I have a dataset where one of the columns are only "#" sign. I used the following code to remove this column.
ia <- as.data.frame(sapply(ia,gsub,pattern="#",replacement=""))
However, after this operation, one of the integer column I had changed to factor.
I wonder what happened and how can i avoid that. Appreciate it.
A more correct version of your code might be something like this:
d <- data.frame(x = as.character(1:5),y = c("a","b","#","c","d"))
> d[] <- lapply(d,gsub,pattern = "#",replace = "")
> d
x y
1 1 a
2 2 b
3 3
4 4 c
5 5 d
But as you'll note, this approach will never actually remove the offending column. It's just replacing the # values with empty character strings. To remove a column of all # you might do something like this:
d <- data.frame(x = as.character(1:5),
y = c("a","b","#","c","d"),
z = rep("#",5))
> d[,!sapply(d,function(x) all(x == "#"))]
x y
1 1 a
2 2 b
3 3 #
4 4 c
5 5 d
Surely if you want to remove an offending column from a data frame, and you know which column it is, you can just subset. So, if it's the first column:
df <- df[,-1]
If it's a later column, increment up.
Related
Thank you good ppl! This must be simple but I'm banging my head against it for a while. Please help. I have a large data set from which I get all kinds of information via table(). I then want to store that information which is essentially different counts, so I also want to store the rownames that were counted. For a reproducible example consider
```
a<-c("a","b","c","d","a","b") #one count, occurring twice for a and
b and once for c and d
b<-c("a","c") # a completly different property from the dataset
occurring once for a and c
x<-table(a)
y<-table(b) #so now x and y hold the information I seek
How can I merge/bind/whatever to get from x and y to this form:
x. y.
a 2. 1
b 2. 0
c 1. 1
d. 1 0
HOWEVER, I need to use the solution iteratively, in a loop that takes x and y and gets the requested form above, and then gets more tables added, each hopefully adding a column. One of my many failed attempts, just to show the logic, is:
`. member<-function (data=dfm,groupvar='group',analysis=kc15 {
res<-matrix(NA,ncol=length(analysis$size)+1)
res[,1]<-table(docvars(data,groupvar))
for (i in 1:length(analysis$size)) {
r<-table(docvars(data,groupvar)[analysis$cluster==i])
res<-cbind(res,r)
}
res
}`
So, to sum, the reproducible example above means to replicate the first column in res and an r, and I'm seeking (I think) a correct solution instead of the cbind, which would allow adding columns of different length but similar names, as in the example above.
Please help its embarrassing how much time I'm wasting on this
In base R, you can use table, stack and full join the two counts.
out <- merge(stack(table(a)), stack(table(b)), by = 'ind', all = TRUE)
out
# ind values.x values.y
#1 a 2 1
#2 b 2 NA
#3 c 1 1
#4 d 1 NA
If you want to replace NA with 0, you can do :
out[is.na(out)] <- 0
One purrr and tidyr solution could be:
map_dfr(lst, ~ stack(table(.)), .id = "ID") %>%
pivot_wider(names_from = "ID", values_from = "values", values_fill = list(values = 0))
ind a b
<chr> <int> <int>
1 a 2 1
2 b 2 0
3 c 1 1
4 d 1 0
lst being:
lst <- list(a = a,
b = b)
I would like to have an equivalent of the Excel function "if". It seems basic enough, but I could not find relevant help.
I would like to assess "NA" to specific cells if two following cells in a different columns are not identical. In Excel, the command would be the following (say in C1): if(A1 = A2, B1, "NA"). I then just need to expand it to the rest of the column.
But in R, I am stuck!
Here is an equivalent of my R code so far.
df = data.frame(Type = c("1","2","3","4","4","5"),
File = c("A","A","B","B","B","C"))
df
To get the following Type of each Type in another column, I found a useful function on StackOverflow that does the job.
# determines the following Type of each Type
shift <- function(x, n){
c(x[-(seq(n))], rep(6, n))
}
df$TypeFoll <- shift(df$Type, 1)
df
Now, I would like to keep TypeFoll in a specific row when the File for this row is identical to the File on the next row.
Here is what I tried. It failed!
for(i in 1:length(df$File)){
df$TypeFoll2 <- ifelse(df$File[i] == df$File[i+1], df$TypeFoll, "NA")
}
df
In the end, my data frame should look like:
aim = data.frame(Type = c("1","2","3","4","4","5"),
File = c("A","A","B","B","B","C"),
TypeFoll = c("2","3","4","4","5","6"),
TypeFoll2 = c("2","NA","4","4","NA","6"))
aim
Oh, and by the way, if someone would know how to easily put the columns TypeFoll and TypeFoll2 just after the column Type, it would be great!
