R: Calculating displacement of variables in order - r
We are developing an online test in which we present 6 images (in a row) that participants are asked to reshuffle from smallest to largest (the actual objective is more challenging, but besides the point here). The test begins with the images being presented in a random order. In the end, I would like to calculate the total displacement (or deviation) of the participant's response from the correct response.
To illustrate:
We begin by presenting six images like so:
[img1] [img2] [img3] [img4] [img5] [img6]
A participant might then reshuffle the images to:
[img2] [img4] [img3] [img1] [img6] [img5]
The correct order for this trial might actually be:
[img1] [img4] [img3] [img2] [img5] [img6]
Thus, we see that the participant has not placed all images at the correct position: img1 is displaced 3 positions to the left, while img2 is displaced 3 positions to the right; and img5 and img6 are each displaced 1 position. Thus the total displacement is 3 + 3 + 1 + 1 = 8
Is there an elegant way in R to calculate this displacement?
You can try this
sum(abs(match(x,y)-match(x,x)))
Data
x=c(2,4,3,1,6,5)
y=c(1,4,3,2,5,6)
I am decomposing this a lot, obviously you can compress this on two lines if you want to.
This the correct output:
> correct_output <- data.frame(img=c(1, 4, 3, 2, 5, 6), rank=1:6)
img rank
1 1 1
2 4 2
3 3 3
4 2 4
5 5 5
6 6 6
This is the user output:
> user_output <- data.frame(img=c(2, 4, 3, 1, 6, 5), user_rank=1:6)
img user_rank
1 2 1
2 4 2
3 3 3
4 1 4
5 6 5
6 5 6
Let us bind them together:
> merge(correct_output, user_output, by="img")
img rank user_rank
1 1 1 4
2 2 4 1
3 3 3 3
4 4 2 2
5 5 5 6
6 6 6 5
From here it is very easy. I am using the dplyr package.
> tmp <- mutate(tmp, penalty=abs(rank-user_rank))
img rank user_rank penalty
1 1 1 4 3
2 2 4 1 3
3 3 3 3 0
4 4 2 2 0
5 5 5 6 1
6 6 6 5 1
> sum(tmp$penalty)
[1] 8
Related
Is there a quick way to transform intervals (Start and End) into a list of number in this interval in R
I have a file with interval values such as this for 50M lines:
>data
start_pos end_pos
1 1 10
2 3 6
3 5 9
4 6 11
And I would like to have a table of position occurrences so that I can compute the coverage on each position in the interval file such as this:
>occurence
position coverage
1 1
2 1
3 2
4 2
5 3
6 4
7 3
8 3
9 3
10 2
11 1
Is there any fast and best way to complete this task in R?
My plan was to loop through the data and concatenate the sequence in each interval into a vector and convert the final vector into a table.
count<-c()
for (row in 1:nrow(data)){
count<-c(count,(data[row,]$start_pos:data[row,]$end_pos))
}
occurence <- table(count)
The problem is that my file is huge and it takes way to much time and memory to do so.
The Bioconductor IRanges package does this fast and efficiently library(IRanges) ir = IRanges(start = c(1, 3, 5, 6), end = c(10, 6, 9, 11)) coverage(ir) with > coverage(ir) |> as.data.frame() value 1 1 2 1 3 2 4 2 5 3 6 4 7 3 8 3 9 3 10 2 11 1
apply conditional numbering to grouped data in R
I have a table like the one below with 100's of rows of data.
ID RANK
1 2
1 3
1 3
2 4
2 8
3 3
3 3
3 3
4 6
4 7
4 7
4 7
4 7
4 7
4 6
I want to try to find a way to group the data by ID so that I can ReRank each group separately. The ReRank column is based on the Rank column and basically renumbering it starting at 1 from least to greatest, but it's important to note that the the number in the ReRank column can be put in more than once depending on the numbers in the Rank column .
In other words, the output needs to look like this
ID Rank ReRANK
1 3 2
1 2 1
1 3 2
2 4 1
2 8 2
3 3 1
3 3 1
3 3 1
For the life of me, I can't figure out how to be able to ReRank the the columns by the grouped columns and the value of the Rank columns.
This has been my best guess so far, but it definitely is not doing what I need it to do
ReRANK = mat.or.vec(length(RANK),1)
ReRANK[1] = counter = 1
for(i in 2:length(RANK)) {
if (RANK[i] != RANK[i-1]) { counter = counter + 1 }
ReRANK[i] = counter
}
Thank you in advance for the help!!
