[R] fminsearch usage
R Help -
I'm attempting to use fminsearch and it continues to give me errors. I have
a normal Rescorla Wagner function that has 2 parameters and returns total
error. for some reason fminsearch keeps saying the type of errors below,
I've put the function and data after it using dput
sol - fminsearch(rescorlaWagner(learningRate,decision_slope), c(-1.2,1))
Error in this$fun(x = x, index = index, fmsfundata = this$costfargument) :
attempt to apply non-function
sol - fminsearch(rescorlaWagner,c(alphaPlus = alphaPlus, decision_slope
= decision_slope), c(-1.2,1))
Error in options[[name]] : subscript out of bounds
rescorlaWagner - function(alphaPlus, decision_slope){
# alphaPlus - parameters[1]
# decision_slope _- parameters[2]
#dat - mainData[mainData$participants == subjects[s],]
dat$choice - ifelse(dat$cueResp.keys == 5, 1, 0)
dat$reward - dat$trialChange
# numberTrials[s] - length(dat$choice)
numberTrials - length(dat$choice)
V1 - 0; V0 - 0; totError - 0; thList = 0; block = 0; valueOne -
0; valueZero
- 0; probOfStim1 - .01; probOfStim2 - .01; logLike - 0;
V1[1] = 0; V0[1] = 0;
for (i in 2:length(dat$choice))
{
if (dat$choice[i-1] == 1) # Chose pattern 1 last time
{
logLike[i] - logLike[i-1] - log(probOfStim1[i-1])
#print(logLike)
delta1 - dat$reward[i] - V1[i-1]
V1[i] - V1[i-1] + alphaPlus * delta1
V0[i] - V0[i-1]
}
else # Chose pattern 0 last time instead of pattern 1
{
logLike[i] - logLike[i-1] - log(probOfStim2[i-1])
delta0 - dat$reward[i] - V0[i-1]
V0[i] - V0[i-1] + alphaPlus * delta0
V1[i] - V1[i-1]
}
probOfStim1[i] - exp(decision_slope * V1[i])/(exp(decision_slope * V1
[i] + exp(decision_slope * V0[i])))
probOfStim2[i] - exp(decision_slope * V0[i])/(exp(decision_slope * V1
[i] + exp(decision_slope * V0[i])))
valueOne[i] - V1[i]
valueZero[i] - V0[i]
newList - list(EVpattern1 = valueOne, EVpattern0 =
valueZero, ProbabilityStim1
= probOfStim1, ProbabilityStim2 = probOfStim2, LogLikelihood = logLike)
logLike[i]
}
#return(newList)
return(logLike[i])
}
dput(dat)
structure(list(highCueImg = c(1.bmp, 1.bmp, 1.bmp, 1.bmp,
1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp,
1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp,
1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp,
1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp,
1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp, 1.bmp,
1.bmp), fixationDuration = c(2.900504833, 2.137709799, 3.314149929,
3.034789701, 3.44018723, 3.559188201, 4, 3.397855819, 2.424957014,
3.010173792, 2.754706831, 2.183118953, 3.328269233, 3.058698132,
3.030312482, 2.948379082, 2.48854601, 2.635627291, 2.62972477,
3.07140, 3.343837986, 3.452638681, 3.044377797, 2, 3.139145494,
2.821634126, 3.463414658, 2.974553193, 2.887472182, 3.276110294,
2.774994095, 2.538595411, 2, 2, 2, 2.937451712, 3.030693282,
3.533087478, 2.821526274, 2), condition = c(gain, gain, gain,
gain, gain, gain, gain, gain, gain, gain, gain,
gain, gain, gain, gain, gain, gain, gain, gain,
gain, gain, gain, gain, gain, gain, gain, gain,
gain, gain, gain, gain, gain, gain, gain, gain,
gain, gain, gain, gain, gain), value = c(1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L), lowCueImg = c(2.bmp, 2.bmp, 2.bmp,
2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp,
2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp,
2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp,
2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp,
2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp, 2.bmp,
2.bmp, 2.bmp), trials.thisRepN = c(0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L), trials.thisTrialN = c(0L, 2L, 4L, 6L, 10L, 12L, 14L,
15L, 18L, 19L, 21L, 24L, 30L, 36L, 0L, 6L, 9L, 10L, 12L, 15L,
16L, 22L, 27L, 31L, 33L, 36L, 37L, 0L, 3L, 4L, 6L, 7L, 8L, 11L,
14L, 23L, 30L, 32L, 33L, 35L), trials.thisN = c(0L, 2L, 4L, 6L,
10L, 12L, 14L, 15L, 18L, 19L, 21L, 24L, 30L, 36L, 0L, 6L, 9L,
10L, 12L, 15L, 16L, 22L, 27L, 31L, 33L, 36L, 37L, 0L, 3L, 4L,
6L, 7L, 8L, 11L, 14L, 23L, 30L, 32L, 33L, 35L), trials.thisIndex = c(0L,
2L, 4L, 6L, 10L, 12L, 14L, 15L, 18L, 19L, 21L, 24L, 30L, 36L,
0L, 6L, 9L, 10L, 12L, 15L, 16L, 22L, 27L, 31L, 33L, 36L, 37L,
0L, 3L, 4L, 6L, 7L, 8L, 11L, 14L, 23L, 30L, 32L, 33L, 35L), high = c(5L,
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L,
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L,
5L, 5L, 5L, 5L, 5L, 5L, 5L), low = c(6L, 6L, 6L, 6L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L,
6L, 6L), cueTime = c(0.119068770004, 20.0149482278, 39.3913808842,
[R] Constructing a matrix of outputs from loop
R -
I would like to construct a matrix from the output of a loop that has 2
values it varies over the course of the loop creating a 20x20 matrix of
output values:
ap = logspace(-3, 0, 20)
am = logspace(-3, .7, 20)
for (ap in apList)
{
for (am in amList)
{
output = func(ap, am)
}
}
i.e. cell 1x1 is -3,-3 and the value is 45 or something
--
*Edward H Patzelt | Research Assistant
Psychology | University of Minnesota | Elliott Hall, 75 East River Road |
Minneapolis, MN 55455
Email: patze...@umn.edu |
Main: 612.626.0072
|
Mobile: 651.315.3410
| Office: S355
**
*
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Changing Order of Factor Levels in Mixed Model (nlme)
R Help - Why is that in the results below, changing the order of the factor (trialType2: levels - DD, SD, DS, SS) changes the estimates in the fixed effects tests? tmp.dat4$trialType2 - sort(tmp.dat4$trialType, decreasing = TRUE) mod2c - lme(proportion.down ~ trialType2, data = tmp.dat4, random = ~ 1 | subject, na.action = na.omit, method = ML) summary(mod2c) Linear mixed-effects model fit by maximum likelihood Data: tmp.dat4 AIC BIClogLik 27.92306 48.23003 -7.961531 Random effects: Formula: ~1 | subject (Intercept) Residual StdDev: 0.3800017 0.1530272 Fixed effects: proportion.down ~ trialType2 Value Std.Error DF t-value p-value (Intercept) 0.6788875 0.08613476 141 7.881690 0. trialType2DS 0.0287062 0.11267357 141 0.254773 0.7993 trialType2SD -0.0197194 0.12142018 141 -0.162406 0.8712 trialType2SS -0.0941918 0.12204707 141 -0.771766 0.4415 Correlation: (Intr) trT2DS trT2SD trialType2DS -0.658 trialType2SD -0.709 0.467 trialType2SS -0.706 0.464 0.571 Standardized Within-Group Residuals: Min Q1 Med Q3 Max -3.67176824 -0.22090663 0.08677971 0.14331603 2.86301670 Number of Observations: 218 Number of Groups: 74 tmp.dat4$trialType2 - sort(tmp.dat4$trialType, decreasing = FALSE) mod2c - lme(proportion.down ~ trialType2, data = tmp.dat4, random = ~ 1 | subject, na.action = na.omit, method = ML) summary(mod2c) Linear mixed-effects model fit by maximum likelihood Data: tmp.dat4 AIC BIClogLik 27.92306 48.23003 -7.961531 Random effects: Formula: ~1 | subject (Intercept) Residual StdDev: 0.3800017 0.1530272 Fixed effects: proportion.down ~ trialType2 Value Std.Error DF t-value p-value (Intercept) 0.5846957 0.08646554 141 6.762181 0. trialType2DS 0.0744724 0.1123 141 0.660347 0.5101 trialType2SD 0.1228981 0.12175741 141 1.009368 0.3145 trialType2SS 0.0941918 0.12204707 141 0.771766 0.4415 Correlation: (Intr) trT2DS trT2SD trialType2DS -0.660 trialType2SD -0.710 0.469 trialType2SS -0.708 0.468 0.573 Standardized Within-Group Residuals: Min Q1 Med Q3 Max -3.67176824 -0.22090663 0.08677971 0.14331603 2.86301670 Number of Observations: 218 Number of Groups: 74 -- *Edward H Patzelt | Research Assistant Psychology | University of Minnesota | Elliott Hall, 75 East River Road | Minneapolis, MN 55455 Email: patze...@umn.edu | Main: 612.626.0072 | Mobile: 651.315.3410 | Office: S355 ** * [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Aggregate Table Data into Cell Frequencies
