If I have a data frame with a column of monotonically increasing values such as:
x
1
2
3
4
1
2
3
1
2
3
4
5
6
1
2
How do I add a column to group each increasing sequence that results in:
x y
1 1
2 1
3 1
4 1
1 2
2 2
3 2
1 3
2 3
3 3
4 3
5 3
6 3
1 4
2 4
I can only think of using a loop which will be slow.
You may choose cumsum function to do it.
> x <- c(1,2,3,4,1,2,3,1,2,4,5,1,2)
> cumsum(x==1)
[1] 1 1 1 1 2 2 2 3 3 3 3 4 4
I would use diff and compute the cumulative sum:
df$y <- c(1, cumsum(diff(df$x) < 0 ) + 1)
> df
x y
1 1 1
2 2 1
3 3 1
4 4 1
5 1 2
6 2 2
7 3 2
8 1 3
9 2 3
10 3 3
11 4 3
12 5 3
13 6 3
14 1 4
15 2 4
Related
I have done an experiment in which participants have solved a task in pairs, with another participant. Each participant has then received a score for how well they did the task. Pairs have gone through different amounts of trials.
I have a data frame similar to the one below:
participant <- c(1,1,2,2,3,3,3,4,4,4,5,6)
pair <- c(1,1,1,1,2,2,2,2,2,2,3,3)
trial <- c(1,2,1,2,1,2,3,1,2,3,1,1)
score <- c(2,3,6,3,4,7,3,1,8,5,4,3)
data <- data.frame(participant, pair, trial, score)
participant pair trial score
1 1 1 2
1 1 2 3
2 1 1 6
2 1 2 3
3 2 1 4
3 2 2 7
3 2 3 3
4 2 1 1
4 2 2 8
4 2 3 5
5 3 1 4
6 3 1 3
I would like to add a new vector to the data frame, where each participant gets the numeric difference between their own score and the other participant's score within each trial.
Does someone have an idea about how one might do that?
It should end up looking something like this:
participant pair trial score difference
1 1 1 2 4
1 1 2 3 0
2 1 1 6 4
2 1 2 3 0
3 2 1 4 3
3 2 2 7 1
3 2 3 3 2
4 2 1 1 3
4 2 2 8 1
4 2 3 5 2
5 3 1 4 1
6 3 1 3 1
Here's a solution that involves first reordering data such that each sequential pair of rows corresponds to a single pair within a single trial. This allows us to make a single call to diff() to extract the differences:
data <- data[order(data$trial,data$pair,data$participant),];
data$diff <- rep(diff(data$score)[c(T,F)],each=2L)*c(-1L,1L);
data;
## participant pair trial score diff
## 1 1 1 1 2 -4
## 3 2 1 1 6 4
## 5 3 2 1 4 3
## 8 4 2 1 1 -3
## 11 5 3 1 4 1
## 12 6 3 1 3 -1
## 2 1 1 2 3 0
## 4 2 1 2 3 0
## 6 3 2 2 7 -1
## 9 4 2 2 8 1
## 7 3 2 3 3 -2
## 10 4 2 3 5 2
I assumed you wanted the sign to capture the direction of the difference. So, for instance, if a participant has a score 4 points below the other participant in the same trial-pair, then I assumed you would want -4. If you want all-positive values, you can remove the multiplication by c(-1L,1L) and add a call to abs():
data$diff <- rep(abs(diff(data$score)[c(T,F)]),each=2L);
data;
## participant pair trial score diff
## 1 1 1 1 2 4
## 3 2 1 1 6 4
## 5 3 2 1 4 3
## 8 4 2 1 1 3
## 11 5 3 1 4 1
## 12 6 3 1 3 1
## 2 1 1 2 3 0
## 4 2 1 2 3 0
## 6 3 2 2 7 1
## 9 4 2 2 8 1
## 7 3 2 3 3 2
## 10 4 2 3 5 2
Here's a solution built around ave() that doesn't require reordering the whole data.frame first:
data$diff <- ave(data$score,data$trial,data$pair,FUN=function(x) abs(diff(x)));
data;
