good day
I don´t understand a topic here, is like it works but I can´t understand why
I have this database
# planets_df is pre-loaded in your workspace
# Use order() to create positions
positions <- order(planets_df$diameter)
positions
# Use positions to sort planets_df
planets_df[positions,]
I don´t understand why if u take the column diameter, then if u want to order it why u put it in a row of the dataframe like for me it should be [ rows, colum] but u put a column in a row and it changes, I really don´t get that.Why it´s not planets_df[,positions].
The exercise is solved I just don´t get it, is a data camp exercise btw.
Sorry if my English is wrong, it is not my native language.
I believe that I have created an example that matches your description. For the mtcars data set, which is pre-loaded in any R session, we can sort based on the variable mpg.
The function order returns the row indices sorted by mpg in this case. The ordering variable indicates the order that the rows should be presented in by storing the row indices based on mpg.
ordering <- order(mtcars$mpg)
This next step indicates that we want the rows of mtcars as specified by ordering. Essentially ordering is the order of the rows we want and so we pass that object to the row portion the call to mtcars.
mtcars[ordering,]
If we instead passed ordering as the columns, we would be reordering the columns of mtcars instead of the rows.
Related
I have a data frame called mydata with multiple columns, one of which is Benefits, which contains information about samples whether they are CB (complete responding), ICB (Intermediate) or NCB (Non-responding at all).
So basically the Benefit column is a vector with three values:
Benefit <- c("CB" , "ICB" , "NCB")
I want to make a histogram/barplot based on the number of each one of those. So basically it's not a numeric column. I tried solving this by the following code :
hist(as.numeric(metadata$Benefit))
tried also
barplot(metadata$Benefit)
didn't work obviously.
The second thing I want to do is to find a relation between the Age column of the same data frame and the Benefit column, like for example do the younger patients get more benefit ? Is there anyway to do that ?
THANKS!
Hi and welcome to the site :)
One nice way to find issues with code is to only run one command at the time.
# lets create some data
metadata <- data.frame(Benefit = c("ICB", "CB", "CB", "NCB"))
now the command 'as.numeric' does not work on character-data
as.numeric(metadata$Benefit) # returns NA's
Instead what we want is to count the number of instances of each unique value of the column Benefit, we do this with 'table'
tabledata <- table(metadata$Benefit)
and then it is the barplot function we want to create the plot
barplot(tabledata)
I'm in a very basic class that introduces R for genetic purposes. I'm encountering a rather peculiar problem in trying to follow the instructions given. Here is what I have along with the instructor's notes:
MangrovesRaw<-read.csv("C:/Users/esteb/Documents/PopGen/MangrovesSites.csv")
#i'm going to make a new dataframe now, with one column more than the mangrovesraw dataframe but the same number of rows.
View(MangrovesRaw)
Mangroves<-data.frame(matrix(nrow = 528, ncol = 23))
#next I want you to name the first column of Mangroves "pop"
colnames(Mangroves)<-c(col1="pop")
#i'm now assigning all values of that column to be 1
Mangroves$pop<-1
#assign the rest of the columns (2 to 23) to the entirety of the MangrovesRaw dataframe
#then change the names to match the mangroves raw names
colnames(Mangroves)[2:23]<-colnames(MangrovesRaw)
I'm not really sure how to assign columns that haven't been named used the $ as we have in the past. A friend suggested I first run
colnames(Mangroves)[2:23]<-colnames(MangrovesRaw)
Mangroves$X338<-MangrovesRaw
#X338 is the name of the first column from MangrovesRaw
But while this does transfer the data from MangrovesRaw, it comes at the cost of having my column names messed up with X338. added to every subsequent column. In an attempt to modify this I found the following "fix"
colnames(Mangroves)[2:23]<-colnames(MangrovesRaw)
Mangroves$X338<-MangrovesRaw[,2]
#Mangroves$X338<-MangrovesRaw[,2:22]
#MangrovesRaw has 22 columns in total
While this transferred all the data I needed for the X338 Column, it didn't transfer any data for the remaining 21 columns. The code in # just results in the same problem of having X388. show up in all my column names.