R-help - I have this set of aggregated tables (sample data below via dput()). And I would like to have delayValue as the column variables with the temp (temp1, temp2, temp3) values as the row variables. However I would like to have the temp variables *aggregated into single rows* so that I have the frequency (Freq | counts) of each time each delayValue occurs in the cells. I've tried this command without luck, it seems to be dropping a bunch of values. tmp - reshape(sampleData, direction = wide,timevar = delayValue, new.row.names = unique(sampleData$id) ) structure(list(delayValue = c(0, NA, 7, 8, 9, 10, NA, NA, 4, 5, 6, 7, 8, NA, NA, 4, 5, NA, 6, 7, 8, 9, 10, 11, 12, NA, NA, 5, 6, 7, 8, 9, NA, 4, 5, 6, 7, 8, 9, 10, NA, 9, 10, NA, NA, NA, 4, 5, 6, NA, 5, 6, 7, NA, 7, 8, 9, 10, 11, NA, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, NA, 6, NA, 4, 7, NA, 4, NA, 2, 3, 4, NA, 5, NA, 5, 6, NA, 6, NA, 6, NA, 6, NA, 2, 3, NA, 5, NA, 3, 4, NA, 4, 5, NA, 4, NA, 3, NA, NA, 2, 3, 4, 5, 6, 7, 8, 9, NA, 9, NA, NA, 2, NA, 3, 4, 5, 6, NA, 4, 5, 6, NA, 5, NA, 2, 3, NA, 4, 5, 6, NA, 3, NA, NA, 2, NA, 2, NA, 6, NA, 2, NA, 4, NA, 2, 3, NA, 2, NA, 2, 3, 4, NA, 4, 5, NA, 2, NA, 3, 4, NA, NA, 18, 19, 20, 21, 22, 23, 24, 25, 26, NA, NA, NA, NA, NA, 4, NA, NA, NA, NA, 5, 6, NA, NA, 7, 8, 9, 10, NA, 11, 12, 13, 14, NA, 10, NA, 11, 12, 13, 14, 15, 16, NA, 10, 11, 12, NA, 13, 14, 15, 16, 17, NA, 2, NA, 2, 3, 4, NA, 2, NA, 3, NA, 2, 3, 4, 5, 6, 5, 6, 7, NA, 6, 8, NA, 7, 8, NA, 6, 7, 8, 9, NA, NA, 6, 7, NA, NA, 5, 6, NA, 4, 5, 6, 7, NA, 6, NA, 7, 8, NA, 6, 7, 8, NA, 5, NA, 2, 3, 4, 5, NA, 7, 7, NA, 7, NA, 7, NA, 7, NA, 7, NA, 7, NA, 7, 8, NA, 7, 8, NA, 5, 6, NA, 4, NA, 7, NA, 4, NA, 7, NA, 7, 8, NA, 8, NA, 7, 8, NA, 7, NA, 7, NA, 7, NA, NA, 16, 17, 18, NA, 19, 20, 21, NA, 21, 22, 23, 24, 25, 25, 26, 27, 28, NA, NA, NA, NA, NA, NA, NA, 4, 6, NA, NA, 4, 5, 6, 7, NA, 7, 8, 9, NA, 10, 11, 12, NA, 12, 13, 14, NA, 13, 14, NA, 12, 13, 14, NA, 15, 16, NA, NA, 2, 3, NA, 2, 3, NA, 4, 5, 6, NA, 5, 6, 7, 8, 9, 10, 11, 12, NA, 9, 10, 11, NA, 12, 13, NA, 9, 10, 11, NA, 12, 13, NA, 13, 14, 15, 16, 17, NA, 6, 7, NA, NA, 6, 7, 8, NA, 9, 10, 11, NA, NA, 5, 7, NA, 5, NA, NA, 2, 3, 4, 5, NA, NA, 8, 9, NA, 8, 9, NA, NA, NA, NA, NA, NA, NA, NA, NA, 6, NA, NA, 6, 7, 8, NA, 8, 9, NA, 9, NA, 9, 10, NA, 9, 10, NA, 9, 10, NA, 9, 10, NA, 7, 2, 3, NA, 2, NA, 2, 3, NA, 2, 3, 4, NA, 2, 3, NA, 2, 3, NA, 3, NA, 2, NA, 3, 4, NA, 3, 4, NA, 5, NA, 2, 3, 4, NA, 3, 5, NA, 3, 4, NA, 2, 3, 4, NA, 2, 3, NA, 2, 3, 4, NA, 2, NA, 3, 4, NA, NA, 18, 19, 20, 21, NA, 21, 22, 23, 24, NA, 23, 25, NA, 23, 24, 25, 26, NA, NA, 22, 23, 24, NA, NA, NA, NA, NA, 6, 7, NA, NA, 7, 8, 9, 10, NA, 11, 12, 13, NA, 14, 15, NA, 13, 14, 15, NA, 14, 15, 16, NA, 15, 16, 17, 18, NA, 16, 17, NA, 7, 20, 21, 22, 23, NA, 22, 23, 24, NA, 25, 26, 27, NA, 27, 28, 29, 30, NA, 31, NA, NA, NA, NA, NA, NA, 7, 8, NA, NA, 8, NA, 8, 9, NA, 10, 11, 12, 13, NA, 13, 14, 15, NA, 15, 16, 17, NA, 15, 16, NA, 17, 18, 19, NA, NA, 12, 13, 14, NA, 13, NA, 9, 10, 11, NA, 12, 13, 14, 15, NA, NA, 9, 10, NA, NA, NA, NA, NA, 6, NA, NA, 4, 5, 6, NA, 3, 4, 5, NA, 6, 7, 8, NA, 9, NA, 3, 4, 5, NA, 6, 7, 8, 9, 10, 11, 12, 13, NA, 7, 13, NA, 11, 12, NA, 13, 14, 15, 16, 17, 14, NA, 13, 15, NA, 14, 15, 16, NA, NA, NA, NA, NA, 6, 7, NA, NA, 7, NA, 7, NA, 8, NA, 8, 9, 10, NA, 10, 11, 12, 13, NA, 13, 14, 15, 16, NA, 14, NA, NA, 2, NA, 2, NA, 2, NA, 2, 3, NA, 2, 3, NA, 2, NA, 2, NA, 2, 3, NA, 2, 3, NA, 4, 5, NA, 5, 6, NA, 4, 5, NA, 4, NA, 2, NA, 2, NA, 2, 3, NA, 2, 3, NA, 3, NA, 2, NA, NA, 4, 5, 6, NA, 6, 7, 8, NA, 8, NA, 8, NA, 3, NA, 2, NA, 2, NA, 2, NA, 2, NA, 2, 3, NA, 6, NA, 2, NA, 3, 4, 5, 6, 7, NA, 4, 5, NA, NA, NA, 2, 3, NA, 4, NA, 3, 4, NA, 7, 2, 3, 4, 5, 6, 4, 5, NA, 2, 3, NA, 3, 4, NA, 5, 6, 7, NA, 8, 9, 10, NA, 11, NA, 10, 12, NA, 9, 10, NA, 10, 11, NA, 6, 7, NA, 8, 9, 10, NA, 4, 5, 6, NA, 6, NA, 3, 4, 5, NA, 2, 3, NA, 3, 4, NA, 2, 3, NA, 3, 4, NA), Freq = c(0, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 5, 5, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 4, 5, 5, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 5, 1, 3, 2, 3, 3, 2, 3, 3, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 2, 2, 3, 3, 3, 3, 2, 2, 3, 1, 1, 4, 1, 1, 4, 1, 4, 1, 1, 2, 1, 3, 2, 1, 2, 2, 2, 3, 3, 2, 1, 4, 1, 1, 4, 2, 3, 2, 1, 3, 2, 1, 3, 2, 3, 2, 3, 5, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 2, 1, 2, 3, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 4, 1, 1, 3, 1, 1, 1, 2, 1, 4, 5, 3, 2, 3, 3, 3, 2, 2, 4, 1, 5, 1, 1, 3, 2, 3, 2, 1, 1, 1, 1, 1, 4, 1, 4, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 5, 5, 5, 5, 1, 4, 5, 5, 6, 2, 1, 2, 6, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 2, 1, 2, 1, 4, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 3, 1, 1, 1, 1, 1, 6, 2, 1, 2, 6, 1, 1, 4, 1, 1, 1, 1, 1, 2, 3, 1, 1, 3, 1, 1, 1, 3, 1, 4, 1, 1, 1, 1, 1, 1, 2, 3, 3, 3, 3, 2, 2, 3, 3, 2, 2, 3, 1, 2, 2, 1, 1, 3, 1,
[R] Axis Breaks with ggplot2
R-help - I'm trying to create axis breaks similar to this : http://www.r-bloggers.com/wp-content/uploads/2010/08/bar-chart-natural-axis-split1.png . Is there a way to do this in R? Here's my code thus far: structure(list(condition = structure(c(2L, 1L, 3L), .Label = c(con, exp, unedit), class = factor), trial.avg = c(4.045833, 4.335417, 4.61875), trial.sd = c(0.928718367573187, 0.851822141963017, 1.03502368980692), s.e. = c(0.0232179591893297, 0.0212955535490754, 0.163651614601074), N = c(40, 40, 40), condition2 = structure(1:3, .Label = c(Interaction Censured, Control Censured, Uncensured), class = factor)), .Names = c(condition, trial.avg, trial.sd, s.e., N, condition2), row.names = c(NA, -3L), class = data.frame) library(ggplot2) none - theme_blank() err1$condition - as.factor(err1$condition) censorA - ggplot() + geom_bar(aes(y = trial.avg, x = as.factor(condition2), fill = as.factor(condition2), position = dodge), data = err1) censorB - censorA + geom_errorbar(aes(x = err1$condition2, ymin = (err1$trial.avg-(err1$trial.sd/sqrt(40))), ymax = (err1$trial.avg+(err1$ trial.sd/sqrt(40))), data = err1, width = .4)) censorC - censorB + opts(panel.background = none) + opts(panel.border = none) + opts(panel.grid.minor = none) + opts(panel.grid.major = none) + opts(axis.line = theme_segment(colour = grey35)) + opts(background.fill = none) censorC + scale_y_continuous(limits = c(0,7), expand = c(0,0), 'Rating') + opts(legend.position = none) Best, -- Edward H. Patzelt Research Assistant TRiCAM Lab University of Minnesota Psychology/Psychiatry VA Medical Center S355 Elliot Hall: 612-626-0072 www.psych.umn.edu/research/tricam [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Softmax Action Selection
R Community -
I'm attempting to apply a softmax action selection to a probability
generated by a hidden Markov model. I'm having difficulties in how to
apply the softmax temperature parameter (beta). Here is my code thus far.
I'm thinking the sigmoid function will work but I need this function to
return the total error instead of the probability so I can't optimize it
with optim().
calc.probs - function(delta,beta)
{
# initial starting probabilities
thList = 0; block = 0
for (i in 1:length(dat$choice))
{
if (dat$RepNum[i] != block)
{
pL = 0.5; pR = 0.5; block = dat$RepNum[i];