## participant pair trial score diff
## 1 1 1 1 2 4
## 2 1 1 2 3 0
## 3 2 1 1 6 4
## 4 2 1 2 3 0
## 5 3 2 1 4 3
## 6 3 2 2 7 1
## 7 3 2 3 3 2
## 8 4 2 1 1 3
## 9 4 2 2 8 1
## 10 4 2 3 5 2
## 11 5 3 1 4 1
## 12 6 3 1 3 1
Here's how you can get the score of the other participant in the same trial-pair:
data$other <- ave(data$score,data$trial,data$pair,FUN=rev);
data;
## participant pair trial score other
## 1 1 1 1 2 6
## 2 1 1 2 3 3
## 3 2 1 1 6 2
## 4 2 1 2 3 3
## 5 3 2 1 4 1
## 6 3 2 2 7 8
## 7 3 2 3 3 5
## 8 4 2 1 1 4
## 9 4 2 2 8 7
## 10 4 2 3 5 3
## 11 5 3 1 4 3
## 12 6 3 1 3 4
Or, assuming the data.frame has been reordered as per the initial solution:
data$other <- c(rbind(data$score[c(F,T)],data$score[c(T,F)]));
data;
## participant pair trial score other
## 1 1 1 1 2 6
## 3 2 1 1 6 2
## 5 3 2 1 4 1
## 8 4 2 1 1 4
## 11 5 3 1 4 3
## 12 6 3 1 3 4
## 2 1 1 2 3 3
## 4 2 1 2 3 3
## 6 3 2 2 7 8
## 9 4 2 2 8 7
## 7 3 2 3 3 5
## 10 4 2 3 5 3
Alternative, using matrix() instead of rbind():
data$other <- c(matrix(data$score,2L)[2:1,]);
data;
## participant pair trial score other
## 1 1 1 1 2 6
## 3 2 1 1 6 2
## 5 3 2 1 4 1
## 8 4 2 1 1 4
## 11 5 3 1 4 3
## 12 6 3 1 3 4
## 2 1 1 2 3 3
## 4 2 1 2 3 3
## 6 3 2 2 7 8
## 9 4 2 2 8 7
## 7 3 2 3 3 5
## 10 4 2 3 5 3
Here is an option using data.table:
library(data.table)
setDT(data)[,difference := abs(diff(score)), by = .(pair, trial)]
data
# participant pair trial score difference
# 1: 1 1 1 2 4
# 2: 1 1 2 3 0
# 3: 2 1 1 6 4
# 4: 2 1 2 3 0
# 5: 3 2 1 4 3
# 6: 3 2 2 7 1
# 7: 3 2 3 3 2
# 8: 4 2 1 1 3
# 9: 4 2 2 8 1
#10: 4 2 3 5 2
#11: 5 3 1 4 1
#12: 6 3 1 3 1
A slightly faster option would be:
setDT(data)[, difference := abs((score - shift(score))[2]) , by = .(pair, trial)]
If we need the value of the other pair:
data[, other:= rev(score) , by = .(pair, trial)]
data
# participant pair trial score difference other
# 1: 1 1 1 2 4 6
# 2: 1 1 2 3 0 3
# 3: 2 1 1 6 4 2
# 4: 2 1 2 3 0 3
# 5: 3 2 1 4 3 1
# 6: 3 2 2 7 1 8
# 7: 3 2 3 3 2 5
# 8: 4 2 1 1 3 4
# 9: 4 2 2 8 1 7
#10: 4 2 3 5 2 3
#11: 5 3 1 4 1 3
#12: 6 3 1 3 1 4
Or using dplyr:
library(dplyr)
data %>%
group_by(pair, trial) %>%
mutate(difference = abs(diff(score)))
# participant pair trial score difference
# <dbl> <dbl> <dbl> <dbl> <dbl>
#1 1 1 1 2 4
#2 1 1 2 3 0
#3 2 1 1 6 4
#4 2 1 2 3 0
#5 3 2 1 4 3
#6 3 2 2 7 1
#7 3 2 3 3 2
#8 4 2 1 1 3
#9 4 2 2 8 1
#10 4 2 3 5 2
#11 5 3 1 4 1
#12 6 3 1 3 1
I try to count triplets; for this I use three vectors that are packed in a dataframe:
X=c(4,4,4,4,4,4,4,4,1,1,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,3,3)
Y=c(1,1,1,1,1,1,1,1,1,1,1,1,2,2,3,4,2,2,2,2,3,4,1,1,2,2,3,3,4,4)
Z=c(4,4,5,4,4,4,4,4,6,1,1,1,1,1,1,1,2,2,2,2,7,2,3,3,3,3,3,3,3,3)
Count_Frame=data.frame(matrix(NA, nrow=(length(X)), ncol=3))
Count_Frame[1]=X
Count_Frame[2]=Y
Count_Frame[3]=Z
Counts=data.frame(table(Count_Frame))
There is the following problem: if I increase the value range in the vectors or use even more vectors the "Counts" dataframe quickly approaches its size limit due to the many 0-counts. Is there a way to exclude the 0-counts while generating "Counts"?
