I want to read a file and calculate the mean of it.
`>list
[1] "book1.csv" "book2.csv".
for book1
observation1
23
24
65
76
34
In books i have a variable observation 1 and observation 2 column for book 1 and 2 respectively. So i want to write a function where i can calculate mean of it.I am new to R and not able subset the variable of books. Can anyone please help me out in writing the function?
Try this. File represents the file to be read in (book1) and the variable represents the variable to take mean over (observation 1)
read.mean<-function(file,variable){
df<-read.csv(file)
mean.df <- mean(df[,variable])
return(mean.df)
}
Make sure to pass your arguments in quotes, i.e. read.mean("book1", "observation1"). There is a way to do it without the quotes (Passing a variable name to a function in R) but it is complicated.
Related
I apologize in advance for the somewhat lack of reproducibility here. I am doing an analysis on a very large (for me) dataset. It is from the CMS Open Payments database.
There are four files I downloaded from that website, read into R using readr, then manipulated a bit to make them smaller (column removal), and then stuck them all together using rbind. I would like to write my pared down file out to an external hard drive so I don't have to read in all the data each time I want to work on it and doing the paring then. (Obviously, its all scripted but, it takes about 45 minutes to do this so I'd like to avoid it if possible.)
So I wrote out the data and read it in, but now I am getting different results. Below is about as close as I can get to a good example. The data is named sa_all. There is a column in the table for the source. It can only take on two values: gen or res. It is a column that is actually added as part of the analysis, not one that comes in the data.
table(sa_all$src)
gen res
14837291 822559
So I save the sa_all dataframe into a CSV file.
write.csv(sa_all, 'D:\\Open_Payments\\data\\written_files\\sa_all.csv',
row.names = FALSE)
Then I open it:
sa_all2 <- read_csv('D:\\Open_Payments\\data\\written_files\\sa_all.csv')
table(sa_all2$src)
g gen res
1 14837289 822559
I did receive the following parsing warnings.
Warning: 4 parsing failures.
row col expected actual
5454739 pmt_nature embedded null
7849361 src delimiter or quote 2
7849361 src embedded null
7849361 NA 28 columns 54 columns
Since I manually add the src column and it can only take on two values, I don't see how this could cause any parsing errors.
Has anyone had any similar problems using readr? Thank you.
Just to follow up on the comment:
write_csv(sa_all, 'D:\\Open_Payments\\data\\written_files\\sa_all.csv')
sa_all2a <- read_csv('D:\\Open_Payments\\data\\written_files\\sa_all.csv')
Warning: 83 parsing failures.
row col expected actual
1535657 drug2 embedded null
1535657 NA 28 columns 25 columns
1535748 drug1 embedded null
1535748 year an integer No
1535748 NA 28 columns 27 columns
Even more parsing errors and it looks like some columns are getting shuffled entirely:
table(sa_all2a$src)
100000000278 Allergan Inc. gen GlaxoSmithKline, LLC.
1 1 14837267 1
No res
1 822559
There are columns for manufacturer names and it looks like those are leaking into the src column when I use the write_csv function.
I have been writing a code to ease performing a discriminant analysis using the lda function. But actually I have a step which I cannot solve. And it is when I have to introduce the name of the categorical column in the code. Imagine we have the next table (called smoke), in which the column Factor represents the groups (in our cases, smoker and nsmok).
smoke
Factor Lung Heart Blood
1 smoker 7 22 15
2 smoker 8 21 12
3 nsmok 22 9 5
This is the code I have been preparing. Please, look at the XXXX's in the code (it appears twice). I want them to write automatically the name of the categorical column, instead of writing directly it twice.
lda=lda(XXXX~.,data=Smoke)
plot(lda)
lda
lda$counts
lda$svd
lda.p=predict(lda)
Tabla=table(Smoke$XXXX,lda.p$class)
Tabla
diag(prop.table(Tabla, 1))
sum(diag(prop.table(Tabla)))
I thought that writing...
colnames(Table)[1]
... would solve it. But actually there still exist some errors when running the code.
Otherwise, I though that introducing directly the name in this way:
Column_Factor-> Factor
and writing Column_Factor in the two places in the code would solve it. But it isn't.
Any ideas?
You could do something like this:
library(MASS)
#gets the column name of the factor, maybe check if there is only one factor column first
Column_Factor <- names(Smoke)[sapply(Smoke, class)=="factor"]
#creates the formula by pasting the name and the RHS
lda <- lda(as.formula(paste(Column_Factor,"~.",sep="")),data=Smoke)
plot(lda)
lda
lda$counts
lda$svd
lda.p=predict(lda)
#selects the column using the variable
Tabla=table(Smoke[,Column_Factor],lda.p$class)
Tabla
diag(prop.table(Tabla, 1))
sum(diag(prop.table(Tabla)))
I'm new to R and I am having some trouble iterating through the unique element of a vector. I have a dataframe "School" with 700 different teachers. Each teacher has around 40 students.
I want to be able to loop through each teacher, create a graphs for the mean score of his/her students' over time, save the graphs in a folder and automatically email that folder to that teacher.