}
# Markov Transitions
pL - pL*(1-delta) + pR*delta
pR - 1-pL
# Apply feedback
pflc - ifelse(dat$choice[i] == dat$reward[i], .8, .2)
pfrc - 1 - pflc
denom - pflc * pL + pfrc * pR
# What's the new belief given observation
posteriorL - pflc * pL/denom
posteriorR - 1-posteriorL
pL - posteriorL; pR - posteriorR
# Insert softmax here
pLlist[i] - pL
}
return(pLlist)
}
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota Psychology/Psychiatry
VA Medical Center
S355 Elliot Hall: 612-626-0072
www.psych.umn.edu/research/tricam
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__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Running Total
I'm am trying to create a vector that has a running total that adds each
time a 1 occurs. here's the code and data
c(1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 1L,
1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L,
0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L,
1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L,
0L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L,
1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L,
0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L)
total - {};
for(i in 1:length(dat$Valid)){
total[i] - ifelse(dat$Valid[i-1]==1, total[i] + 1, total[i])
}
Cheers,
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota Psychology/Psychiatry
VA Medical Center
S355 Elliot Hall: 612-626-0072
www.psych.umn.edu/research/tricam
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Running Total
Actually in looking at this I need it to only add if a 0 occurs instead of
a 1.
On Mon, Mar 5, 2012 at 12:57 PM, jim holtman jholt...@gmail.com wrote:
cumsum
is probably what you want:
x - c(1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 1L,
+ 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L,
+ 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L,
+ 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L,
+ 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L,
+ 0L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L,
+ 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L,
+ 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L)
cbind(x, cumsum(x))
x
[1,] 1 1
[2,] 1 2
[3,] 0 2
[4,] 1 3
[5,] 1 4
[6,] 1 5
[7,] 1 6
[8,] 1 7
[9,] 0 7
[10,] 1 8
[11,] 0 8
[12,] 1 9
[13,] 1 10
[14,] 1 11
[15,] 1 12
[16,] 1 13
[17,] 0 13
On Mon, Mar 5, 2012 at 1:51 PM, Edward Patzelt patze...@umn.edu wrote:
I'm am trying to create a vector that has a running total that adds each
time a 1 occurs. here's the code and data
c(1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 1L,
1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L,
0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L,
1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L,
0L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L,
1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L,
0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L)
total - {};
for(i in 1:length(dat$Valid)){
total[i] - ifelse(dat$Valid[i-1]==1, total[i] + 1, total[i])
}
Cheers,
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota Psychology/Psychiatry
VA Medical Center
S355 Elliot Hall: 612-626-0072
www.psych.umn.edu/research/tricam
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota Psychology/Psychiatry
VA Medical Center
S355 Elliot Hall: 612-626-0072
www.psych.umn.edu/research/tricam
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Cubic Gradient Descent Package
R - Does anyone know of a cubic gradient descent package? I found grad.desc() but that only allows for a 2d function. I have 3 free parameters and thus am looking for a 3d function. Thank you, -- Edward H. Patzelt Research Assistant TRiCAM Lab University of Minnesota Psychology/Psychiatry VA Medical Center S355 Elliot Hall: 612-626-0072 www.psych.umn.edu/research/tricam [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Counting Elements Conditionally
R - I have 3 variables with data below. Variable Rev is a vector that changes from 1 to 2, 2 to 3, etc Variable FF is a binary variable with 1's and 0's. Variable bin is a different binary variable with 1's and 0's. I want to calculate the number of elements: 1. Starting with the first element where Rev switches (i.e. 1 to 2) 2. The number of elements between the transition and the first 0 in the FF vector; *when bin is also a 0.* * * 3. I want to do this for each transition in Rev, but ignore all elements after calculating #2 until another transition has occurred. structure(list(Rev = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), FF = c(0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L), bin = c(NA, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1)), .Names = c(Rev, FF, bin ), row.names = c(NA, -40L), class = data.frame) -- Edward H. Patzelt Research Assistant TRiCAM Lab University of Minnesota Psychology/Psychiatry VA Medical Center Office: S355 Elliot Hall - Twin Cities Campus Phone: 612-626-0072 Email: patze...@umn.edu Please consider the environment before printing this email www.psych.umn.edu/research/tricam [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Counting Elements Conditionally
Awesome, this is close, couple changes. Below is full data set for 1 person. I want the code to look at the first time it sees a 0 in FF after the transition in Rev. I then want it to test whether bin is also a 0. If and only if this is the first 0 in FF after the transition, and bin = 0, then count the number of elements between the transition in Rev and the 0 in FF. structure(list(Rev = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), FF = c(1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L), bin = c(NA, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)), .Names = c(Rev, FF, bin), row.names = c(NA, -125L), class = data.frame) On Mon, Aug 22, 2011 at 3:57 PM, Jean V Adams jvad...@usgs.gov wrote: Re: [R] Counting Elements Conditionally Jean V Adams to: Edward Patzelt 08/22/2011 03:53 PM [R] Counting Elements Conditionally Edward Patzelt to: r-help 08/22/2011 02:33 PM R - I have 3 variables with data below. Variable Rev is a vector that changes from 1 to 2, 2 to 3, etc Variable FF is a binary variable with 1's and 0's. Variable bin is a different binary variable with 1's and 0's. I want to calculate the number of elements: 1. Starting with the first element where Rev switches (i.e. 1 to 2) 2. The number of elements between the transition and the first 0 in the FF vector; *when bin is also a 0.* * * 3. I want to do this for each transition in Rev, but ignore all elements after calculating #2 until another transition has occurred. structure(list(Rev = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), FF = c(0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L), bin = c(NA, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1)), .Names = c(Rev, FF, bin ), row.names = c(NA, -40L), class = data.frame) -- Edward H. Patzelt Research Assistant ? TRiCAM Lab University of Minnesota ? Psychology/Psychiatry VA Medical Center Office: S355 Elliot Hall - Twin Cities Campus Phone: 612-626-0072 Email: patze...@umn.edu Try this (I'm assuming your data.frame is called df: uR - unique(df$Rev) uR0 - uR*10L first.Rev - match(uR, df$Rev) first.Rev0 - match(uR0, df$Rev * 10L + df$FF) no.elements - first.Rev0 - first.Rev + 1 This starts with Rev=1 (rather than Rev=2), but you can get rid of that by dropping the first result ... no.elements[-1] Jean Ooops. Too hasty in my reply. I missed the part about bin also being zero. Try this instead. uR - unique(df$Rev) uR00 - uR*100L first.Rev - match(uR, df$Rev) first.Rev00 - match(uR00, df$Rev * 100L + df$FF * 10L + df$bin) no.elements - first.Rev00 - first.Rev + 1 Jean -- Edward H. Patzelt Research Assistant TRiCAM Lab University of Minnesota Psychology/Psychiatry VA Medical Center Office: S355 Elliot Hall - Twin Cities Campus Phone: 612-626-0072 Email: patze...@umn.edu Please consider the environment before printing this email www.psych.umn.edu/research/tricam [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Counting Elements Conditionally
