I would like to create a dataframe that essentially would look something like this
Repeating the period from 1 to 10 and assigning the ID 42,574 times
so that I would end up with a 425,740 row dataframe.
I tried to create a dataframe using the following code
periodstring <- as.numeric(gl(10, 42574))
periods <- as.data.frame(periodstring)
but that sorts the numbers and other approaches did not quiete work. Is there a simple way to do this?
Thanks in advance.
Another option using rep:
data.frame(Period=rep(1:10,times=42574),
ID=rep(1:42574,each=10))
Output sample:
Period ID
1 1 1
2 2 1
3 3 1
4 4 1
5 5 1
6 6 1
7 7 1
8 8 1
9 9 1
10 10 1
11 1 2
12 2 2
13 3 2
14 4 2
15 5 2
16 6 2
17 7 2
18 8 2
19 9 2
20 10 2
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)
This question already has answers here:
Extract row corresponding to minimum value of a variable by group
(9 answers)
Closed 5 years ago.
I have a table which contains multiple rows of the different data for a key of multiple columns.
Table looks like this:
A B C
1 1 1 2
2 1 1 3
3 2 1 4
4 1 2 4
5 2 2 3
6 2 3 1
7 2 3 2
8 2 3 2
I also discovered how to remove all of the duplicate elements using unique command for multiple colums, so the data duplication is not a problem.
I would like to know how to for every key(columns A and B in example) in the table to find only the minimum value in third column(C column in table)
At the end table should look like this
A B C
1 1 1 2
3 2 1 4
4 1 2 4
5 2 2 3
6 2 3 1
Thanks for any help. It is really appreciated
In any question, feel free to ask
con <- textConnection(" A B C
1 1 1 2
2 1 1 3
3 2 1 4
4 1 2 4
5 2 2 3
6 2 3 1
7 2 3 2
8 2 3 2")
df <- read.table(con, header = T)
df[with(df, order(A, B, C)), ]
df[!duplicated(df[1:2]),]
# A B C
# 1 1 1 2
# 3 2 1 4
# 4 1 2 4
# 5 2 2 3
# 6 2 3 1
I want repeat a sequence for specific length:
Sequence is 1:4 and I want to repeat the sequence till number of rows in a data frame.
Lets say length of the data frame is 24
I tried following:
test <- rep(1:4, each=24/4)
1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4
Lengthwise this is fine but i want to retain the sequence
1 2 3 4 1 2 3 4 1 2 3 4.....
You need to use times instead of each
rep(1:4, times=24/4)
[1] 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
We can just pass it without any argument and it takes the times by default
rep(1:4, 24/4)
#[1] 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
Suppose I have a vector of size n=8 v=(5,8,2,7,9,12,2,1). I would like to know how to build a N x N matrix that compares every pair of values of v and returns the minimum value of each comparation. In this example, it would be like this:
5 5 2 5 5 5 2 1
5 8 2 7 8 8 2 1
2 2 2 2 2 2 2 1
5 7 2 7 7 7 2 1
5 8 2 7 9 9 2 1
5 8 2 7 9 12 2 1
2 2 2 2 2 2 2 1
1 1 1 1 1 1 1 1
Could you help me with this, please?
outer(v, v, pmin)
Notice the use of pmin, not min, as the former is vectorised but not the latter.