Tried to create your data using the following: df <- data.frame(col1 = c(1,1,2,2,1,1,2,2), col2 = c(34,34,342,23,34,34,342,23), col3 = c(3,4,3,4,3,4,3,4)) And, if you wish to subset based on only one column, you can use @kyle-marsh solution > df[df$col1 == 1, ] col1 col2 col3 1 1 34 3 2 1 34 4 5 1 34 3 … Read more
Here is one way to accomplish this: #make a data frame df <- data.frame(a = c(1,2,3), b = c(1,NA,3), c = c(1,2,3)) #use rowSums on a dataframe made with each of your columns and specify na.rm = TRUE df$mySum <- rowSums(cbind(df$a,df$b), na.rm = TRUE) 0 solved Row-wise sum for select columns while ignoring NAs [closed]
We may use either of the following: do.call(function (…) paste(…, sep = “-“), rld[groups]) do.call(paste, c(rld[groups], sep = “-“)) We can consider a small, reproducible example: rld <- mtcars[1:5, ] groups <- names(mtcars)[c(1,3,5,6,8)] do.call(paste, c(rld[groups], sep = “-“)) #[1] “21-160-3.9-2.62-0” “21-160-3.9-2.875-0” “22.8-108-3.85-2.32-1” #[4] “21.4-258-3.08-3.215-1” “18.7-360-3.15-3.44-0” Note, it is your responsibility to ensure all(groups %in% names(rld)) … Read more
This is just a direct translation of your plyr call in dplyr: library(dplyr) DF2 = summarise(group_by(DF, id, feature), value=value[which(date == max(date))]) 2 solved summarizing a field based on the value of another field in dplyr
The following should work, I have stored your example input as a dataframe called ‘dat’: library(ggplot2) library(cowplot) plt = ggplot(dat, aes(Income_group, percentage)) + geom_bar(stat=”identity”) + facet_grid(Airport ~.) + background_grid(major=”y”, minor = “none”) + panel_border() plt + theme(axis.text.x = element_text(angle = 90, hjust = 1)) 2 solved In ggplot, how can I plot multiple graphs in … Read more
You were almost there. you just had to specify n and return the right object at the end. matrix_a = function(n) { A = matrix(data = 0, nrow = n, ncol = n) for (i in 1:n) { for (j in 1:n) { if (i>=j) {A[i,j] = (i^2)} if (i<j) {A[i,j] = (j^2)} } } … Read more
This answer assumes reading of the earlier question and comments so is not “code only”. qt = expand.grid(Month=c(“G”,”J”,”M”,”Q”,”V”), Year=1975:2016) query_names_vec <- apply(qt, 1, function(x) paste0(“CME/GC”, paste0( x, collapse=””) ) ) > head( query_names_vec ) [1] “CME/GCG1975” “CME/GCJ1975” “CME/GCM1975” “CME/GCQ1975” [5] “CME/GCV1975” “CME/GCG1976” 10 solved Loop R with Quandl as optional
One solution is as follows: df2[, (df2[‘Row’, ] %in% df1$Row) & (df2[‘Column’, ] %in% df1$Column)] Head of the output is as follows: V1 V2 V3 Row 1 2 3 Column 5 4 3 3 49 29 34 4 45 42 18 5 9 15 45 6 34 35 19 2 solved Matching information from different … Read more
You can do this using the popular stringr and dplyr libraries. library(dplyr) library(stringr) df <- tibble( sentence = c( “Time & tide waits for none”, ” Tit for tat”, “Eyes are mirror of person’s thoughts”, “Some Other Sentence”, “Odd sentences failure” ) ) df <- df %>% # Split the sentence and store it in … Read more
I don’t think data.frame() can handle vectors as individual elements, so I used data_frame() from the tibble package: df <- tibble::data_frame(var1 = list(c(“apple”, “www”), c(“dog”, “cat”, “kkk”)), var2 = list(c(“apple”, “zzz”), c(“cat”, “kkk”))) apply function by row, where the function takes the intersection of the first and second list elements: apply(df, 1, function(x) intersect(x[[1]], x[[2]])) … Read more
I’ll make some fake data (I won’t try to transcribe yours) first: set.seed(2) x <- data.frame( Date = rep(Sys.Date() + 0:1, each = 24), # Year, Month, Day … are not used here Hour = rep(0:23, times = 2), Value = sample(1e2, size = 48, replace = TRUE) ) This is a straight-forward ggplot2 plot: … Read more
You seem to be asking how to use a column name passed as an argument to a function?? myfunction <- function(df, col) mean(df[,col], na.rm=T) # test set.seed(1) df <- data.frame(x=rnorm(10),y=rnorm(10)) myfunction(df,”x”) # [1] 0.1322028 This also works if you pass a column number. myfunction(df,1) # [1] 0.1322028 1 solved R function calculating statistic of variable … Read more
Converting a number of seconds (such as 75) to a decimal format such as minutes.seconds is a bit odd, but you could do it this way: a <- 75 hour <- floor(a / 60) # floor() rounds down to the last full hour minute <- a %% 60 * 0.01 # the %% operator gives … Read more
We can convert those factor columns to numeric library(dplyr) library(magrittr) mydata %<>% mutate_if(is.factor, funs(as.numeric(as.character(.)))) Or using base R i1 <- sapply(mydata, is.factor) mydata[i1] <- lapply(mydata[i1], function(x) as.numeric(as.character(x))) and then the code should work 2 solved R: Not meaningful for factors even though its data type is numeric [duplicate]
If the data frames are DF1, DF2, …, DF20 then assuming you want the data frame DF such that DF[i, j] = min(DF1[i, j], DF2[i, j], …, DF20[i, j]) then: L <- list(DF1, DF2, DF3, DF4, DF5, DF6, DF7, DF8, DF9, DF10, DF11, DF12, DF13, DF14, DF15, DF16, DF17, DF18, DF19, DF20) Reduce(pmin, L) or … Read more