I am trying to read and transpose a data set within more than 18,000 rows and 90 columns into R. (Because the data set is actually including 18,000 variables, and 90 samples.) I tried read.transpose but does not work. Any suggestion? Many thanks.
Actually that is a pretty average/small data set. Just read it in like you would any data frame. Then the function you are looking for is t(), use ?t for more information
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I am a beginner in R program.
I imported a csv file. This file only contains one column with 50 characters, but R classifies it as a dataframe. I need all possible combinations within elements of this column. I think I need to work with a vector not with a data frame, how can I do it?
Thank you!
Actually your data frame already contains the vector you need. You can call it with
dataframe$column_name
The text before the $ operator specifies your data frame, and after is your vector, which is a column in your data frame. So when you run your calculations you can just write
function(dataframe$column_name)
In your specific case with a single vector, it may be simplest to change the dataframe into a 2d vector. But when you start manipulating your data, you'll likely store more vectors of variables. You'll want to keep those vectors organized within data frames.
Do you mean unlist?
You can use it to change a data frame into a vector, then you can use combn to get combination.
I have a LARGE dataset with over 100 Million rows. I only want to read part of the data corresponds to one particular level of a factor, say column1 == A. How do I accomplish this in R using read.csv?
Thank you
You can't filter rows using read.csv. You might try sqldf::read.csv.sql as outlined in answers to this question.
But I think most people would process the file using another tool first. For example, csvkit allows filtering by rows.
I am trying to transfer data from one data frame to other. I want to copy all 8 columns from a huge data frame to a smaller one and name the columns n1, n2, etc..
first I am trying to find the column number from which I need to copy by using this
x=as.numeric(which(colnames(old_df)=='N1_data'))
Then I am pasting it in new data frame this way
new_df[paste('N',1:8,'new',sep='')]=old_df[x:x+7]
However, when I run this, all the new 8 columns have exactly same data. However, instead if I directly use the value of x, then I get what I want like
new_df[paste('N',1:8,'new',sep='')]=old_df[10:17]
So my questions are
Why I am not able to use the variable x. I added as.numeric just to make sure it is a number not a list. However, that does not seem to help.
Is there any better or more efficient way to achieve this?
If I'm understanding your question correctly, you may be overthinking the problem.
library(dplyr);
new_df <- select(old_df, N1_data, N2_data, N3_data, N4_data,
N5_data, N6_data, N7_data, N8_data);
colnames(new_df) <- sub("N(\\d)_data", "n\\\\1", colnames(new_df));
I have a huge data frame df in R. Now,if I invoke View(df) then Rstudio does not respond since it's too big. So, I am wondering, if there is any way to view, say first 500 lines, of a data frame as spreadsheet.
(I know it's possible to view using head but I want to see it as spreadsheet since it has too many columns and using head to see too many columns is not really user friendly)
If you want to see first 100 lines of the data frame df as spreadsheet, use
View(head(df,100))
You can look at the dataframe as a matrix such as df[rowfrom:rowto, columnfrom:columnto], for example in your case:
df[1:500,]
I am producing a script where I have done many manipulations to a bunch of data and, I do these same manipulations to another dataset. Both data sets have the same rows, columns, and headers. I would like to be able to join the two data sets together where I place dataset A above dataset B. I wouldn't need to headers for dataset B and would instead just clump all of the data together as if they were never really separated in the first place. Is there a simply way to do this?
Yes. Use rbind() command.
combineddataset = rbind(dataset1, dataset2)
Hope that helps.
And for completeness, you could also use the rbind.fill function found in the plyr package.