I have a list of data frame (lets call that "data") that I have generated which goes something like this:
$"something.csv"
x y z
1 1 1 1
2 2 2 2
3 3 3 3
$"something else.csv"
x y z
1 1 1 1
2 2 2 2
3 3 3 3
I would like to output from the table "something.csv" one number within column x.
So far I have used:
data$"something.csv"$x[2]
This coding works and I am happy that it does, but my problem is that I want to automate this process and so i have put all the table titles into a list (filename) which goes:
[1] "something.csv", "something else.csv"
So i made a for loop which should allow me to do so but when I put in:
data$filename[1]$x[2]
it gives me back NULL.
When i print filename[1], I get [1] "something.csv" and if I type
data$"something.csv"$x[2]
I get the result I want. so if filename[1] = "something.csv", why does it not give me the same results?
I just want my code to out put the second row of column x and automate by using filename[i] in a for loop.
The way you have tried to approach the problem tries to find a column 'filename[1]' from the list, but it is not found. Hence, the NULL gets returned.
You need to use square brackets, and subset the object data. Here's an example:
# Generate data
data<-vector("list", 2)
names(data)<-c("something.csv", "something else.csv")
data[[1]]<-data.frame(x=1:3, y=1:3, z=1:3)
data[[2]]<-data.frame(x=1:3, y=1:3, z=1:3)
filename<-names(l)
# Subset the data
# The first data frame, notice the square brackets for subsetting lists!
data[[filename[1]]]
# column x
data[[filename[1]]]$x
# Second observation of x
data[[filename[1]]]$x[2]
The above uses for subsetting the names of the objects in the list. You can also use the number-based subsetting suggested by #Jeremy.
you can also use [ and [[ to call data$"something.csv"$x[2] try
data[[1]][2,1]
where [[1]] is the first list element and [2,1] is the data frame reference element
Related
How can I get an array form a column in data frame satisfying a condition?
example:
x=data.frame(pn=c('a','b','c','d','e','f'),price=c(1,2,3,4,5,6))
Then, for a given list of pn (an array that can have any size), like this:
y=c('a','b','f','a','a','b','b','a','f','f')
I want an array of prices regarding y. The expected output is:
1,2,6,1,1,2,2,1,6,6
(No loop or lambda function)
Use a named vector to match
unname(setNames(x$price, x$pn)[y])
#[1] 1 2 6 1 1 2 2 1 6 6
I have 20 excel files containing city level data for each year. I imported them in a list because I thought it will be easier to loop over them.
The first task that I wanted to do is to change the name of the second column of each file.
If, for a single file I do:
#data is a list of data tables/frames. Example:
data<-list(a = data.frame(1:2,3:4),b = data.frame(5:8,15:18) )
#renaming first column of a (works)
names(data[[1]])[2]<-"ABC"
I am able to rename the column.
To do batch editing I wanted to write a function to be used in lapply. The function should be a simple version of the above thing:
rename <-function(df){
names(df)[2]<-"XYZ"}
Rename(data[[1]]) however, does nothing to the second column. Any ideas why?
You need to return the full modified object at each iteration:
data <- lapply( data, function(x) {names(x)[2]<-"ABC"; x})
data
#---------
[[1]]
X1.2 ABC
1 1 3
2 2 4
[[2]]
X5.8 ABC
1 5 15
2 6 16
3 7 17
4 8 18
I'm sure this is a duplicate but I don't know what the right search terms might be, so I'm just answering it .... again.
I have got the following problem. I have a data.frame with an x and y column representing some points in space:
X<-c(18.25743,18.25783,18.25823,18.25850,18.25863,18.25878,
18.25885,18.25912,18.25943,18.25962,18.25978,18.26000,
18.26022,18.26051,18.26070,18.26095,18.26118,18.26140,
18.26189,18.26250,18.26310,18.26390)
Y<-c(44.69561,44.69564,44.69567,44.69567,44.69586,
44.69600,44.69637,44.69671,44.69691,44.69701,44.69720,
44.69740,44.69763,44.69774,44.69787,44.69790,44.69791,
44.69795,44.69812,44.69802,44.69812,44.69834)
eDF<-data.frame(X,Y)
Now my problem is they are "sorted" wrong for plotting.So what I need is a function to write together the rows of the two points which belong together (in a list of lists):
1 and 12 is ID1
2 and 13 is ID2
3 and 14 is ID3
...
11 and 22 is ID11
Every so created list within the list of lists should have its unique ID (just numerating from 1 to the end). Well because I got this problem in all my data with different length.
It would be great if the starting point of the second consecutive row selecting (the 12) is flexible always taking the first row after half of the data.((rownumber/2)+1) in this example
12.
Well I have tried some things and i think Im on the right way but I cant figure out a solution by myself.
This function is pretty near but i cant manage to make it start at different rows(1 and 12):
lapply(2:nrow(eDF), function(x) eDF[(x-1):x,])
I also tried to figure it out with seq and it would do what i need if i could make a list of lists by connecting both code samples. Well I also need to change the concrete start and end numbers to a dynamic solution.
eDF[(seq(1,to=11,by=1)),] # selecting rows 1 to 11
eDF[(seq(12,to=nrow(eDF),by=1)),] #selecting rows 12 to end
Anyone any ideas?
I don't know if you needed an ID column inside of the new list but another way would be:
#create the IDs
eDF$ID <- rep(1:11,2)
#split the data.frame according to those
mylist <- split(eDF, eDF$ID)
Output:
mylist
$`1`
X Y ID
1 18.25743 44.69561 1
12 18.26000 44.69740 1
$`2`
X Y ID
2 18.25783 44.69564 2
13 18.26022 44.69763 2
$`3`
X Y ID
3 18.25823 44.69567 3
14 18.26051 44.69774 3
$`4`
X Y ID
4 18.2585 44.69567 4
15 18.2607 44.69787 4
#and so on...
