I have a dataframe as described below. Now I want to reverse the order of column B without hampering the total order of the dataframe. So now the column B has 5,4,3,2,1. I want to change it to 1,2,3,4,5. I don't want to sort as it will hamper the total ordering.
A B C
1 5 6
2 4 8
3 3 5
4 2 5
5 1 3
You can replace just that column:
x$B <- rev(x$B)
On your data:
> x$B <- rev(x$B)
> x
A B C
1 1 1 6
2 2 2 8
3 3 3 5
4 4 4 5
5 5 5 3
transform is also handy for this:
> transform(x, B = rev(B))
A B C
1 1 1 6
2 2 2 8
3 3 3 5
4 4 4 5
5 5 5 3
This doesn't modify x so you need to assign the result to something (perhaps back to x).
Related
If I have a dataframe like the one below which has numerical column names
example = data.frame(1=c(1,8,3,9), 2=c(3,2,3,3), 3=c(5,2,5,4), 4=c(1,2,3,4), 5=c(2,5,7,8))
Which looks like this:
1 2 3 4 5
1 3 5 1 2
8 2 2 2 5
3 3 5 3 7
9 3 4 4 8
And I want to arrange it so that the column names start with three and proceed through five and back to one, like this:
3 4 5 1 2
5 1 2 1 3
2 2 5 8 2
5 3 7 3 3
4 4 8 9 3
I know how to rearrange the position of a single column in a dataset, but I'm not sure how to do this with more than one column in this particular order.
We can use the column index concatenated (c) based on the sequence (:) on a range of values
example[c(3:5, 1:2)]
# 3 4 5 1 2
#1 5 1 2 1 3
#2 2 2 5 8 2
#3 5 3 7 3 3
#4 4 4 8 9 3
As the column names are all numeric, just convert to numeric and use that for ordering
v1 <- as.numeric(names(example))
example[c(v1[3:5], v1[1:2])]
Or simply do
example[c(names(example)[3:5], names(example)[1:2])]
Or another way is with head and tail
example[c(tail(names(example), 3), head(names(example), 2))]
data
example <- data.frame(`1`=c(1,8,3,9), `2`=c(3,2,3,3),
`3`=c(5,2,5,4), `4`=c(1,2,3,4), `5`=c(2,5,7,8), check.names = FALSE)
R will not easily let you create columns with numbers as name. If somehow, you are able to create columns with numbers you can use match to get order in which you want the column names.
example[match(c(3:5, 1:2), names(example))]
# 3 4 5 1 2
#1 5 1 2 1 3
#2 2 2 5 8 2
#3 5 3 7 3 3
#4 4 4 8 9 3
I have a factor variable with 6 levels, which simplified looks like:
1 1 2 2 2 3 3 3 4 4 4 4 5 5 5 6 6 6 1 1 1 2 2 2 2... 1 1 1 2 2... (with n = 78)
Note, that each number is repeated mostly but not always three times.
I need to transform this variable into the following pattern:
1 1 2 2 2 3 3 3 4 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 8...
where each repetition of the 6 levels continuous counting ascending.
Is there any way / any function that lets me do that?
Sorry for my bad description!
Assuming that you have a numerical vector that represents your simplified version you posted. i.e. x = c(1,1,1,2,2,3,3,3,1,1,2,2), you can use this:
library(dplyr)
cumsum(x != lag(x, default = 0))
# [1] 1 1 1 2 2 3 3 3 4 4 5 5
which compares each value to its previous one and if they are different it adds 1 (starting from 1).
Maybe you can try rle, i.e.,
v <- rep(seq_along((v<-rle(x))$values),v$lengths)
Example with dummy data
x = c(1,1,1,2,2,3,3,3,4,4,5,6,1,1,2,2,3,3,3,4,4)
then we can get
> v
[1] 1 1 1 2 2 3 3 3 4 4 5 6 7 7 8 8 9 9
[19] 9 10 10
In base you can use diff and cumsum.
c(1, cumsum(diff(x)!=0)+1)
# [1] 1 1 2 2 2 3 3 3 4 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 8
Data:
x <- c(1,1,2,2,2,3,3,3,4,4,4,4,5,5,5,6,6,6,1,1,1,2,2,2,2)
I want to subset rows which do not contain special values. For example:
df <- data.frame(a=c(1,2,2,3,4,4),b=c(-9999,2,3,4,5,6),c=c(2,3,4,-9999,2,4))
a b c
1 1 -9999 2
2 2 2 3
3 2 3 4
4 3 4 -9999
5 4 5 2
6 4 6 4
df has many rows and columns , I want to subset the rows which don't contain -9999. Expect result as follow codes:
df[which(df$a!=-9999,df$b!=-9999,df$c!=-9999),]
a b c
2 2 2 3
3 2 3 4
5 4 5 2
6 4 6 4
when columns are to many to type above logical judge, how to subset it?
You can try this one:
temp <- which(df == "-9999",arr.ind = T)
df[-unique(temp[,1]),]
a b c
2 2 2 3
3 2 3 4
5 4 5 2
6 4 6 4
I am trying to remove couple elements from a dataset. It has A,B,C,1,2,3,4,5 as its contents:
>dataset
[1] A 4 3 C 3 3 3 C 3 B 3 4 3 3 3 B 3 3 5 3 3 4 A 3 3 5 3 3 4 3 2 3 C 6 A 3 3
[38] 3 A 3 3 A 3 3 3 3 3 A 3 C B 3 B 3 A 3 1 8 1 1 C 1 1 3 3 3 3 B 3 A A 3 5 3
I want to remove all "A"s and "B"s from the dataset.
The expected dataset should only have 1,2,3,4,5,C as its elements.
I have tried with following codes but could not succeed:
>rm(dataset$"B") # to remove "B"s
> x.sub <- subset(dataset, "B" > 1) #to remove Bs appearing more than once
Do you know how can I remove them?
dataset <- dataset[!(dataset %in% c('A','B'))]
After looking at another post about column names and combn function here consider the same data.frame. We make a combn with all 2 possible vectors:
foo <- data.frame(x=1:5,y=4:8,z=10:14, w=8:4)
all_comb <- combn(foo,2)
Is there a way to keep column names after the combn call so in this case we could get "x y" instead of "X1.5 X4.8" as shown below ?
comb_df <- data.frame(all_comb[1,1],all_comb[2,1])
print(comb_df)
X1.5 X4.8
1 1 4
2 2 5
3 3 6
4 4 7
5 5 8
I suspect you really want to use expand.grid() instead.
Try this:
head(expand.grid(foo))
x y z w
1 1 4 10 8
2 2 4 10 8
3 3 4 10 8
4 4 4 10 8
5 5 4 10 8
6 1 5 10 8
or
head(expand.grid(foo[, 1:2]))
x y
1 1 4
2 2 4
3 3 4
4 4 4
5 5 4
6 1 5