after it sees the first occurrence of 0 in FF following a transition, I want it to ignore all further elements until the next transition. On Mon, Aug 22, 2011 at 3:58 PM, Edward Patzelt patze...@umn.edu wrote: Awesome, this is close, couple changes. Below is full data set for 1 person. I want the code to look at the first time it sees a 0 in FF after the transition in Rev. I then want it to test whether bin is also a 0. If and only if this is the first 0 in FF after the transition, and bin = 0, then count the number of elements between the transition in Rev and the 0 in FF. structure(list(Rev = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), FF = c(1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L), bin = c(NA, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)), .Names = c(Rev, FF, bin), row.names = c(NA, -125L), class = data.frame) On Mon, Aug 22, 2011 at 3:57 PM, Jean V Adams jvad...@usgs.gov wrote: Re: [R] Counting Elements Conditionally Jean V Adams to: Edward Patzelt 08/22/2011 03:53 PM [R] Counting Elements Conditionally Edward Patzelt to: r-help 08/22/2011 02:33 PM R - I have 3 variables with data below. Variable Rev is a vector that changes from 1 to 2, 2 to 3, etc Variable FF is a binary variable with 1's and 0's. Variable bin is a different binary variable with 1's and 0's. I want to calculate the number of elements: 1. Starting with the first element where Rev switches (i.e. 1 to 2) 2. The number of elements between the transition and the first 0 in the FF vector; *when bin is also a 0.* * * 3. I want to do this for each transition in Rev, but ignore all elements after calculating #2 until another transition has occurred. structure(list(Rev = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), FF = c(0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L), bin = c(NA, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1)), .Names = c(Rev, FF, bin ), row.names = c(NA, -40L), class = data.frame) -- Edward H. Patzelt Research Assistant ? TRiCAM Lab University of Minnesota ? Psychology/Psychiatry VA Medical Center Office: S355 Elliot Hall - Twin Cities Campus Phone: 612-626-0072 Email: patze...@umn.edu Try this (I'm assuming your data.frame is called df: uR - unique(df$Rev) uR0 - uR*10L first.Rev - match(uR, df$Rev) first.Rev0 - match(uR0, df$Rev * 10L + df$FF) no.elements - first.Rev0 - first.Rev + 1 This starts with Rev=1 (rather than Rev=2), but you can get rid of that by dropping the first result ... no.elements[-1] Jean Ooops. Too hasty in my reply. I missed the part about bin also being zero. Try this instead. uR - unique(df$Rev) uR00 - uR*100L first.Rev - match(uR, df$Rev) first.Rev00 - match(uR00, df$Rev * 100L + df$FF * 10L + df$bin) no.elements - first.Rev00 - first.Rev + 1 Jean -- Edward H. Patzelt Research Assistant TRiCAM Lab University of Minnesota Psychology/Psychiatry VA Medical Center Office: S355 Elliot Hall - Twin Cities Campus Phone: 612-626-0072 Email: patze...@umn.edu Please consider the environment before printing this email www.psych.umn.edu/research/tricam -- Edward H. Patzelt Research Assistant TRiCAM Lab University of Minnesota Psychology/Psychiatry VA Medical Center Office: S355 Elliot Hall - Twin Cities
Re: [R] Counting Elements Conditionally
that is exactly correct, assuming we did not start at the beginning, but started at the first transition (this is the correct way to think about it) On Mon, Aug 22, 2011 at 4:08 PM, Jean V Adams jvad...@usgs.gov wrote: So, using the full data set, what should the result look like? c(NA, NA, NA, 3, NA,NA, NA, 2) ? Jean Edward Patzelt patze...@umn.edu wrote on 08/22/2011 03:58:38 PM: [image removed] Re: [R] Counting Elements Conditionally Edward Patzelt to: Jean V Adams 08/22/2011 03:58 PM Cc: r-help Awesome, this is close, couple changes. Below is full data set for 1 person. I want the code to look at the first time it sees a 0 in FF after the transition in Rev. I then want it to test whether bin is also a 0. If and only if this is the first 0 in FF after the transition, and bin = 0, then count the number of elements between the transition in Rev and the 0 in FF. structure(list(Rev = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), FF = c(1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L), bin = c(NA, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)), .Names = c(Rev, FF, bin), row.names = c(NA, -125L), class = data.frame) On Mon, Aug 22, 2011 at 3:57 PM, Jean V Adams jvad...@usgs.gov wrote: Re: [R] Counting Elements Conditionally Jean V Adams to: Edward Patzelt 08/22/2011 03:53 PM [R] Counting Elements Conditionally Edward Patzelt to: r-help 08/22/2011 02:33 PM R - I have 3 variables with data below. Variable Rev is a vector that changes from 1 to 2, 2 to 3, etc Variable FF is a binary variable with 1's and 0's. Variable bin is a different binary variable with 1's and 0's. I want to calculate the number of elements: 1. Starting with the first element where Rev switches (i.e. 1 to 2) 2. The number of elements between the transition and the first 0 in the FF vector; *when bin is also a 0.* * * 3. I want to do this for each transition in Rev, but ignore all elements after calculating #2 until another transition has occurred. structure(list(Rev = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), FF = c(0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L), bin = c(NA, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1)), .Names = c(Rev, FF, bin ), row.names = c(NA, -40L), class = data.frame) -- Edward H. Patzelt Research Assistant ? TRiCAM Lab University of Minnesota ? Psychology/Psychiatry VA Medical Center Office: S355 Elliot Hall - Twin Cities Campus Phone: 612-626-0072 Email: patze...@umn.edu Try this (I'm assuming your data.frame is called df: uR - unique(df$Rev) uR0 - uR*10L first.Rev - match(uR, df$Rev) first.Rev0 - match(uR0, df$Rev * 10L + df$FF) no.elements - first.Rev0 - first.Rev + 1 This starts with Rev=1 (rather than Rev=2), but you can get rid of that by dropping the first result ... no.elements[-1] Jean Ooops. Too hasty in my reply. I missed the part about bin also being zero. Try this instead. uR - unique(df$Rev) uR00 - uR*100L first.Rev - match(uR, df$Rev) first.Rev00 - match(uR00, df$Rev * 100L + df$FF * 10L + df$bin) no.elements - first.Rev00 - first.Rev + 1 Jean
[R] Indexing Permutation Values
R-help -
This code iterates over a function with 2 free parameters to find a list of
values (which are the number of incorrect predictions for a computational
model). I want to find the values of i,e when there is the minimum number
of incorrect predictions. In other words, the value of i and e when
variable thesum is at the lowest possible number. The bestDeltas variable
at the bottom worked with 1 free parameter.