We can use data.table. Convert the 'data.frame' to 'data.table' (setDT(Count_Frame)), grouped by all the columns (.(X, Y, Z)), we get the number or rows (.N).
library(data.table)
setDT(Count_Frame)[,.N ,.(X, Y, Z)]
# X Y Z N
# 1: 4 1 4 7
# 2: 4 1 5 1
# 3: 1 1 6 1
# 4: 1 1 1 3
# 5: 1 2 1 2
# 6: 1 3 1 1
# 7: 1 4 1 1
# 8: 2 2 2 4
# 9: 2 3 7 1
#10: 2 4 2 1
#11: 3 1 3 2
#12: 3 2 3 2
#13: 3 3 3 2
#14: 3 4 3 2
Instead of naming all the columns, we can use names(Count_Frame) as well (if there are many columns)
setDT(Count_Frame)[,.N , names(Count_Frame)]
You can accomplish this with aggregate:
Count_Frame$one <- 1
aggregate(one ~ X1 + X2 + X3, data=Count_Frame, FUN=sum)
This will calculate the positive instances of table, but will not list the zero counts.
One solution is to create a combination of the column values and count those instead:
library(tidyr)
as.data.frame(table(unite(Count_Frame, tmp, X1, X2, X3))) %>%
separate(Var1, c('X1', 'X2', 'X3'))
Resulting output is:
X1 X2 X3 Freq
1 1 1 1 3
2 1 1 6 1
3 1 2 1 2
4 1 3 1 1
5 1 4 1 1
6 2 2 2 4
7 2 3 7 1
8 2 4 2 1
9 3 1 3 2
10 3 2 3 2
11 3 3 3 2
12 3 4 3 2
13 4 1 4 7
14 4 1 5 1
Or using plyr:
library(plyr)
count(Count_Frame, colnames(Count_Frame))
output
# > count(Count_Frame, colnames(Count_Frame))
# X1 X2 X3 freq
# 1 1 1 1 3
# 2 1 1 6 1
# 3 1 2 1 2
# 4 1 3 1 1
# 5 1 4 1 1
# 6 2 2 2 4
# 7 2 3 7 1
# 8 2 4 2 1
# 9 3 1 3 2
# 10 3 2 3 2
# 11 3 3 3 2
# 12 3 4 3 2
# 13 4 1 4 7
# 14 4 1 5 1
Brain afunctional today: How do I tell acast to return different aggregations?