You could only do split(eDF, rep(1:11,2) if you don't need the ID column.
We can modify the OP's lapply code
lapply(1:11, function(i) eDF[c(i, i+11),])
I am working with two CSV files. They are formatted like this:
File 1
able,2
gobble,3
highway,3
test,6
zoo,10
File 2
able,6
gobble,10
highway,3
speed,7
test,8
upper,3
zoo,10
In my program I want to do the following:
Create a keyword list by combining the values from two CSV files and keeping only unique keywords
Compare that keyword list to each individual CSV file to determine the maximum number of occurences of a given keyword, then append that information to the keyword list.
The first step I have done already.
I am getting confused by R reading things as vectors/factors/data frames etc...and "coercion to lists". For example in my files given above, the maximum occurrence for the word "gobble" should be 10 (its value is 3 in file 1 and 10 in file 2)
So basically two things need to happen. First, I need to create a column in "keywords" that holds information about the maximum number of occurrences of a word from the CSV files. Second, I need to populate that column with the maximum value.
Here is my code:
# Read in individual data sets
keywordset1=as.character(read.csv("set1.csv",header=FALSE,sep=",")$V1)
keywordset2=as.character(read.csv("set2.csv",header=FALSE,sep=",")$V1)
exclude_list=as.character(read.csv("exclude.csv",header=FALSE,sep=",")$V1)
# Sort, capitalize, and keep unique values from the two keyword sets
keywords <- sapply(unique(sort(c(keywordset1, keywordset2))), toupper)
# Keep keywords greater than 2 characters in length (basically exclude in at etc...)
keywords <- keywords[nchar(keywords) > 2]
# Keep keywords that are not in the exclude list
keywords <- setdiff(keywords, sapply(exclude_list, toupper))
# HERE IS WHERE I NEED HELP
# Compare the read keyword list to the master keyword list
# and keep the frequency column
key1=read.csv("set1.csv",header=FALSE,sep=",")
key1$V1=sapply(key1[[1]], toupper)
keywords$V2=key1[which(keywords[[1]] %in% key1$V1),2]
return(keywords)
The reason that your last commmand fails is that you try to use the $ operator on a vector. It only works on lists or data frames (which are a special case of lists).
A remark regarding toupper (and many other functions in R): it works on vectors, such that you don't need to use sapply. toupper(c(keywordset1, keywordset2)) is perfectly fine.
But I would like to propose an entirely different solution to your problem. First, I create the data as follows:
keywords1 <- read.table(text="able,2
gobble,3
highway,3
test,6
zoo,10",sep=",",stringsAsFactors=FALSE)
keywords2 <- read.table(text="gobble,10
highway,3
speed,7
test,8
upper,3
zoo,10",sep=",",stringsAsFactors=FALSE)
Note that I use stringsAsFactors=FALSE. This prevents read.table from converting characters to factors, such that there is no need to call as.character later.
The next steps are to capitalize the keyword columns in both tables. At the same time, I put both tables in a list. This is often a good way to simplify calculations in R, because you can use lapply to apply a function on all the list elements. Then I put both tables into a single table.
keyword_list <- lapply(list(keywords1,keywords2),function(kw)
transform(kw,V1=toupper(V1)))
keywords_all <- do.call(rbind,keyword_list)
The next step is to sort the data frame in decreasing order by the number in the second column:
keywords_sorted <- keywords_all[order(keywords_all$V2,decreasing=TRUE),]
keywords_sorted looks as follows:
V1 V2
5 ZOO 10
6 GOBBLE 10
11 ZOO 10
9 TEST 8
8 SPEED 7
4 TEST 6
2 GOBBLE 3
3 HIGHWAY 3
7 HIGHWAY 3
10 UPPER 3
1 ABLE 2
As you notice, some keywords appear only once and for those that appear twice, the first appearance is the one you want to keep. There is a function in R that can be used to extract exactly these elements: duplicated() (run ?duplicated to learn more). Basically, the function returns TRUE, if an element appears for the at least second time in a vector. These are the elements you don't want. To convert TRUE to FALSE (and vice versa), you use the operator !. So the following gives your desired result:
keep <- !duplicated(keywords_sorted$V1)
keywords_max <- keywords_sorted[keep,]
V1 V2
5 ZOO 10
6 GOBBLE 10
9 TEST 8
8 SPEED 7
3 HIGHWAY 3
10 UPPER 3
1 ABLE 2
I have done lot of googling but I didn't find satisfactory solution to my problem.
Say we have data file as:
Tag v1 v2 v3
A 1 2 3
B 1 2 2
C 5 6 1
A 9 2 7
C 1 0 1
The first line is header. The first column is Group id (the data have 3 groups A, B, C) while other column are values.
I want to read this file in R so that I can apply different functions on the data.
For example I tried to read the file and tried to get column mean
dt<-read.table(file_name,head=T) #gives warnings
apply(dt,2,mean) #gives NA NA NA
I want to read this file and want to get column mean. Then I want to separate the data in 3 groups (according to Tag A,B,C) and want to calculate mean(column wise) for each group. Any help
apply(dt,2,mean) doesn't work because apply coerces the first argument to an array via as.matrix (as is stated in the first paragraph of the Details section of ?apply). Since the first column is character, all elements in the coerced matrix object will be character.
Try this instead:
sapply(dt,mean) # works because data.frames are lists
To calculate column means by groups:
# using base functions
grpMeans1 <- t(sapply(split(dt[,c("v1","v2","v3")], dt[,"Tag"]), colMeans))
# using plyr
library(plyr)
grpMeans2 <- ddply(dt, "Tag", function(x) colMeans(x[,c("v1","v2","v3")]))