I have a data set that has 655 Rows, and 21 Columns. I'm currently looping through each column and need to find the top ten of each, but when I use the head() function, it doesn't keep the labels (they are names of bacteria, each column is a sample). Is there a way to create sorted subset of data that sorts the row name along with it?
right now I am doing
topten <- head(sort(genuscounts[,c(1,i)], decreasing = TRUE) n = 10)
but I am getting an error message since column 1 is the list of names.
Thanks!
Because sort() applies to vectors, it's not going to work with your subset genuscounts[,c(1,i)], because the subset has multiple columns. In base R, you'll want to use order():
thisColumn <- genuscounts[,c(1,i)]
topten <- head(thisColumn[order(thisColumn[,2],decreasing=T),],10)
You could also use arrange_() from the dplyr package, which provides a more user-friendly interface:
library(dplyr)
head(arrange_(genuscounts[,c(1,i)],desc(names(genuscounts)[i])),10)
You'd need to use arrange_() instead of arrange() because your column name will be a string and not an object.
Hope this helps!!
Related
I'm trying to wirte a function to process multiple similar dataset, here I want to subtract scores obtained by subject in the second interview by scores obtained by the same subject in the previous interview. In all dataset I want to process, interested score will be stored in the second column. Writing for each specific dataset is simple, simply use the exact column name, everything will go fine.
d <- a %>%
arrange(by_group=interview_date) %>%
dplyr::group_by(subjectkey) %>%
dplyr::mutate(score_change = colname_2nd-lag(colname_2nd))
But since I need a generic function that can be used to process multiple dataset, I can not use exact column name. So I tried 3 approaches, both of them only altered the last line
Approach#1:
dplyr::mutate(score_change = dplyr::vars(2)-lag(dplyr::vars(2)))
Approach#2：
Second column name of interested dataset contains a same string ,so I tried
dplyr::mutate(score_change = dplyr::vars(matches('string'))-lag(dplyr::vars(matches('string'))))
Error messages of the above 2 approaches will be
Error in dplyr::vars(2) - lag(dplyr::vars(2)) :
non-numeric argument to binary operator
Approach#3:
dplyr::mutate(score_change = .[[2]]-lag(.[[2]]))
Error message:
Error: Column `score_change` must be length 2 (the group size) or one, not 10880
10880 is the row number of my sample dataset, so it look like group_by does not work in this approach
Does anyone know how to make the function perform in the desired way?
If you want to use position of the column names use cur_data()[[2]] to refer the 2nd column of the dataframe.
library(dplyr)
d <- a %>%
arrange(interview_date) %>%
dplyr::group_by(subjectkey) %>%
dplyr::mutate(score_change = cur_data()[[2]]-lag(cur_data()[[2]]))
Also note that cur_data() doesn't count the grouped column so if subjectkey is first column in your data and colname_2nd is the second one you may need to use cur_data()[[1]] instead when you group_by.
I recently moved from common dataframe manipulation in R to the tidyverse. But I got a problem regarding scaling of columns with the scale()function.
My data consists of columns of whom some are numerical and some categorical features. Also the last column is the y value of data. So I want to scale all numerical columns but not the last column.
With the select()function i am able to write a very short line of code and select all my numerical columns that need to be scaled if i add the ends_with("...") argument. But I can't really make use of that with scaling. There I have to use transmute(feature1=scale(feature1),feature2=scale(feature2)...)and name each feature individually. This works fine but bloats up the code.
So my question is:
Is there a smart solution to manipulate column by column without the need to address every single column name with
transmute?
I imagine something like:
transmute(ends_with("...")=scale(ends_with("..."),featureX,featureZ)
(well aware that this does not work)
Many thanks in advance
library(tidyverse)
data("economics")
# add variables that are not numeric
economics[7:9] <- sample(LETTERS[1:10], size = dim(economics)[1], replace = TRUE)
# add a 'y' column (for illustration)
set.seed(1)
economics$y <- rnorm(n = dim(economics)[1])
economics_modified <- economics %>%
select(-y) %>%
transmute_if(is.numeric, scale) %>%
add_column(y = economics$y)
If you want to keep those columns that are not numeric replace transmute_if with modify_if. (There might be a smarter way to exclude column y from being scaled.)
I have two vectors:
qing.emperors = c("Shunzhi","Kangxi","Yongzheng","Qianlong","Jiaqing")
reign.length = c(18,61,13,60,25)
I want to display the names of the emperors in order of decreasing reign length.
I'm supposed to use the order function, and I'm having trouble using this to get to a vector with strings as a result. Any ideas? Thanks!
another way with pipes:
library(dplyr)
data <-data.frame(qing.emperors, reign.length)
ordered <- data %>% arrange(desc(reign.length)) %>% select(qing.emperors, reign.length)
Im am trying to split a column of a dataframe into 2 columns using transform and colsplit from reshape package. I don't get what I am doing wrong. Here's an example...
library(reshape)
df1 <- data.frame(col1=c("x-1","y-2","z-3"))
Now I am trying to split the col1 into col1.a and col1.b at the delimiter '-'. the following is my code...
df1 <- transform(df1,col1 = colsplit(col1,split='-',names = c('a','b')))
Now in my RStudio when I do View(df1) I do get to see col1.a and col1.b split the way I want to.
But when I run...
df1$col1.a or head(df1$col1.a) I get NULL. Apparently I am not able to make any further operations on these split columns. What exactly is wrong with this?
colsplit returns a list, the easiest (and idiomatic) way to assign these to multiple columns in the data frame is to use [<-
eg
df1[c('col1.a','col1.b')] <- colsplit(df1$col1,'-',c('a','b'))
it will be much harder to do this within transform (see Assign multiple new variables on LHS in a single line in R)
I want to split a large dataframe into a list of dataframes according to the values in two columns. I then want to apply a common data transformation on all dataframes (lag transformation) in the resulting list. I'm aware of the split command but can only get it to work on one column of data at a time.
You need to put all the factors you want to split by in a list, eg:
split(mtcars,list(mtcars$cyl,mtcars$gear))
Then you can use lapply on this to do what else you want to do.
If you want to avoid having zero row dataframes in the results, there is a drop parameter whose default is the opposite of the drop parameter in the "[" function.
split(mtcars,list(mtcars$cyl,mtcars$gear), drop=TRUE)
how about this one:
library(plyr)
ddply(df, .(category1, category2), summarize, value1 = lag(value1), value2=lag(value2))
seems like an excelent job for plyr package and ddply() function. If there are still open questions please provide some sample data. Splitting should work on several columns as well:
df<- data.frame(value=rnorm(100), class1=factor(rep(c('a','b'), each=50)), class2=factor(rep(c('1','2'), 50)))
g <- c(factor(df$class1), factor(df$class2))
split(df$value, g)
You can also do the following:
split(x = df, f = ~ var1 + var2...)
This way, you can also achieve the same split dataframe by many variables without using a list in the f parameter.