Thanks in advance
I would do it as follows (not keeping the result from the shift function)
df = data.frame(Type = c("1","2","3","4","4","5"),
File = c("A","A","B","B","B","C"), stringsAsFactors = FALSE)
# This is your shift function
len=nrow(df)
A1 <- df$File[1:(len-1)]
A2 <- df$File[2:len]
# Why do you save the result of the shift function in the df?
Then assign if(A1 = A2, B1, "NA"). As akrun mentioned ifelse is vectorised: Btw. this is how you append a column to a data.frame
df$TypeFoll2 <- c(ifelse(A1 == A2, df$Type, NA), 6) #Why 6?
As 6 is hardcoded here something like:
df$TypeFoll2 <- c(ifelse(A1 == A2, df$Type, NA), max(df$Type)+1)
Is more generic.
First off, 'for' loops are pretty slow in R, so try to think of this as vector manipulation instead.
df = data.frame(Type = c("1","2","3","4","4","5"),
File = c("A","A","B","B","B","C"));
Create shifted types and files and put it in new columns:
df$TypeFoll = c(as.character(df$Type[2:nrow(df)]), "NA");
df$FileFoll = c(as.character(df$File[2:nrow(df)]), "NA");
Now, df looks like this:
> df
Type File TypeFoll FileFoll
1 1 A 2 A
2 2 A 3 B
3 3 B 4 B
4 4 B 4 B
5 4 B 5 C
6 5 C NA NA
Then, create TypeFoll2 by combining these:
df$TypeFoll2 = ifelse(df$File == df$FileFoll, df$TypeFoll, "NA");
And you should have something that looks a lot like what you want:
> df;
Type File TypeFoll FileFoll TypeFoll2
1 1 A 2 A 2
2 2 A 3 B NA
3 3 B 4 B 4
4 4 B 4 B 4
5 4 B 5 C NA
6 5 C NA NA NA
If you want to remove the FileFoll column:
df$FileFoll = NULL;
I am working in R with a data frame d:
ID <- c("A","A","A","B","B")
eventcounter <- c(1,2,3,1,2)
numberofevents <- c(3,3,3,2,2)
d <- data.frame(ID, eventcounter, numberofevents)
> d
ID eventcounter numberofevents
1 A 1 3
2 A 2 3
3 A 3 3
4 B 1 2
5 B 2 2
where numberofevents is the highest value in the eventcounter for each ID.
Currently, I am trying to create an additional vector z <- c(6,6,6,3,3).
If the numberofevents == 3, it is supposed to calculate sum(1:3), equally to 3 + 2 + 1 = 6.
If the numberofevents == 2, it is supposed to calculate sum(1:2) equally to 2 + 1 = 3.
Working with a large set of data, I thought it might be convenient to create this additional vector
by using the sum function in R d$z<-sum(1:d$numberofevents), i.e.
sum(1:3) # for the rows 1-3
and
sum(1:2) # for the rows 4-5.
However, I always get this warning:
Numerical expression has x elements: only the first is used.
You can try ave
d$z <- with(d, ave(eventcounter, ID, FUN=sum))
Or using data.table
library(data.table)
setDT(d)[,z:=sum(eventcounter), ID][]
Try using apply sapply or lapply functions in R.
sapply(numberofevents, function(x) sum(1:x))
It works for me.
I want to delete the header from a dataframe that I have. I read in the data from a csv file then I transposed it, but it created a new header that is the name of the file and the row that the data is from in the file.
Here's an example for a dataframe df:
a.csv.1 a.csv.2 a.csv.3 ...
x 5 6 1 ...
y 2 3 2 ...
I want to delete the a.csv.n row, but when I try df <- df[-1,] it deletes row x and not the top.
If you really, really, really don't like column names, you may convert your data frame to a matrix (keeping possible coercion of variables of different class in mind), and then remove the dimnames.
dd <- data.frame(x1 = 1:5, x2 = 11:15)
mm1 <- as.matrix(dd)
mm2 <- matrix(mm1, ncol = ncol(dd), dimnames = NULL)
I add my previous comment here as well:
?data.frame: "The column names should be non-empty, and attempts to use empty names will have unsupported results.".