structure(list(grid = c(3162L, 3162L, 3162L, 3162L, 3162L, 3162L,
3162L, 3162L, 3162L, 3162L, 3162L, 3162L, 3162L, 3162L, 3162L,
3162L, 3162L, 3162L, 3162L, 3162L), RepNum = c(1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L
), stimbinary = c(1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 1), choice = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 1, 1, 1, 1, 1), accuracy = c(0L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 0L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 0L), actualcorrect = c(1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1), correctstim = c(0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1), choseright = c(0,
1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1), reward = c(0L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 0L, 0L, 0L, 1L, 1L,
1L, 1L, 0L), resp = structure(c(5L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
5L, 3L, 5L, 3L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 3L), .Label = c(f11.bmp,
f13.bmp, f14.bmp, f4.bmp, f7.bmp, f9.bmp), class =
factor)), .Names = c(grid,
RepNum, stimbinary, choice, accuracy, actualcorrect,
correctstim, choseright, reward, resp), row.names = c(NA,
20L), class = data.frame)
install.packages(Hmisc, dependencies = T)
library(Hmisc)
bestDeltas= 0; betterThanChance = 0; minErr = 0
fitProp- 0
thresholded_test - 0
calc.probs - function(delta,gamma){
# initial starting probabilities
thList = 0
block = 0
# finMatrix - 0
for (i in 1:length(dat2$choice))
{
if (dat2$RepNum[i] != block)
{
pL = 0.5
pR = 0.5
block = dat2$RepNum[i]
certState = 0.5
}
# Markov Transitions
pL - pL*(1-delta) + pR*delta
pR - 1-pL
# Apply feedback
#denom - p(F|L,C) * p(L) + p(F|R,C) * p(R)
pflc - ifelse(dat2$choice[i] == dat2$reward[i], .8, .2)
pfrc - 1 - pflc
denom - pflc * pL + pfrc * pR
# What's the new belief given observation
posteriorL - pflc * pL/denom
posteriorR - 1-posteriorL
pL - posteriorL
pR - posteriorR
certState = gamma * certState + (1 - gamma) * pL
thList[i] = ifelse(certState 0.55, 1, ifelse(certState 0.45, 0,
ifelse(pL 0.5, 1, 0)))
}
return(thList)
}
thesum=0
j = 1
minx = 0.01; maxx = 0.5; incx = 0.05;
x = seq(minx, maxx, incx)
miny = 0.0; maxy = 1; incy = 0.05;
y = seq(miny, maxy, incy)
for (i in x)
{
for (e in y)
{
thresholded = calc.probs(i, e)
diff - abs(dat2$choice - Lag(thresholded))
diff = ifelse(is.na(diff), 0, diff)
thesum[j] = sum(diff, na.rm = T)
j = j + 1
}
}
minSum = min(thesum)
minErr[s] = min(thesum)
bestDeltas[s] = min(which(thesum==minSum)) * incx + minx
# bestDeltasGammas[s] - matrix(min(which(thsum==minSum))*inc + min)
betterThanChance[s] = sum(pbinom(0:minSum, length(dat2$choice), 0.5))
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota Psychology/Psychiatry
VA Medical Center
Office: S355 Elliot Hall - Twin Cities Campus
Phone: 612-626-0072 Email: patze...@umn.edu
Please consider the environment before printing this email
www.psych.umn.edu/research/tricam
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] If find x, then y, else do nothing
Thanks Josh for the code to post I have been trying to figure out how to do
that. Your code works except that it changes subjects that responded with
1 2 to all 1's. What does the ave argument mean in the execution of
the function?
library(car)
foo - function(x) {
if (any(grepl(4, x))) {
x - recode(x, 2 = 1; 4 = 2)
}
return(x)
}
## do it
dat$test - with(dat, *ave*(Slide1_RESP, Subject, FUN = foo))
On Fri, Aug 5, 2011 at 4:56 PM, Joshua Wiley jwiley.ps...@gmail.com wrote:
On Fri, Aug 5, 2011 at 2:34 PM, Edward Patzelt patze...@umn.edu wrote:
The problem is that we were using a task where some subjects responded
with 1 2 and some responded with 2 4. So there is overlap for 2
because it means stimulus 1 for subject 1 and it means stimulus 2 for
subject 2.
subject
subject_1
1
subject_1
1
subject_1
1
subject_1
2
subject_1
2
subject_2
4
subject_2
2
subject_2
2
subject_2
4
subject_2
2
subject_2
2
subject_2
2
subject_2
4
FYI providing data in the format above (this list is plain text) makes
the job of those trying to help substantially harder. You can use
dput(). For example, if I wanted to share the first 10 rows of the
built in mtcars data set, I would just copy and paste the output from
running:
dput(mtcars[1:10, ])
Anyway, here you go, this should be directly executable as long as you
have installed the 'car' package.
## your data in a form easily copied and pasted into the console
## created using dput() (highly recommended for future posts)
dat - structure(list(subject = c(subject_1, subject_1, subject_1,
subject_1, subject_1, subject_2, subject_2, subject_2,
subject_2, subject_2, subject_2, subject_2, subject_2
), val = c(1, 1, 1, 2, 2, 4, 2, 2, 4, 2, 2, 2, 4)), .Names = c(subject,
val), row.names = c(NA, -13L), class = data.frame)
## load the required package for recoding
## though it is overkill for only two levels
require(car)
## define a function to do the recoding
foo - function(x) {
if (any(grepl(4, x))) {
x - recode(x, 2 = 1; 4 = 2)
}
return(x)
}
## do it
dat$altval - with(dat, ave(val, subject, FUN = foo))
Cheers,
Josh
On Fri, Aug 5, 2011 at 4:25 PM, Joshua Wiley jwiley.ps...@gmail.com
wrote:
Hi Edward,
You can try something like:
u.ppl - unique(init.dat1$grid)
l.ppl - ifelse(grepl(4, init.dat1$Slide1_RESP), 2,
init.dat1$Slide1_RESP)
Note that this is not exact as you have not provided a reproducible
example. I am not exactly sure how you are putting 1 for 2 and 2 for
4, if the value is equal to 4, but presumably it is clearer with data.
In any event, look at ?ifelse it is something like a vectorized if
statement and is, I believe, preferable to your use of a for loop. I
can probably give you a runnable solution if you can give the first
few rows of the relevant data.
Cheers,
Josh
On Fri, Aug 5, 2011 at 2:15 PM, Edward Patzelt patze...@umn.edu
wrote:
I want to write code that says If you find an element equal to 4 in
this
vector for each person in the data set tested separately, then put in
1 for
2 and 2 for 4, else leave the variable as is
u.ppl - (unique(init.dat1$grid))
l.ppl - length(u.ppl)
for (i in 1:l.ppl)
{
if (grep(4,init.dat1$Slide1_RESP)) {2 == 1, 4 == 2}; else
init.dat1$Slide1_RESP
}
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota Psychology/Psychiatry
VA Medical Center
Office: S355 Elliot Hall - Twin Cities Campus
Phone: 612-626-0072 Email: patze...@umn.edu
Please consider the environment before printing this email
www.psych.umn.edu/research/tricam
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Ph.D. Student, Health Psychology
Programmer Analyst II, ATS Statistical Consulting Group
University of California, Los Angeles
https://joshuawiley.com/
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota Psychology/Psychiatry
VA Medical Center
Office: S355 Elliot Hall - Twin Cities Campus
Phone: 612-626-0072 Email: patze...@umn.edu
Please consider the environment before printing this email
www.psych.umn.edu/research/tricam
--
Joshua Wiley
Ph.D. Student, Health Psychology
Programmer Analyst II, ATS Statistical Consulting Group
University of California, Los Angeles
https://joshuawiley.com/
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota Psychology/Psychiatry
VA Medical Center
Office: S355
Re: [R] If find x, then y, else do nothing
Here's the code. I don't want it to even touch the vector if there are
already 1's 2's
structure(list(subject = c(8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L,
8L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), resp = c(2, 1, 1, 2, 1, 2, 1,
1, 1, 1, 4, 4, 2, 2, 4, 4, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2,
1, 2)), .Names = c(subject, resp), row.names = c(1L, 2L,
3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 4161L, 4162L, 4163L, 4164L,
4165L, 4166L, 4167L, 4168L, 4169L, 4170L, 166L, 167L, 168L, 169L,
170L, 171L, 172L, 173L, 174L, 175L), class = data.frame)
R version 2.12.2 (2011-02-25)
Platform: x86_64-pc-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United
States.1252LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C LC_TIME=English_United
States.1252
attached base packages:
[1] splines stats graphics grDevices utils datasets
methods base
other attached packages:
[1] car_2.0-10 nnet_7.3-1 MASS_7.3-11 Hmisc_3.8-3
survival_2.36-5 RODBC_1.3-2
loaded via a namespace (and not attached):
[1] cluster_1.13.3 grid_2.12.2 lattice_0.19-17 tools_2.12.2
On Mon, Aug 8, 2011 at 10:43 AM, Joshua Wiley jwiley.ps...@gmail.comwrote:
On Mon, Aug 8, 2011 at 8:23 AM, Edward Patzelt patze...@umn.edu wrote:
Thanks Josh for the code to post I have been trying to figure out how to
do
that. Your code works except that it changes subjects that responded
with
1 2 to all 1's. What does the ave argument mean in the execution
of
Not in the example data you provided on my system. If you can provide
data (preferablly using dput() or uploading a txt file on a file
hosting service) that reproduces this issue, I will be happy to look
at it for you. You might also try reporting your sessionInfo() ---
this may be related to the version of R or the packages you are using,
but at present I have no information.
