Calculate post error slowing in R - r

Question

For my research, I would like to calculate the post-error slowing in the stop signal task to find out whether people become slower after they failed to inhibit their response. Here is some data and I would like to do the following:
For each subject determine first if it was a stop-trial (signal = 1)
For each stop-trial, determine if it is correct (signal = 1 & correct = 2) and then determine whether the next trial (thus the trial directly after the stop-trial) is a go-trial (signal = 0)
Then calculate the average reaction time for all these go-trials that directly follow a stop trial when the response is correct (signal = 0 & correct = 2).
For each incorrect stop trial (signal = 1 & correct = 0) determine whether the next trial (thus the trial directly after the stop-trial) is a go-trial (signal = 0)
Then calculate the average reaction time for all these go-trials that directly follow a stop-trial when the response is correct (correct = 2).
Then calculate the difference between the RTs calculated in step 2 and 3 (= post-error slowing).
I'm not that experienced in R to achieve this. I hope someone can help me with this script.
subject trial signal correct RT
1 1 0 2 755
1 2 0 2 543
1 3 1 0 616
1 4 0 2 804
1 5 0 2 594
1 6 0 2 705
1 7 1 2 0
1 8 1 2 0
1 9 0 2 555
1 10 1 0 604
1 11 0 2 824
1 12 0 2 647
1 13 0 2 625
1 14 0 2 657
1 15 1 0 578
1 16 0 2 810
1 17 1 2 0
1 18 0 2 646
1 19 0 2 574
1 20 0 2 748
1 21 0 0 856
1 22 0 2 679
1 23 0 2 738
1 24 0 2 620
1 25 0 2 715
1 26 1 2 0
1 27 0 2 675
1 28 0 2 560
1 29 1 0 584
1 30 0 2 564
1 31 0 2 994
1 32 1 2 0
1 33 0 2 715
1 34 0 2 644
1 35 0 2 545
1 36 0 2 528
1 37 1 2 0
1 38 0 2 636
1 39 0 2 684
1 40 1 2 0
1 41 0 2 653
1 42 0 2 766
1 43 0 2 747
1 44 0 2 821
1 45 0 2 612
1 46 0 2 624
1 47 0 2 665
1 48 1 2 0
1 49 0 2 594
1 50 0 2 665
1 51 1 0 658
1 52 0 2 800
1 53 1 2 0
1 54 1 0 738
1 55 0 2 831
1 56 0 2 815
1 57 0 2 776
1 58 0 2 710
1 59 0 2 842
1 60 1 0 516
1 61 0 2 758
1 62 1 2 0
1 63 0 2 628
1 64 0 2 713
1 65 0 2 835
1 66 1 0 791
1 67 0 2 871
1 68 0 2 816
1 69 0 2 769
1 70 0 2 930
1 71 0 2 676
1 72 0 2 868
2 1 0 2 697
2 2 0 2 689
2 3 0 2 584
2 4 1 0 788
2 5 0 2 448
2 6 0 2 564
2 7 0 2 587
2 8 1 0 553
2 9 0 2 706
2 10 0 2 442
2 11 1 0 245
2 12 0 2 601
2 13 0 2 774
2 14 1 0 579
2 15 0 2 652
2 16 0 2 556
2 17 0 2 963
2 18 0 2 725
2 19 0 2 751
2 20 0 2 709
2 21 0 2 741
2 22 1 0 613
2 23 0 2 781
2 24 1 2 0
2 25 0 2 634
2 26 1 2 0
2 27 0 2 487
2 28 1 2 0
2 29 0 2 692
2 30 0 2 745
2 31 1 2 0
2 32 0 2 610
2 33 0 2 836
2 34 1 0 710
2 35 0 2 757
2 36 0 2 781
2 37 0 2 1029
2 38 0 2 832
2 39 1 0 626
2 40 1 2 0
2 41 0 2 844
2 42 0 2 837
2 43 0 2 792
2 44 0 2 789
2 45 0 2 783
2 46 0 0 0
2 47 0 0 468
2 48 0 2 686