I'm just getting started and am having trouble setting up the for-loop. In Stata, I know how to loop through each unique element in a list, but am having trouble doing that in R. Any help would be appreciated.
School$Teacher School$Student School$ScoreNovember School$ScoreDec School$TeacherEmail
A 1 35 45 A#school.org
A 2 43 65 A#school.org
B 1 66 54 B#school.org
A 3 97 99 A#school.org
C 1 23 45 C#school.org
Your question seems a bit vague and it looks like you want us to write your whole project. Could you share what you have done so far and where exactly you are struggling?
see ?subset
School=data.frame(Teacher=c("A","B"), ScoreNovember=10:11, ScoreDec=13:14)
for (teacher in unique(School$Teacher)) {
teacher_df=subset(School, Teacher==teacher)
MeanScoreNovember=mean(teacher_df$ScoreNovember)
MeanScoreDec =mean(teacher_df$ScoreDec)
# do your plot
# send your email
}
I think you have 3 questions, which will need separate questions, how do I:
Create graphs
Automatically email output
Compute a subset mean based on group
For the 3rd one, I like using the plyr package, other people will recommend data.table or dplyrpackages. You can also use aggregate from base. To get a teacher's mean:
library(plyr)
ddply(School,.(Teacher),summarise,Nov_m=mean(ScoreNovember))
If you want per student per teacher, etc. just add between the columns, like:
library(plyr)
ddply(School,.(Teacher,Student),summarise,Nov_m=mean(ScoreNovember))
You could do that for each score column (and then chart it) if your data was long rather than wide you could also add the date ('November', 'Dec') as a group in the brackets, or:
library(plyr)
ddply(School,.(Teacher,Student),summarise,Nov_m=mean(ScoreNovember),Dec_m=mean(ScoreDec))
See if that helps with the 3rd, but look at splitting your questions up too.
thank you for the help. I am attempting to write an equation that uses values selected from an .csv file. It looks something like this, let's call it df.
df<-read.csv("SiteTS.csv", header=TRUE,sep=",")
df
Site TS
1 H4A1 -42.75209
2 H4A2 -43.75101
3 H4A3 -41.75318
4 H4C3 -46.76770
5 N1C1 -42.68940
6 N1C2 -36.95200
7 N1C3 -43.16750
8 N2A2 -38.58040
9 S4C1 -35.32000
10 S4C2 -34.52420
My equation requires the value in the TS column for each site. I am attempting to create a new column called SigmaBS with the results of the equation using TS.
df["SigmaBS"]<-10^(subset(df, Site=="H4A1"/10)
Which is where I am running into issues, as the subset function returns all columns that correlate with the Site column = H4A1
subset(df, Site =="H4A1")
Site TS
1 2411 -42.75209
But again, I only need the value -42.75209.
I apologize if this is a simple question, but I would very much appreciate any help you may be able to offer.
If you insist on using the subset function, it has a select argument:
subset(df, Site=="H4A1", select="TS")
A better option is to use [] notation:
df[df$Site=="H4A1", "TS"]
Or the $ operator:
subset(df, Site=="H4A1")$TS
You can use this simple command:
df$SigmaBS <- 10 ^ (df$TS / 10)
It sounds like you're trying to create a new column called SigmaBS where the values in each row are 10^(value of TS) / 10
If so, this code should work:
SigmaBS <- sapply(df$TS, function(x) 10^(x/10))
df$SigmaBS <- SigmaBS
I am trying to plot histograms with long term (several years) mean precipitation (pp) for each day of the month from a series of files. Each file has data collected from a different place (and has a different code). Each of my files looks like this:
X code year month day pp
1 2867 1945 1 1 0.0
2 2867 1945 1 2 0.0
...
And I am using the following code:
files <- list.files(pattern=".csv")
par(mfrow=c(4,6))
for (i in 1:24) {
obs <- read.table(files[i],sep=",", header=TRUE)
media.dia <- ddply(obs, .(day), summarise, daily.mean<-mean(pp))
codigo <- unique(obs$code)
hist(daily.mean, main=c("hist per day of month", codigo))
}
I get 24 histograms with 24 different codes in the title, but instead of 24 DIFFERENT histograms from 24 different locations, I get the same histogram 24 times (with 24 different titles). Can anybody tell me why? Thanks!
There are at least two errors I can see in your code.
There is an error in your ddply statement.
You are passing the wrong variable to hist, thus plotting something that may or may not exist depending on previous session actions.
The problem in your ddply statement is that you are doing an invalid assign (using <- ). Fix this by using =:
media.dia<- ddply(obs, .(day),summarise, daily.mean = mean(pp))
Then edit your hist statement:
hist(media.dia$daily.mean,main=c("hist per day of month",codigo))
I suspect the problem is that you are not passing the correct parameter to hist. The reason that your code actually produces a plot at all, is because in some previous step in your session you must have created a variable called daily.mean (as Brandon points out in the comment.)
I think the daily.mean calculated in the ddply function is assigned in a separate environment, and does not exist in an environment hist can see.
Try daily.mean<<-mean(pp)