the function?
'ave' is not an argument in the function. ave() *is* a function. I
call with() to have the ave() function evaluated in an environment
created from dat (the data). See ?ave and ?with I could have
equivalently (though more cumbersomely) written:
ave(dat$Slide1_RESP, dat$Subject, FUN = foo))
because ave will now be evaluated in the global environment, it will
not have access to the variables stored in 'dat' unless explicitly
told that they are in dat (as above).
Cheers,
Josh
library(car)
foo - function(x) {
if (any(grepl(4, x))) {
x - recode(x, 2 = 1; 4 = 2)
}
return(x)
}
## do it
dat$test - with(dat, ave(Slide1_RESP, Subject, FUN = foo))
On Fri, Aug 5, 2011 at 4:56 PM, Joshua Wiley jwiley.ps...@gmail.com
wrote:
On Fri, Aug 5, 2011 at 2:34 PM, Edward Patzelt patze...@umn.edu
wrote:
The problem is that we were using a task where some subjects responded
with 1 2 and some responded with 2 4. So there is overlap for
2
because it means stimulus 1 for subject 1 and it means stimulus 2 for
subject 2.
subject
subject_1
1
subject_1
1
subject_1
1
subject_1
2
subject_1
2
subject_2
4
subject_2
2
subject_2
2
subject_2
4
subject_2
2
subject_2
2
subject_2
2
subject_2
4
FYI providing data in the format above (this list is plain text) makes
the job of those trying to help substantially harder. You can use
dput(). For example, if I wanted to share the first 10 rows of the
built in mtcars data set, I would just copy and paste the output from
running:
dput(mtcars[1:10, ])
Anyway, here you go, this should be directly executable as long as you
have installed the 'car' package.
## your data in a form easily copied and pasted into the console
## created using dput() (highly recommended for future posts)
dat - structure(list(subject = c(subject_1, subject_1, subject_1,
subject_1, subject_1, subject_2, subject_2, subject_2,
subject_2, subject_2, subject_2, subject_2, subject_2
), val = c(1, 1, 1, 2, 2, 4, 2, 2, 4, 2, 2, 2, 4)), .Names =
c(subject,
val), row.names = c(NA, -13L), class = data.frame)
## load the required package for recoding
## though it is overkill for only two levels
require(car)
## define a function to do the recoding
foo - function(x) {
if (any(grepl(4, x))) {
x - recode(x, 2 = 1; 4 = 2)
}
return(x)
}
## do it
dat$altval - with(dat, ave(val, subject, FUN = foo))
Cheers,
Josh
On Fri, Aug 5, 2011 at 4:25 PM, Joshua Wiley jwiley.ps...@gmail.com
wrote:
Hi Edward,
You can try something like:
u.ppl - unique(init.dat1$grid)
l.ppl - ifelse(grepl(4, init.dat1$Slide1_RESP), 2,
init.dat1$Slide1_RESP)
Note that this is not exact as you have not provided a reproducible
example. I am not exactly sure how you
Re: [R] If find x, then y, else do nothing
H, I pulled out a portion of the data set to create the code for
posting. When I execute this on the data frame I get the following for
subject 8 which is clearly incorrect. I get this for all subjects who
originally had 1 2.
dat$respalt - with(dat, ave(Slide1_RESP, factor(grid), FUN = foo))
head(dat$respalt,20+ ) [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
On Mon, Aug 8, 2011 at 11:14 AM, Joshua Wiley jwiley.ps...@gmail.comwrote:
Hmm, well I suppose it technically does touch in some sense still if
there are 1 2s, but it should just return it as is, not changed.
Thanks for the data, very easy! Here is what I get:
dat - structure(list(subject = c(8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L,
8L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), resp = c(2, 1, 1, 2, 1, 2, 1,
1, 1, 1, 4, 4, 2, 2, 4, 4, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2,
1, 2)), .Names = c(subject, resp), row.names = c(1L, 2L,
3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 4161L, 4162L, 4163L, 4164L,
4165L, 4166L, 4167L, 4168L, 4169L, 4170L, 166L, 167L, 168L, 169L,
170L, 171L, 172L, 173L, 174L, 175L), class = data.frame)
require(car)
foo - function(x) {
if (any(grepl(4, x))) {
x - recode(x, 2 = 1; 4 = 2)
}
return(x)
}
dat$respalt - with(dat, ave(resp, factor(subject), FUN = foo))
## which at least for me gives:
structure(list(subject = c(8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L,
8L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), resp = c(2, 1, 1, 2, 1, 2, 1,
1, 1, 1, 4, 4, 2, 2, 4, 4, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2,
1, 2), respalt = c(2, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1,
2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 1, 2)), .Names = c(subject,
resp, respalt), row.names = c(1L, 2L, 3L, 4L, 5L, 6L, 7L,
8L, 9L, 10L, 4161L, 4162L, 4163L, 4164L, 4165L, 4166L, 4167L,
4168L, 4169L, 4170L, 166L, 167L, 168L, 169L, 170L, 171L, 172L,
173L, 174L, 175L), class = data.frame)
Does that work for you? I am running:
R Under development (unstable) (2011-07-30 r56564)
Platform: x86_64-pc-mingw32/x64 (64-bit)
with car_2.0-10
HTH,
Josh
On Mon, Aug 8, 2011 at 9:02 AM, Edward Patzelt patze...@umn.edu wrote:
Here's the code. I don't want it to even touch the vector if there are
already 1's 2's
structure(list(subject = c(8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L,
8L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), resp = c(2, 1, 1, 2, 1, 2, 1,
1, 1, 1, 4, 4, 2, 2, 4, 4, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2,
1, 2)), .Names = c(subject, resp), row.names = c(1L, 2L,
3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 4161L, 4162L, 4163L, 4164L,
4165L, 4166L, 4167L, 4168L, 4169L, 4170L, 166L, 167L, 168L, 169L,
170L, 171L, 172L, 173L, 174L, 175L), class = data.frame)
R version 2.12.2 (2011-02-25)
Platform: x86_64-pc-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United
States.1252LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C LC_TIME=English_United
States.1252
attached base packages:
[1] splines stats graphics grDevices utils datasets methods
base
other attached packages:
[1] car_2.0-10 nnet_7.3-1 MASS_7.3-11 Hmisc_3.8-3
survival_2.36-5 RODBC_1.3-2
loaded via a namespace (and not attached):
[1] cluster_1.13.3 grid_2.12.2 lattice_0.19-17 tools_2.12.2
On Mon, Aug 8, 2011 at 10:43 AM, Joshua Wiley jwiley.ps...@gmail.com
wrote:
On Mon, Aug 8, 2011 at 8:23 AM, Edward Patzelt patze...@umn.edu
wrote:
Thanks Josh for the code to post I have been trying to figure out how
to
do
that. Your code works except that it changes subjects that responded
with
1 2 to all 1's. What does the ave argument mean in the
execution
of
Not in the example data you provided on my system. If you can provide
data (preferablly using dput() or uploading a txt file on a file
hosting service) that reproduces this issue, I will be happy to look
at it for you. You might also try reporting your sessionInfo() ---
this may be related to the version of R or the packages you are using,
but at present I have no information.
the function?