Set names to NULL
names(df) <- NULL
You can also use the header option in read.csv
You can use names(df) to change the names of header or col names. If newnames is a list of names as newname<-list("col1","col2","col3"), then names(df)<-newname will give you a data with col names as col1 col2 col3.
As # Henrik said, the col names should be non-empty. Setting the names(df)<-NULLwill give NA in col names.
If your data is csv file and if you use header=TRUE to read the data in R then the data will have same colnames as csv file, but if you set the header=FALSE, R will assign the colnames as V1,V2,...and your colnames in the original csv file appear as a first row.
anydata.csv
a b c d
1 1 2 3 13
2 2 3 1 21
read.csv("anydata.csv",header=TRUE)
a b c d
1 1 2 3 13
2 2 3 1 21
read.csv("anydata.csv",header=FALSE)
V1 V2 V3 V4
1 a b c d
2 1 2 3 13
3 2 3 1 21
You could use
setNames(dat, rep(" ", length(dat)))
where dat is the name of the data frame. Then all columns will have the name " " and hence will be 'invisible'.
It comes with some years of delay but you can simply use a vector renaming de columns:
## if you want to delete all column names:
colnames(df)[] <- ""
## if you want to delete let's say column 1:
colnames(df)[1] <- ""
## if you want to delete 1 to 3 and 7:
colnames(df)[c(1:3,7)] <- ""
As already mentioned not having column names just isn't something that is going to happen with a data frame, but I'm kind of guessing that you don't care so much if they are there you just don't want to see them when you print your data frame? If so, you can write a new print function to get around that, like so:
> dat <- data.frame(var1=c("A","B","C"),var2=rnorm(3),var3=rnorm(3))
> print(dat)
var1 var2 var3
1 A 1.2771777 -0.5726623
2 B -1.5000047 1.3249348
3 C 0.1989117 -1.4016253
> ncol.print <- function(dat) print(matrix(as.matrix(dat),ncol=ncol(dat),dimnames=NULL),quote=F)
> ncol.print(dat)
[,1] [,2] [,3]
[1,] A 1.2771777 -0.5726623
[2,] B -1.5000047 1.3249348
[3,] C 0.1989117 -1.4016253
Your other option it set your variable names to unique amounts of whitespace, for example:
> names(dat) <- c(" ", " ", " ")
> dat
1 A 1.2771777 -0.5726623
2 B -1.5000047 1.3249348
3 C 0.1989117 -1.4016253
You can also write a function do this:
> blank.names <- function(dat){
+ for(i in 1:ncol(dat)){
+ names(dat)[i] <- paste(rep(" ",i),collapse="")
+ }
+ return(dat)
+ }
> dat <- data.frame(var1=c("A","B","C"),var2=rnorm(3),var3=rnorm(3))
> dat
var1 var2 var3
1 A -1.01230289 1.2740237
2 B -0.13855777 0.4689117
3 C -0.09703034 -0.4321877
> blank.names(dat)
1 A -1.01230289 1.2740237
2 B -0.13855777 0.4689117
3 C -0.09703034 -0.4321877
But generally I don't think any of this should be done.
A function that I use in one of my R scripts:
read_matrix <- function (csvfile) {
a <- read.csv(csvfile, header=FALSE)
matrix(as.matrix(a), ncol=ncol(a), dimnames=NULL)
}
How to call this:
iops_even <- read_matrix('even_iops_Jan15.csv')
iops_odd <- read_matrix('odd_iops_Jan15.csv')
You can simply do:
print(df.to_string(header=False))
if you want to remove the line indexes as well, you can do:
print(df.to_string(index=False,header=False))
I have a dataset like this:
x
A B
1 x 2
2 y 4
3 z 4
4 x 4
5 x 4
6 x 3
......
I want to know if in this dataset are present a same number of "A" upper than some value(for example 3).
Probably i will need to group this value in a temporary table getting this:
X Y z
4 1 1
and after this i will call another method (that i don't know) that gives me this result
X
because only the value X is present more than 3 times in my previous table.
Can R optimise this operation?
data<-data.frame(factor(c("x","y","z","x","x","x")),c(2,4,4,4,4,3))
To get the count of each letter, do
table(data[,1])
and to get the name of the factors with > 3
names(table(data[,1]))[table(data[,1]) > 3]
Don´t know if I understand you right... whats with this B column?
Is this working for you?
set.seed(1234)
A <- sample(c("x", "y", "z"), 20, replace = TRUE)
Ad <- data.frame(table(A))
with(Ad, A[Freq >= 7])
[1] x y