# the rows and columns have integer names
Rgames> foo
1 2
1 1 1
2 2 2
3 3 3
4 4 4
1 1 4
2 2 8
3 3 2
4 4 1
Rgames> mfoo<-melt(foo)
Rgames> mfoo
Var1 Var2 value
1 1 1 1
2 2 1 2
3 3 1 3
4 4 1 4
5 1 1 1
6 2 1 2
7 3 1 3
8 4 1 4
9 1 2 1
10 2 2 2
11 3 2 3
12 4 2 4
13 1 2 4
14 2 2 8
15 3 2 2
16 4 2 1
Rgames> acast(mfoo,Var1~Var2,function(x)x[1]-x[2])
1 2
1 0 -3
2 0 -6
3 0 1
4 0 3
# what I would like is the casting formula to return
1 2
1 1 -3
2 2 -6
3 3 1
4 4 3
With the caveat that this is a simple example. In the general case, there will be rows with unique names -- but never more than two rows with a given name, so my x[1]-x[2] won't ever fail.
Or should I just use this:
aggregate(foo[,2],by=list((foo[,1])),function(x)x[1]-x[2])
I've got two different data frames, lets call them "Months" and "People".
Months looks like this:
Month Site X
1 1 4
2 1 3
3 1 5
1 2 10
2 2 7
3 2 5
and People looks like this:
ID Month Site
1 1 1
2 1 2
3 1 1
4 2 2
5 2 2
6 2 2
7 3 1
8 3 2
I'd like to combine them so essentially each time an entry in "People" has a particular Month and Site combination, it's added to the appropriate aggregated data frame, so I'd get something like the following:
Month Site X People
1 1 4 2
2 1 3 0
3 1 5 1
1 2 10 1
2 2 7 3
3 2 5 1
But I haven't the foggiest idea of how to go about doing that. Any suggestions?
Using base packages
> aggregate( ID ~ Month + Site, data=People, FUN = length )
Month Site ID
1 1 1 2
2 3 1 1
3 1 2 1
4 2 2 3
5 3 2 1
> res <- merge(Months, aggdata, all.x = TRUE)
> res
Month Site X ID
1 1 1 4 2
2 1 2 10 1
3 2 1 3 NA
4 2 2 7 3
5 3 1 5 1
6 3 2 5 1
> res[is.na(res)] <- 0
> res
Month Site X ID
1 1 1 4 2
2 1 2 10 1
3 2 1 3 0
4 2 2 7 3
5 3 1 5 1
6 3 2 5 1
Assuming your data.frames are months and people, here's a data.table solution:
require(data.table)
m.dt <- data.table(months, key=c("Month", "Site"))
p.dt <- data.table(people, key=c("Month", "Site"))
# one-liner
dt.f <- p.dt[m.dt, list(X=X[1], People=sum(!is.na(ID)))]
> dt.f
# Month Site X People
# 1: 1 1 4 2
# 2: 1 2 10 1
# 3: 2 1 3 0
# 4: 2 2 7 3
# 5: 3 1 5 1
# 6: 3 2 5 1
I'm sure this has been asked before but for the life of me I can't figure out what to search for!
I have the following data:
x y
1 3
1 3
1 3
1 2
1 2
2 2
2 4
3 4
3 4
And I would like to output a running count that resets everytime either x or y changes value.
x y o
1 3 1
1 3 2
1 3 3
1 2 1
1 2 2
2 2 1
2 4 1
3 4 1
3 4 2
Try something like
df<-read.table(header=T,text="x y
1 3
1 3
1 3
1 2
1 2
2 2
2 4
3 4
3 4")
cbind(df,o=sequence(rle(paste(df$x,df$y))$lengths))
> cbind(df,o=sequence(rle(paste(df$x,df$y))$lengths))
x y o
1 1 3 1
2 1 3 2
3 1 3 3
4 1 2 1
5 1 2 2
6 2 2 1
7 2 4 1
8 3 4 1
9 3 4 2
After seeing #ttmaccer's I see my first attempt with ave was wrong and this is perhaps what is needed:
> dat$o <- ave(dat$y, list(dat$y, dat$x), FUN=seq )
# there was a warning but the answer is corect.
> dat
x y o
1 1 3 1
2 1 3 2
3 1 3 3
4 1 2 1
5 1 2 2
6 2 2 1
7 2 4 1
8 3 4 1
9 3 4 2