'ave' is not an argument in the function. ave() *is* a function. I
call with() to have the ave() function evaluated in an environment
created from dat (the data). See ?ave and ?with I could have
equivalently (though more cumbersomely) written:
ave(dat$Slide1_RESP, dat$Subject, FUN = foo))
because ave will now be evaluated in the global environment, it will
not have access to the variables stored in 'dat' unless explicitly
told that they are in dat (as above).
Cheers,
Josh
library(car)
foo - function(x) {
if (any(grepl(4, x))) {
x - recode(x, 2 = 1; 4 = 2)
}
return(x)
}
## do it
dat$test - with(dat, ave(Slide1_RESP, Subject, FUN = foo))
On Fri
Re: [R] If find x, then y, else do nothing
Here's a thought, when I execute it on a data frame with only the subject
and resp variables I have the same problem. But not on the subset of data
we've been working with. The difference between the full vector and the
subset vector is that the full vector contains NA's. Only 35 out of 17,000
trials, but could this screw up the function?
unique(blah$resp)
[1] 2 1 NA 4
On Mon, Aug 8, 2011 at 11:46 AM, Joshua Wiley jwiley.ps...@gmail.comwrote:
These are shots in the dark, but you could try: running it in a clean
R session with only the necessary packages and data loaded; testing on
subsets of your data; examing the classes of all your variables and
make sure they are what is expexted; upgrading your version of R.
I just retested this on:
R version 2.12.1 (2010-12-16)
Platform: x86_64-pc-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
attached base packages:
[1] grid splines stats graphics grDevices utils datasets
[8] methods base
other attached packages:
[1] car_2.0-9 survival_2.36-5 nnet_7.3-1 MASS_7.3-11
again without issue, so I am thinking there must be something
particular perhaps to your session going on, but it is difficult to
say.
On Mon, Aug 8, 2011 at 9:35 AM, Edward Patzelt patze...@umn.edu wrote:
H, I pulled out a portion of the data set to create the code for
posting. When I execute this on the data frame I get the following for
subject 8 which is clearly incorrect. I get this for all subjects who
originally had 1 2.
dat$respalt - with(dat, ave(Slide1_RESP, factor(grid), FUN = foo))
head(dat$respalt,20
+ )
[1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
On Mon, Aug 8, 2011 at 11:14 AM, Joshua Wiley jwiley.ps...@gmail.com
wrote:
Hmm, well I suppose it technically does touch in some sense still if
there are 1 2s, but it should just return it as is, not changed.
Thanks for the data, very easy! Here is what I get:
dat - structure(list(subject = c(8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L,
8L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), resp = c(2, 1, 1, 2, 1, 2, 1,
1, 1, 1, 4, 4, 2, 2, 4, 4, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2,
1, 2)), .Names = c(subject, resp), row.names = c(1L, 2L,
3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 4161L, 4162L, 4163L, 4164L,
4165L, 4166L, 4167L, 4168L, 4169L, 4170L, 166L, 167L, 168L, 169L,
170L, 171L, 172L, 173L, 174L, 175L), class = data.frame)
require(car)
foo - function(x) {
if (any(grepl(4, x))) {
x - recode(x, 2 = 1; 4 = 2)
}
return(x)
}
dat$respalt - with(dat, ave(resp, factor(subject), FUN = foo))
## which at least for me gives:
structure(list(subject = c(8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L,
8L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), resp = c(2, 1, 1, 2, 1, 2, 1,
1, 1, 1, 4, 4, 2, 2, 4, 4, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2,
1, 2), respalt = c(2, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1,
2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 1, 2)), .Names = c(subject,
resp, respalt), row.names = c(1L, 2L, 3L, 4L, 5L, 6L, 7L,
8L, 9L, 10L, 4161L, 4162L, 4163L, 4164L, 4165L, 4166L, 4167L,
4168L, 4169L, 4170L, 166L, 167L, 168L, 169L, 170L, 171L, 172L,
173L, 174L, 175L), class = data.frame)
Does that work for you? I am running:
R Under development (unstable) (2011-07-30 r56564)
Platform: x86_64-pc-mingw32/x64 (64-bit)
with car_2.0-10
HTH,
Josh
On Mon, Aug 8, 2011 at 9:02 AM, Edward Patzelt patze...@umn.edu
wrote:
Here's the code. I don't want it to even touch the vector if there
are
already 1's 2's
structure(list(subject = c(8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L,
8L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 42L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), resp = c(2, 1, 1, 2, 1, 2, 1,
1, 1, 1, 4, 4, 2, 2, 4, 4, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2,
1, 2)), .Names = c(subject, resp), row.names = c(1L, 2L,
3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 4161L, 4162L, 4163L, 4164L,
4165L, 4166L, 4167L, 4168L, 4169L, 4170L, 166L, 167L, 168L, 169L,
170L, 171L, 172L, 173L, 174L, 175L), class = data.frame)
R version 2.12.2 (2011-02-25)
Platform: x86_64-pc-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United
States.1252LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C LC_TIME=English_United
States.1252
attached base packages:
[1] splines stats graphics grDevices utils datasets
methods
base
other attached packages:
[1] car_2.0-10 nnet_7.3-1 MASS_7.3-11 Hmisc_3.8-3
survival_2.36-5 RODBC_1.3-2
loaded via a namespace (and not attached):
[1] cluster_1.13.3 grid_2.12.2
Re: [R] If find x, then y, else do nothing
I think the problem may be there are multiple subjects with the same subject number. I'll have to index by subject date. I put together a zip file with the original data set, the cleaned data set, and the code (with your code integrated). It's called clean.zip netfiles.umn.edu/users/patze003/www/ 'data.frame': 16275 obs. of 48 variables: $ Experiment : chr Rev-Fractals_2008_1 Rev-Fractals_2008_1 Rev-Fractals_2008_1 Rev-Fractals_2008_1 ... $ grid: int 8 8 8 8 8 8 8 8 8 8 ... $ Session : int 1 1 1 1 1 1 1 1 1 1 ... $ Age : int 22 22 22 22 22 22 22 22 22 22 ... $ Group : int 1 1 1 1 1 1 1 1 1 1 ... $ Handedness : chr left left left left ... $ Name: chr 2008_1 2008_1 2008_1 2008_1 ... $ SessionDate : chr 04-14-2008 04-14-2008 04-14-2008 04-14-2008 ... $ SessionTime : chr 11:07:31 11:07:31 11:07:31 11:07:31 ... $ Sex : chr female female female female ... $ Block : int 24 25 26 27 28 29 30 31 32 33 ... $ ABCorr : int 0 1 2 3 4 1 2 3 4 5 ... $ ABRev : int 1 1 1 1 1 1 1 1 1 1 ... $ acc : int 0 1 1 1 1 1 1 1 1 1 ... $ ansr1 : int 1 1 2 2 1 2 1 1 1 2 ... $ BeginInstruction: int NA NA NA NA NA NA NA NA NA NA ... $ BeginInstruction.Cycle : int NA NA NA NA NA NA NA NA NA NA ... $ BeginInstruction.Sample : int NA NA NA NA NA NA NA NA NA NA ... $ BlockInst : int NA NA NA NA NA NA NA NA NA NA ... $ BlockInst.Cycle : int NA NA NA NA NA NA NA NA NA NA ... $ BlockInst.Sample: int NA NA NA NA NA NA NA NA NA NA ... $ incorrect : int 2 2 1 1 2 1 2 2 2 1 ... $ LastCorr: int 0 0 0 0 0 4 4 4 4 4 ... $ LeftExample : chr ... $ LeftStim: chr f12.bmp f12.bmp f12.bmp f5.bmp ... $ Procedure : chr Trainb Trainb Trainb Trainb ... $ RepNum : int 1 1 1 1 1 1 1 1 1 1 ... $ ReqCorr : int 11 11 11 11 11 11 11 11 11 11 ... $ RightExample: chr ... $ RightStim : chr f5.bmp f5.bmp f5.bmp f12.bmp ... $ Running : chr TrainRule TrainRule TrainRule TrainRule ... $ Slide1_ACC : int 0 1 0 1 1 1 1 1 1 0 ... $ Slide1_CRESP: int 1 1 2 2 1 2 1 1 1 2 ... $ Slide1_RESP : num 2 1 1 2 1 2 1 1 1 1 ... $ Slide1_RT : int 1059 467 535 649 621 567 613 562 618 482 ... $ TrainInstructions : int NA NA NA NA NA NA NA NA NA NA ... $ TrainInstructions.Cycle : int NA NA NA NA NA NA NA NA NA NA ... $ TrainInstructions.Sample: int NA NA NA NA NA NA NA NA NA NA ... $ TrainListAB : int 1 1 2 1 1 1 1 1 1 2 ... $ TrainReversal : int NA NA NA NA NA NA NA NA NA NA ... $ TrainReversal.Cycle : int NA NA NA NA NA NA NA NA NA NA ... $ TrainReversal.Sample: int NA NA NA NA NA NA NA NA NA NA ... $ TrainRule : int 1 1 1 1 1 1 1 1 1 1 ... $ TrainRule.Cycle : int 31 32 33 34 35 36 37 38 39 40 ... $ TrainRule.Sample: int 21 22 23 24 25 26 27 28 29 30 ... $ TrainTestBlock : int 1 1 1 1 1 1 1 1 1 1 ... $ TrialType : chr AB AB AB BA ... $ Valid : int 1 1 0 1 1 1 1 1 1 0 ... On Mon, Aug 8, 2011 at 12:35 PM, Edward Patzelt patze...