut as Integer While the IF statements in there are a nice idea, VBA will not allow you to do that. It doesn't do conditional 'compilation' since it isn't a compiled language. When VBA runs your code, all your variables are declared (no...
r,string-split,stemming,text-analysis
Given a list of English words you can do this pretty simply by looking up every possible split of the word in the list. I'll use the first Google hit I found for my word list, which contains about 70k lower-case words: wl <- read.table("http://www-personal.umich.edu/~jlawler/wordlist")$V1 check.word <- function(x, wl) {...
It's because your moving forward through the rows - if you delete row 4 then row 5 becomes row 4 and the code will jump to the new row 5 - which is in fact row 6. Hope that made sense. :) The solution will be to use a For...
If you only have 4 GBs of RAM you cannot put 5 GBs of data 'into R'. You can alternatively look at the 'Large memory and out-of-memory data' section of the High Perfomance Computing task view in R. Packages designed for out-of-memory processes such as ff may help you. Otherwise...
If data in your first table starts at A2, and your other column starts at D2, then use in E2 =VLOOKUP(D2,$A$2:$B$17,2,0) Copy down as needed....
The ###### is shown in MS Excel when the data in a cell is too long for the column width.... the data inside the cell is still correct, as you can see if you select one of those cells and look at the value displayed in the cell content bar...
r,data.table,stata,code-translation
Your intuition is correct. collapse is the Stata equivalent of R's aggregate function, which produces a new dataset from an input dataset by applying an aggregating function (or multiple aggregating functions, one per variable) to every variable in a dataset.
some reproducible code would allow me to give you some example code, but in the absence of that... wrap what you currently have in another if(), checking for length = 0 (or just && it, with the NULL check first), and display your favorite placeholder message....
string,excel,if-statement,comparison
We need an Array formula. In G2 enter: =NOT(ISERROR(MATCH(1,--EXACT(F$2:F$7,E2),0))) and copy down. Array formulas must be entered with Ctrl + Shift + Enter rather than just the Enter key. Note: The curly brackets that appear in the Formula Bar should not be typed....
r,conditional,subset,find-occurrences
Here's another possible data.table solution library(data.table) setDT(df1)[, list(Value = c("uncensored", "censored"), Time = c(Time[match("uncensored", Value)], Time[(.N - match("uncensored", rev(Value))) + 2L])), by = ID] # ID Value Time # 1: 1 uncensored 3 # 2: 1 censored 5 # 3: 2 uncensored 2 # 4: 2 censored 5 Or similarly,...
copy() is for copying data.table's. You are using it to copy a list. Try.. zz <- lapply(z,copy) zz[[1]][ , newColumn := 1 ] Using your original code, you will see that applying copy() to the list does not make a copy of the original data.table. They are still referenced by...
User the formula =LEFT(TEXT(A1,"+0.000000000;-0.000000000"),12) where A1 contains your number. This will be a text cell type but will format the cell exactly how you want it....
You can try with difftime df1$time.diff <- with(df1, difftime(time.stamp2, time.stamp1, unit='min')) df1 # time.stamp1 time.stamp2 time.diff #1 2015-01-05 15:00:00 2015-01-05 16:00:00 60 mins #2 2015-01-05 16:00:00 2015-01-05 17:00:00 60 mins #3 2015-01-05 18:00:00 2015-01-05 20:00:00 120 mins #4 2015-01-05 19:00:00 2015-01-05 20:00:00 60 mins #5 2015-01-05 20:00:00 2015-01-05 22:00:00 120...
The problem is that you pass the condition as a string and not as a real condition, so R can't evaluate it when you want it to. if you still want to pass it as string you need to parse and eval it in the right place for example: cond...
This should get you headed in the right direction, but be sure to check out the examples pointed out by @Jaap in the comments. library(ggmap) map <- get_map(location = "Mumbai", zoom = 12) df <- data.frame(location = c("Airoli", "Andheri East", "Andheri West", "Arya Nagar", "Asalfa", "Bandra East", "Bandra West"), values...
Use the LEFT() function, as shown below: lastRow = Range("A" & Rows.Count).End(xlUp).Row colNum = WorksheetFunction.Match("Number", Range("A1:CC1"), 0) For Each c In Range(Cells(2, colNum), Cells(lastRow, colNum)) If LEFT(c.Value,3) = "614" _ Or LEFT(c.Value,3) = "626" _ Or LEFT(c.Value,3) = "618" _ Or LEFT(c.Value,3) = "609" _ Or LEFT(c.Value,3) = "605" Then...
The logic here is: (1) Find the date for each subject that is the principal date, and return it for each row; and (2) subtract this date from the current date in col B. (2) is easy, but (1) requires a way to match the value in B on both...
You could loop through the rows of your data, returning the column names where the data is set with an appropriate number of NA values padded at the end: `colnames<-`(t(apply(dat == 1, 1, function(x) c(colnames(dat)[x], rep(NA, 4-sum(x))))), paste("Impair", 1:4)) # Impair1 Impair2 Impair3 Impair4 # 1 "A" NA NA NA...
if (length(z) %% 2) { z[-c(1, ceiling(length(z)/2), length(z))] } else z[-c(1, c(1,0) + floor(length(z)/2), length(z))] ...
You are just saving a map into variable and not displaying it. Just do library(ggmap) map <- qmap('Anaheim', zoom = 10, maptype = 'roadmap') map Or library(ggmap) qmap('Anaheim', zoom = 10, maptype = 'roadmap') ...
Assuming that you want to get the rowSums of columns that have 'Windows' as column names, we subset the dataset ("sep1") using grep. Then get the rowSums(Sub1), divide by the rowSums of all the numeric columns (sep1[4:7]), multiply by 100, and assign the results to a new column ("newCol") Sub1...
