t( apply(mydf[-1], 1, function(x) as.numeric ( c( # need the t() to change columns to rows any( grepl(“POLLUTION|EMISSION|WASTE”, x) ), any(grepl(“OIL\\sSPILL”, x) ), any(grepl(“MERGER|ACQUI”, x) ), any(grepl(“MERGER|ACQUI”, x) ) ) ) ) ) #——- [,1] [,2] [,3] [,4] [1,] 1 0 0 0 [2,] 1 0 1 1 [3,] 1 1 0 0 [4,] 1 … Read more
One way is to spread your data table using Country.Name as key: dat.spread <- dat %>% spread(key=”Country.Name”, value=”inflation”) dat.spread %>% str ‘data.frame’: 50 obs. of 31 variables: $ year : chr “1967” “1968” “1969” “1970” … $ Albania : num NA NA NA NA NA NA NA NA NA NA … $ Armenia : num … Read more
Here’s a one-liner, base solution: out <- t(apply(mydf, 2, function(x) row.names(mydf)[which(as.logical(x))])) The result is a matrix: > out [,1] [,2] name1 “word1” “word3” name2 “word2” “word3” name3 “word1” “word2” which is easily made into a dataframe: > as.data.frame(out) V1 V2 name1 word1 word3 name2 word2 word3 name3 word1 word2 Here’s your data as I read … Read more
Let’s simply read and think about the error message: Error: variable ‘dummygen’ was fitted with type “numeric” but type “factor” was supplied This error occurs after the line: ooslogit <- predict.glm(logit, newlogit, se.fit=TRUE) (Presumably, at least, because you’re question isn’t very clear about this and provides lots of code that doesn’t seem related.) So R … Read more
It’s not clear how strict you want to be with the no package requirement. If you are willing to use the stats package (which most people don’t really even think of as a package because it comes standard with base R and you don’t have to load it with library()), then you can just call … Read more
First, make two lists from your database. Something like: cursor = con.cursor() cursor.execute(“SELECT * FROM traffic”) #Retrieves data from SQL rows = cursor.fetchall() Month = list() Traffic = list() for row in rows: Month.append(row[‘Month’]) # guesswork – what does a row look like? Traffic.append(row[‘Traffic’]) Then, now you have two python lists, you can make a … Read more
You want the merge function. Since your column names that you want to match on already have the same name you don’t even need to do anything special. If that wasn’t the case you would want to look into the by.x and by.y parameters that merge takes. df1 = data.frame(Site.1=c(“A”,”A”,”B”),Site.2=c(“B”,”C”,”C”),Score1=c(60,70,80)) df2 = data.frame(Site.1=c(“B”,”A”,”A”),Site.2=c(“C”,”B”,”C”), Score2=c(10,20,30)) df3 … Read more
R: sort barplot colors solved R: sort barplot colors
Basically, a minimal reproducible example (MRE) should enable others to exactly reproduce your issue on their machines. Please do not post images of your data, code, or console output! tl;dr A MRE consists of the following items: a minimal dataset, necessary to demonstrate the problem the minimal runnable code necessary to reproduce the issue, which … Read more
First convert your dates into a date-time class using asPOSIXct df = data.frame(x = c(“2012-03-01T00:05:55+00:00”, “2012-03-01T00:06:23+00:00”, “2012-03-01T00:06:52+00:00”)) df$times = as.POSIXct(df$x, format = “%Y-%m-%dT00:%H:%M+%S”) Then extract just the hour part using format df$hour = format(df$times, ‘%H’) This give you : x times hour 1 2012-03-01T00:05:55+00:00 2012-03-01 05:55:00 05 2 2012-03-01T00:06:23+00:00 2012-03-01 06:23:00 06 3 2012-03-01T00:06:52+00:00 2012-03-01 … Read more