Here is a base R method using ave and rank:
df$ReRank <- ave(df$Rank, df$ID, FUN=function(i) rank(i, ties.method="min"))
The min argument in rank assures that the minimum ranking will occur when there are ties. the default is to take the mean of the ranks.
In the case that you have ties lower down in the groups, rank will count those lower values and then add continue with the next lowest value as the count of the lower values + 1. These values wil still be ordered and distinct. If you really want to have the count be 1, 2, 3, and so on rather than 1, 3, 6 or whatever depending on the number of duplicate values, here is a little hack using factor:
df$ReRank <- ave(df$Rank, df$ID, FUN=function(i) {
as.integer(factor(rank(i, ties.method="min"))))
Here, we use factor to build values counting from upward for each level. We then coerce it to be an integer.
For example,
temp <- c(rep(1, 3), 2,5,1,4,3,7)
[1] 2.5 2.5 2.5 5.0 8.0 2.5 7.0 6.0 9.0
rank(temp, ties.method="min")
[1] 1 1 1 5 8 1 7 6 9
as.integer(factor(rank(temp, ties.method="min")))
[1] 1 1 1 2 5 1 4 3 6
data
df <- read.table(header=T, text="ID Rank
1 2
1 3
1 3
2 4
2 8
3 3
3 3
3 3 ")
Identify repetitive pattern in numeric vector in R with fuzzy search [duplicate]
This question already has answers here: Find and break on repeated runs (3 answers) Closed 6 years ago. Imagine a vector of integers like so: > rep(c(1,4,2),10) [1] 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 For us human beings it seems easy to identify the pattern 1 - 4 - 2 even without knowing the function how the vector was created. But how would you identify this pattern using R? Edit As this question was marked as a dupe I'm going to specify it a bit. The above example was an easy one to explain the idea. The main goal would be to identify more hidden patterns like 1 4 2 5 6 7 1 4 2 9 1 4 2 3 4 5 1 4 2 and also patterns that are approximately the same like 1 4 2 1 4 1.99 1 4 2 1.01 4 2 1 4.01 2. What are the ideas to always Identify the pattern 1 4 2 in those cases?
Assuming that the subpattern must start at the beginning and repeat to the end of the input try it for a subpattern length of k = 1, 2, 3, ... We have assumed that only patterns that are half the length of the input or less are to be considered:
for(k in seq_len(length(x)/2)) {
pat <- x[1:k]
if (identical(rep(pat, length = length(x)), x)) {
print(pat)
break
}
}
## [1] 1 4 2
Note: This was used as the input x:
x <- rep(c(1, 4, 2), 10)
Select max or equal value from several columns in a data frame
I'm trying to select the column with the highest value for each row in a data.frame. So for instance, the data is set up as such. > df <- data.frame(one = c(0:6), two = c(6:0)) > df one two 1 0 6 2 1 5 3 2 4 4 3 3 5 4 2 6 5 1 7 6 0 Then I'd like to set another column based on those rows. The data frame would look like this. > df one two rank 1 0 6 2 2 1 5 2 3 2 4 2 4 3 3 3 5 4 2 1 6 5 1 1 7 6 0 1 I imagine there is some sort of way that I can use plyr or sapply here but it's eluding me at the moment.
There might be a more efficient solution, but ranks <- apply(df, 1, which.max) ranks[which(df[, 1] == df[, 2])] <- 3 edit: properly spaced!
Calculating the occurrences of numbers in the subsets of a data.frame
I have a data frame in R which is similar to the follows. Actually my real ’df’ dataframe is much bigger than this one here but I really do not want to confuse anybody so that is why I try to simplify things as much as possible.
So here’s the data frame.
id <-c(1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3)
a <-c(3,1,3,3,1,3,3,3,3,1,3,2,1,2,1,3,3,2,1,1,1,3,1,3,3,3,2,1,1,3)
b <-c(3,2,1,1,1,1,1,1,1,1,1,2,1,3,2,1,1,1,2,1,3,1,2,2,1,3,3,2,3,2)
c <-c(1,3,2,3,2,1,2,3,3,2,2,3,1,2,3,3,3,1,1,2,3,3,1,2,2,3,2,2,3,2)
d <-c(3,3,3,1,3,2,2,1,2,3,2,2,2,1,3,1,2,2,3,2,3,2,3,2,1,1,1,1,1,2)
e <-c(2,3,1,2,1,2,3,3,1,1,2,1,1,3,3,2,1,1,3,3,2,2,3,3,3,2,3,2,1,3)
df <-data.frame(id,a,b,c,d,e)
df
Basically what I would like to do is to get the occurrences of numbers for each column (a,b,c,d,e) and for each id group (1,2,3) (for this latter grouping see my column ’id’).