Answer 1

This may be too late to be useful but here's my solution: (i.e. I first split the data frame by subject, and then apply the same algorithm to each subject; the result is:
# 1 2
# -74.60317 23.39286
X <- read.table(
text=" subject trial signal correct RT
1 1 0 2 755
1 2 0 2 543
1 3 1 0 616
1 4 0 2 804
1 5 0 2 594
1 6 0 2 705
1 7 1 2 0
1 8 1 2 0
1 9 0 2 555
1 10 1 0 604
1 11 0 2 824
1 12 0 2 647
1 13 0 2 625
1 14 0 2 657
1 15 1 0 578
1 16 0 2 810
1 17 1 2 0
1 18 0 2 646
1 19 0 2 574
1 20 0 2 748
1 21 0 0 856
1 22 0 2 679
1 23 0 2 738
1 24 0 2 620
1 25 0 2 715
1 26 1 2 0
1 27 0 2 675
1 28 0 2 560
1 29 1 0 584
1 30 0 2 564
1 31 0 2 994
1 32 1 2 0
1 33 0 2 715
1 34 0 2 644
1 35 0 2 545
1 36 0 2 528
1 37 1 2 0
1 38 0 2 636
1 39 0 2 684
1 40 1 2 0
1 41 0 2 653
1 42 0 2 766
1 43 0 2 747
1 44 0 2 821
1 45 0 2 612
1 46 0 2 624
1 47 0 2 665
1 48 1 2 0
1 49 0 2 594
1 50 0 2 665
1 51 1 0 658
1 52 0 2 800
1 53 1 2 0
1 54 1 0 738
1 55 0 2 831
1 56 0 2 815
1 57 0 2 776
1 58 0 2 710
1 59 0 2 842
1 60 1 0 516
1 61 0 2 758
1 62 1 2 0
1 63 0 2 628
1 64 0 2 713
1 65 0 2 835
1 66 1 0 791
1 67 0 2 871
1 68 0 2 816
1 69 0 2 769
1 70 0 2 930
1 71 0 2 676
1 72 0 2 868
2 1 0 2 697
2 2 0 2 689
2 3 0 2 584
2 4 1 0 788
2 5 0 2 448
2 6 0 2 564
2 7 0 2 587
2 8 1 0 553
2 9 0 2 706
2 10 0 2 442
2 11 1 0 245
2 12 0 2 601
2 13 0 2 774
2 14 1 0 579
2 15 0 2 652
2 16 0 2 556
2 17 0 2 963
2 18 0 2 725
2 19 0 2 751
2 20 0 2 709
2 21 0 2 741
2 22 1 0 613
2 23 0 2 781
2 24 1 2 0
2 25 0 2 634
2 26 1 2 0
2 27 0 2 487
2 28 1 2 0
2 29 0 2 692
2 30 0 2 745
2 31 1 2 0
2 32 0 2 610
2 33 0 2 836
2 34 1 0 710
2 35 0 2 757
2 36 0 2 781
2 37 0 2 1029
2 38 0 2 832
2 39 1 0 626
2 40 1 2 0
2 41 0 2 844
2 42 0 2 837
2 43 0 2 792
2 44 0 2 789
2 45 0 2 783
2 46 0 0 0
2 47 0 0 468
2 48 0 2 686", header=TRUE)
sapply(split(X, X["subject"]), function(D){
PCRT <- with(D, RT[which(c(signal[-1],NA)==1 & c(correct[-1], NA)==2 & signal==0) ])
PERT <- with(D, RT[which(c(signal[-1],NA)==1 & c(correct[-1], NA)==0 & signal==0) ])
mean(PERT) - mean(PCRT)
})
This is ok if you can be sure that every respondent has at least 1 correct and 1 incorrect "stop" trial followed by a "go" trial. A more general case would be (giving NA if they are either always correct or always mistaken):
sapply(split(X, X["subject"]), function(D){
PCRT <- with(D, RT[which(c(signal[-1],NA)==1 & c(correct[-1], NA)==2 & signal==0) ])
PERT <- with(D, RT[which(c(signal[-1],NA)==1 & c(correct[-1], NA)==0 & signal==0) ])
if(length(PCRT)>0 & length(PERT)>0) mean(PERT) - mean(PCRT) else NA
})

Answer 2

Does that help you? A little bit redundant maybe, but I tried to follow your steps as best as possible (not sure whether I mixed something up, please check for yourself looking at the table). The idea is to put the data in a csv file first and treat it as a data frame. Find the csv raw file here: http://pastebin.com/X5b2ysmQ
data <- read.csv("datatable.csv",header=T)
data[,"condition1"] <- data[,"signal"] == 1
data[,"condition2"] <- data[,"condition1"] & data[,"correct"] == 2
data[,"RT1"] <- NA
for(i in which(data[,"condition2"])){
if( nrow(data)>i && !data[i+1,"condition1"] && data[i+1,"correct"] == 2 )
# next is a go trial
data[i+1,"RT1"] <- data[i+1,"RT"]
}
averageRT1 <- mean( data[ !is.na(data[,"RT1"]) ,"RT1"] )
data[,"RT2"] <- NA
for(i in which(data[,"condition1"] & data[,"correct"] == 0)){
if( nrow(data)>i && !data[i+1,"condition1"] && data[i+1,"correct"] == 2 )
# next is a go trial
data[i+1,"RT2"] <- data[i+1,"RT"]
}
averageRT2 <- mean( data[ !is.na(data[,"RT2"]) ,"RT2"] )
postErrorSlowing <- abs(averageRT2-averageRT1)

Answer 3

#Nilsole I just tried it and it is almost perfect. How could the code be improved that for each subject the postErrorSlowing is calculated and placed in a dataframe? Thus that a new data frame is created which consists of subject number (1,2,3 etc.) and the postErrorSlowing variable? Something like this (postErrorSlowing are made up numbers)
subject postErrorSlowing
1 50
2 75
....