You could try something like this in data.table data <- data.table(yourdataframe) bar <- data[,.N,by=y] foo <- data[x==1 & z==1,list(mean.t=mean(t,na.rm=T),median.t=median(t,na.rm=T)),by=y] merge(bar[,list(y)],foo,by=”y”,all.x=T) y mean.t median.t 1: 1 12.5 12.5 2: 2 NA NA 3: 3 NA NA 4: 4 NA NA You probably could do the same in aggregate, but I am not sure you can do … Read more
Item 1: I’m not versed in k-mer counting and the definition of x is a bit shaky. If you could update your question, then I’ll try to address point 1. But, to me it seems as if you just need to use an algorithm that converts from base 10 back to base 4? Item 2: … Read more
Here’s an option. It’s more or less handcrafted as I’m not aware of any single command to do this exactly as you’ve specified. # the working directory contains d1.csv, d2.csv, d3.csv with # matching first two columns. The third column is random data read.csv(“./d1.csv”) #> a b c #> 1 1 A 0.5526777 #> 2 … Read more
Here is one approach. I created new lat for text annotation using transform in base R. I used geom_text to add the labels you wanted. I used scale_colour_discrete to change the name of the legend. library(ggmap) library(ggplot2) ### Get a map map <- get_map(location=c(lon=34.832, lat=0.852), color=”color”, source=”google”, maptype=”terrain”, zoom=9) ### Create new lat for annotation … Read more
Turns out the phrase I was looking for was “list slicing”. And I get proper access to the data using group[[i]] solved How do I gain programmatic access to names in R?
I’ll take one stab at this with two implementations. First, I’ll use a character vector. If yours is in a frame, replace it with myframe$mycolumn. v <- c(“110231 validation 108871 validation 85933”, “21102 validation 93442 21232 validation 73769 26402 validation 127221 26402”, “99763 99763 validation 99763 validation 99763”, “validation 199022 validation 122099 validation 12209 validation … Read more
You can use paste to paste the day with the ‘Date’ column, convert to Date class using as.Date with the appropriate format argument. df1$Date <- as.Date(paste(df1$Date, ’01’), ‘%Y %B %d’) For more info, you can check ?as.Date, ?strptime, ?as.POSIXct 1 solved Converting a year-month column to date type [duplicate]
A possible solution with base R by using a combination of colSums, which, toString and apply: strs$colids <- apply(strs, 1, function(x) toString(which(colSums(lut == x, na.rm=TRUE) > 0))) which gives: > strs strings colids 1 O75663 1, 3 2 O95400 1, 3 3 O95433 1, 3 4 O95456 2, 3, 4 5 O95670 2, 3, 4 … Read more
We can use rbindlist library(data.table) rbindlist(mget(paste0(“df”, 1:3)))[, .N, Id]\ # Id N #1: 1 2 #2: 2 3 #3: 3 1 #4: 4 2 #5: 7 1 #6: 6 1 #7: 9 1 #8: 11 1 solved How to count occurrence of a value across multiple data frames [duplicate]
I opted to code it in Julia Logic same just put the curly R braces in: signal = [0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 ] n_day = zeros(signal) i = 1 j = 1 for i in 1:length(signal) if … Read more
I would use indexing # save all the “transactions” where iteamtype equals “a” to an object called F f <- df[ df$itemType %in% “a” , “transaction” ] #optional (look as f) print(f) print( unique(f)) # Subset to all the rows which equals one of the “transactions” stored in the object f df[ df$transaction %in% unique( … Read more
Is this what you want? mat <- as.matrix(data.frame(V1 = sample(1:2, 100, replace = TRUE), V2 = sample(1:5, 100, replace = TRUE), V3 = sample(10:50, 100, replace = TRUE), V4 = sample(1:3, 100, replace = TRUE), V5 = sample(1:5, 100, replace = TRUE))) 1 solved Creating a matrix with random generated variables in R [closed]
Here’s a base R one-liner use ave: df$A <- ave(df$X, df$Subject, FUN = function(x) if (max(x) == 3) 1 else 0) > df Subject Year X A 1 A 1990 1 0 2 A 1991 1 0 3 A 1992 2 0 4 A 1993 3 0 5 A 1994 4 0 6 A 1995 … Read more
As I understand it, you want a column that checks if the temperature deviates by 5 – 6 degrees (if the temp is greater than 10) and another column to check if it is differs by 7 degrees or more. The current code you are using seems to identify the coldwave values correctly, although is … Read more
We can use pivot_longer library(dplyr) library(tidyr) # 1.0.0 df1 %>% pivot_longer(cols = matches(“^Q\\d+_\\d+”), values_to = ‘Q1’) %>% select(-name) 4 solved Surveydata Package Question, merge columns [closed]