So, for column ’a’ and for id number ’1’ (for the latter see column ’id’) the code would be something like this:
as.numeric(table(df[1:10,2]))
##The results are:
[1] 3 7
Just to briefly explain my results: in column ’a’ (and regarding only those records which have number ’1’ in column ’id’) we can say that number '1' occured 3 times and number '3' occured 7 times.
Again, just to show you another example. For column ’a’ and for id number ’2’ (for the latter grouping see again column ’id’):
as.numeric(table(df[11:20,2]))
##After running the codes the results are:
[1] 4 3 3
Let me explain a little again: in column ’a’ and regarding only those observations which have number ’2’ in column ’id’) we can say that number '1' occured 4 times, number '2' occured 3 times and number '3' occured 3 times.
So this is what I would like to do. Calculating the occurrences of numbers for each custom-defined subsets (and then collecting these values into a data frame). I know it is not a difficult task but the PROBLEM is that I’m gonna have to change the input ’df’ dataframe on a regular basis and hence both the overall number of rows and columns might change over time…
What I have done so far is that I have separated the ’df’ dataframe by columns, like this:
for (z in (2:ncol(df))) assign(paste("df",z,sep="."),df[,z])
So df.2 will refer to df$a, df.3 will equal df$b, df.4 will equal df$c etc. But I’m really stuck now and I don’t know how to move forward…
Is there a proper, ”automatic” way to solve this problem?
How about - > library(reshape) > dftab <- table(melt(df,'id')) > dftab , , value = 1 variable id a b c d e 1 3 8 2 2 4 2 4 6 3 2 4 3 4 2 1 5 1 , , value = 2 variable id a b c d e 1 0 1 4 3 3 2 3 3 3 6 2 3 1 4 5 3 4 , , value = 3 variable id a b c d e 1 7 1 4 5 3 2 3 1 4 2 4 3 5 4 4 2 5 So to get the number of '3's in column 'a' and group '1' you could just do > dftab[3,'a',1] [1] 4
A combination of tapply and apply can create the data you want: tapply(df$id,df$id,function(x) apply(df[id==x,-1],2,table)) However, when a grouping doesn't have all the elements in it, as in 1a, the result will be a list for that id group rather than a nice table (matrix). $`1` $`1`$a 1 3 3 7 $`1`$b 1 2 3 8 1 1 $`1`$c 1 2 3 2 4 4 $`1`$d 1 2 3 2 3 5 $`1`$e 1 2 3 4 3 3 $`2` a b c d e 1 4 6 3 2 4 2 3 3 3 6 2 3 3 1 4 2 4 $`3` a b c d e 1 4 2 1 5 1 2 1 4 5 3 4 3 5 4 4 2 5
I'm sure someone will have a more elegant solution than this, but you can cobble it together with a simple function and dlply from the plyr package.
ColTables <- function(df) {
counts <- list()
for(a in names(df)[names(df) != "id"]) {
counts[[a]] <- table(df[a])
}
return(counts)
}
results <- dlply(df, "id", ColTables)
This gets you back a list - the first "layer" of the list will be the id variable; the second the table results for each column for that id variable. For example:
> results[['2']]['a']
$a
1 2 3
4 3 3
For id variable = 2, column = a, per your above example.
A way to do it is using the aggregate function, but you have to add a column to your dataframe
> df$freq <- 0
> aggregate(freq~a+id,df,length)
a id freq
1 1 1 3
2 3 1 7
3 1 2 4
4 2 2 3
5 3 2 3
6 1 3 4
7 2 3 1
8 3 3 5
Of course you can write a function to do it, so it's easier to do it frequently, and you don't have to add a column to your actual data frame
> frequency <- function(df,groups) {
+ relevant <- df[,groups]
+ relevant$freq <- 0
+ aggregate(freq~.,relevant,length)
+ }
> frequency(df,c("b","id"))
b id freq
1 1 1 8
2 2 1 1
3 3 1 1
4 1 2 6
5 2 2 3
6 3 2 1
7 1 3 2
8 2 3 4
9 3 3 4
You didn't say how you'd like the data. The by function might give you the output you like. by(df, df$id, function(x) lapply(x[,-1], table))