What am I doing wrong?
There are a few ways to solve this problem. It may be that your instructor wants it done a certain way, but here's one simple solution: just cbind() the Mangroves$pop column with the real data. Then the data and column names are already added.
Mangroves <- cbind(Mangroves$pop, MangrovesRaw)
Here's another way:
Mangroves[, 2:23] <- MangrovesRaw
colnames(Mangroves)[2:23] <- colnames(MangrovesRaw)
What I'm trying to do right now is to get, from the mtcars dataset on R, the car names that are that have an hp column value of less than 200. The thing is that I want the car names and I don't want to go back and rematch all the numbers with the corresponding car names. How might I go about doing that? Here's what I have so far:
> carvector<-mtcars["hp"]
> while (i<=32)
+ i<-i+1
if (mtcars[i,1]<200)
p<-mtcars[i,1]
Its not quite working and I guess I'm closer to storing the values than the names of the cars. How can I make sure the names are not lost? How do I dispose of the values I don't want?
I have a df with over 30 columns and over 200 rows, but for simplicity will use an example with 8 columns.
X1<-c(sample(100,25))
B<-c(sample(4,25,replace=TRUE))
C<-c(sample(2,25,replace =TRUE))
Y1<-c(sample(100,25))
Y2<-c(sample(100,25))
Y3<-c(sample(100,25))
Y4<-c(sample(100,25))
Y5<-c(sample(100,25))
df<-cbind(X1,B,C,Y1,Y2,Y3,Y4,Y5)
df<-as.data.frame(df)
I wrote a function that melts the data generates a plot with X1 giving the x-axis values and faceted using the values in B and C.
plotdata<-function(l){
melt<-melt(df,id.vars=c("X1","B","C"),measure.vars=l)
plot<-ggplot(melt,aes(x=X1,y=value))+geom_point()
plot2<-plot+facet_grid(B ~ C)
ggsave(filename=paste("X_vs_",l,"_faceted.jpeg",sep=""),plot=plot2)
}
I can then manually input the required Y variable
plotdata("Y1")
I don't want to generate plots for all columns. I could just type the column of interest into plotdata and then get the result, but this seems quite inelegant (and time consuming). I would prefer to be able to manually specify the columns of interest e.g. "Y1","Y3","Y4" and then write a loop function to do all those specified.
However I am new to writing for loops and can't find a way to loop in the specific column names that are required for my function to work. A standard for(i in 1:length(df)) wouldn't be appropriate because I only want to loop the user specified columns
Apologies if there is an answer to this is already in stackoverflow. I couldn't find it if there was.
Thanks to Roland for providing the following answer:
Try
for (x in c("Y1","Y3","Y4")) {plotdata(x)}
The index variable doesn't have to be numeric
This probably has a very simple answer, but I'm having trouble figuring it out...
What is a vector-based way to take one value in the cell of one column in a dataframe, conditional on some criterion in a given row being satisfied, and assign it to a cell along the same row but in a different column? I've done it with loops over if-else statements, but I'm working with pretty big data sets, and my little laptop freezes for many minutes going through the looping conditionals.
Eg. if I have sometihng like this:
Results$TResponseCorrect[Results$rownum %in% CorrectTs$rownum] <- 1
that works fine. But what doesn't work is something like
Results$TResponseCorrect[Results$rownum %in% CorrectTs$rownum] <- Results$TCorrect
In that case I get a warning saying, "number of items to replace is not a multiple of replacement length", which I basically take to mean that it can't figure out which cell of the Results$Subject column to take.
Since your problem statement implies that all these are in the same data frame you may want:
Results$TResponseCorrect[Results$rownum %in% CorrectTs$rownum] <-
Results$TCorrect[Results$rownum %in% CorrectTs$rownum]
It will then have the same number of items on the LHS and the RHS of the assignment.