Using IRanges, you should use findOverlaps or mergeByOverlaps instead of countOverlaps. It, by default, doesn't return no matches though. I'll leave that to you. Instead, will show an alternate method using foverlaps() from data.table package: require(data.table) subject <- data.table(interval = paste("int", 1:4, sep=""), start = c(2,10,12,25), end = c(7,14,18,28)) query...
You can do it with rJava package. install.packages('rJava') library(rJava) .jinit() jObj=.jnew("JClass") result=.jcall(jObj,"[D","method1") Here, JClass is a Java class that should be in your ClassPath environment variable, method1 is a static method of JClass that returns double[], [D is a JNI notation for a double array. See that blog entry for...
You can try cSplit library(splitstackshape) setnames(cSplit(mergedDf, 'PROD_CODE', ','), paste0('X',1:4))[] # X1 X2 X3 X4 #1: PRD0900033 PRD0900135 PRD0900220 PRD0900709 #2: PRD0900097 PRD0900550 NA NA #3: PRD0900121 NA NA NA #4: PRD0900353 NA NA NA #5: PRD0900547 PRD0900614 NA NA Or using the devel version of data.table i.e. v1.9.5 library(data.table) setDT(mergedDf)[,...
In order to filter for "any" column, you could combine a Find result and Filter like this: Sub DateFilter() Dim nRow As Range Dim toSearch As Range 'hide dialogs Application.ScreenUpdating = False 'filter for records that have June 11, 2012 in column 3 Set toSearch = Range("A1:C4") 'detect row that...
You can try library(data.table)#v1.9.4+ setDT(yourdf)[, .N, by = A] ...
Using data.table library(data.table) setDT(df1)[, list(pages=paste(page, collapse="_")), list(user_id, date=as.Date(date, '%m/%d/%Y'))] Or using dplyr library(dplyr) df1 %>% group_by(user_id, date=as.Date(date, '%m/%d/%Y')) %>% summarise(pages=paste(page, collapse='_')) ...
It looks like you're trying to grab summary functions from each entry in a list, ignoring the elements set to -999. You can do this with something like: get_scalar <- function(name, FUN=max) { sapply(mydata[,name], function(x) if(all(x == -999)) NA else FUN(as.numeric(x[x != -999]))) } Note that I've changed your function...
It's easier to think of it in terms of the two exposures that aren't used, rather than the five that are. Let's limit the number of times an exposure can be excluded: draw_exc <- function(exposures,nexp,ng,max_excluded = 10){ nexc <- length(exposures)-nexp exp_rem <- exposures exc <- matrix(,ng,nexc) for (i in 1:ng){...
It's generally not a good idea to try to add rows one-at-a-time to a data.frame. it's better to generate all the column data at once and then throw it into a data.frame. For your specific example, the ifelse() function can help list<-c(10,20,5) data.frame(x=list, y=ifelse(list<8, "Greater","Less")) ...
In your calling code: ThinksCommerciallyInt = 1 should be ThinksCommerciallyInt := 1 similarly for the other parameters...
You can get the values with get or mget (for multiple objects) lst <- mget(myvector) lapply(seq_along(lst), function(i) write.csv(lst[[i]], file=paste(myvector[i], '.csv', sep='')) ...
Change the panel.margin argument to panel.margin = unit(c(-0.5,0-0.5,0), "lines"). For some reason the top and bottom margins need to be negative to line up perfectly. Here is the result: ...
r,if-statement,recursion,vector,integer
Your sapply call is applying fun across all values of x, when you really want it to be applying across all values of i. To get the sapply to do what I assume you want to do, you can do the following: sapply(X = 1:length(x), FUN = fun, x =...
In linux, you could use awk with fread or it can be piped with read.table. Here, I changed the delimiter to , using awk pth <- '/home/akrun/file.txt' #change it to your path v1 <- sprintf("awk '/^(ID_REF|LMN)/{ matched = 1} matched {$1=$1; print}' OFS=\",\" %s", pth) and read with fread library(data.table)...
If you have two columns of the shirt numbers and the corresponding player names then vlookup() will do this, but a warning : are shirt numbers unique i.e. one player one number... With a list of numbers in D2 to D8 and corresponding names in E2 to E8, then =VLOOKUP(A2,D2:E8,2,0)...
You have to find a suitable formula for entering in the target cell. Then you would build such formula with string concatenation, etc., for entering it via VBA. One option for the formula is to use OFFSET, as in =SUM(OFFSET($A$1,D3-1,COLUMN()-1):OFFSET($A$1,ROW()-3-1,COLUMN()-1)) This sums all values from Cell1 to Cell2, in the...
You can create a similar plot in ggplot, but you will need to do some reshaping of the data first. library(reshape2) #ggplot needs a dataframe data <- as.data.frame(data) #id variable for position in matrix data$id <- 1:nrow(data) #reshape to long format plot_data <- melt(data,id.var="id") #plot ggplot(plot_data, aes(x=id,y=value,group=variable,colour=variable)) + geom_point()+ geom_line(aes(lty=variable))...
I'm going with the assumption you meant "to the right" since you said "Another solution might be to drawn a polygon around the Baltic Sea and only to select the points within this polygon" # your sample data pts <- read.table(text="lat long 59.979687 29.706236 60.136177 28.148186 59.331383 22.376234 57.699154 11.667305...
Or you could place a rectangle on the region of interest: rect(xleft=1994,xright = 1998,ybottom=range(CVD$cvd)[1],ytop=range(CVD$cvd)[2], density=10, col = "blue") ...