@umn.edu wrote: Here's a thought, when I execute it on a data frame with only the subject and resp variables I have the same problem. But not on the subset of data we've been working with. The difference between the full vector and the subset vector is that the full vector contains NA's. Only 35 out of 17,000 trials, but could this screw up the function? unique(blah$resp) [1] 2 1 NA 4 On Mon, Aug 8, 2011 at 11:46 AM, Joshua Wiley jwiley.ps...@gmail.comwrote: These are shots in the dark, but you could try: running it in a clean R session with only the necessary packages and data loaded; testing on subsets of your data; examing the classes of all your variables and make sure they are what is expexted; upgrading your version of R. I just retested this on: R version 2.12.1 (2010-12-16) Platform: x86_64-pc-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=English_United States.1252 [2] LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] grid splines stats graphics grDevices utils datasets [8] methods base other attached packages: [1] car_2.0-9 survival_2.36-5 nnet_7.3-1 MASS_7.3-11 again without issue, so I am thinking there must be something particular perhaps to your session going on, but it is difficult to say. On Mon, Aug 8, 2011 at 9:35 AM, Edward Patzelt patze...@umn.edu wrote: H, I
[R] If find x, then y, else do nothing
I want to write code that says If you find an element equal to 4 in this
vector for each person in the data set tested separately, then put in 1 for
2 and 2 for 4, else leave the variable as is
u.ppl - (unique(init.dat1$grid))
l.ppl - length(u.ppl)
for (i in 1:l.ppl)
{
if (grep(4,init.dat1$Slide1_RESP)) {2 == 1, 4 == 2}; else
init.dat1$Slide1_RESP
}
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota Psychology/Psychiatry
VA Medical Center
Office: S355 Elliot Hall - Twin Cities Campus
Phone: 612-626-0072 Email: patze...@umn.edu
Please consider the environment before printing this email
www.psych.umn.edu/research/tricam
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] If find x, then y, else do nothing
The problem is that we were using a task where some subjects responded with
1 2 and some responded with 2 4. So there is overlap for 2 because
it means stimulus 1 for subject 1 and it means stimulus 2 for subject 2.
subject
subject_1
1
subject_1
1
subject_1
1
subject_1
2
subject_1
2
subject_2
4
subject_2
2
subject_2
2
subject_2
4
subject_2
2
subject_2
2
subject_2
2
subject_2
4
On Fri, Aug 5, 2011 at 4:25 PM, Joshua Wiley jwiley.ps...@gmail.com wrote:
Hi Edward,
You can try something like:
u.ppl - unique(init.dat1$grid)
l.ppl - ifelse(grepl(4, init.dat1$Slide1_RESP), 2,
init.dat1$Slide1_RESP)
Note that this is not exact as you have not provided a reproducible
example. I am not exactly sure how you are putting 1 for 2 and 2 for
4, if the value is equal to 4, but presumably it is clearer with data.
In any event, look at ?ifelse it is something like a vectorized if
statement and is, I believe, preferable to your use of a for loop. I
can probably give you a runnable solution if you can give the first
few rows of the relevant data.
Cheers,
Josh
On Fri, Aug 5, 2011 at 2:15 PM, Edward Patzelt patze...@umn.edu wrote:
I want to write code that says If you find an element equal to 4 in this
vector for each person in the data set tested separately, then put in 1
for
2 and 2 for 4, else leave the variable as is
u.ppl - (unique(init.dat1$grid))
l.ppl - length(u.ppl)
for (i in 1:l.ppl)
{
if (grep(4,init.dat1$Slide1_RESP)) {2 == 1, 4 == 2}; else
init.dat1$Slide1_RESP
}
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota Psychology/Psychiatry
VA Medical Center
Office: S355 Elliot Hall - Twin Cities Campus
Phone: 612-626-0072 Email: patze...@umn.edu
Please consider the environment before printing this email
www.psych.umn.edu/research/tricam
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Ph.D. Student, Health Psychology
Programmer Analyst II, ATS Statistical Consulting Group
University of California, Los Angeles
https://joshuawiley.com/
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota Psychology/Psychiatry
VA Medical Center
Office: S355 Elliot Hall - Twin Cities Campus
Phone: 612-626-0072 Email: patze...@umn.edu
Please consider the environment before printing this email
www.psych.umn.edu/research/tricam
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Hidden Markov Models in R
R Community - I am attempting to fit a model as described in Hampton, Bossaerts, and O'doherty (J. Neuroscience) 2006. They use a bayesian hidden markov model to model the Reversal Learning data. I have tried using HMM and depmixS4 with no success. My data is a Reversal Learning Task in which there are 3 sets of patterns over 3 blocks. The participant receives incorrect or correct feedback. 20% of the time they receive false feedback (they are told incorrect when they were in fact correct). Once the person achieves the criterion of 9/10 correct responses the contigencies reverse. I am confused on how to set up my states, symbols, starting probabilities, transition probabilities, and emission probabilities in R. This is what I have so far. hmm - initHMM(c(stay, switch), c(correct, incorrect), c(.5, .5), matrix(c(.9, .1, .1, .9),2), matrix(c(.2, .8, .8, .2), 2)) dat$test - ifelse(dat$Slide1_ACC == 0, incorrect, correct) viterbi(hmm, dat$test) The sequence of observations I run through the model is the feedback the participant receives. Any help would be greatly appreciated. I think what I want to do is run the model on each participant to generate the most probable path and then compare that to the actual path to see if they match up. Best, -- Edward H. Patzelt Research Assistant TRiCAM Lab University of Minnesota Psychology/Psychiatry VA Medical Center Office: S355 Elliot Hall - Twin Cities Campus Phone: 612-626-0072 Email: patze...@umn.edu Please consider the environment before printing this email www.psych.umn.edu/research/tricam [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Loop through each subject
R help -
I am attempting to write a script that has multiple subjects in 1 data file.
Each subject has multiple rows with columns as variables. Here is my code,
I am having problem executing it on each unique subject id (dat$Subject).
getwd()
setwd(/Users/edwardpatzelt/Desktop/Neuroimaging/MERIT/SRRT/merge)
dat - read.table(test2.txt, header = TRUE, na.strings = NA,
stringsAsFactors = FALSE, sep = \t)
for(i in 1:length(dat))
{
for (i in 1:)dat[(unique(dat$Subject)),)]
{
colg - dat[grep(Green, dat$CueProbe),]
colg - data.frame(colg$SRRTCue.OnsetTime/1000, (colg$SRRTFix2.OnsetTime-
colg$SRRTCue.OnsetTime)/1000, (ifelse((colg$SRRTProbe.ACC == 1 | colg$Probe==
+), 1, 0)))
colr - dat[grep(Red, dat$CueProbe),]
colr - data.frame(colr$SRRTCue.OnsetTime/1000, (colr$SRRTFix2.OnsetTime-
colr$SRRTCue.OnsetTime)/1000, (ifelse((colr$SRRTProbe.ACC == 1 | colr$Probe==
+), 1, 0)))
write.table(colg, file = paste(dat$Subject[[1]], sep = \t, append =
green.txt), col.names = FALSE, row.names = FALSE)
write.table(colr, file = paste(dat$Subject[[1]], sep = \t, append =
red.txt), col.names = FALSE, row.names = FALSE)
}
}
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota Psychology/Psychiatry
VA Medical Center
Office: S355 Elliot Hall - Twin Cities Campus
Phone: 612-626-0072 Email: patze...@umn.edu
Please consider the environment before printing this email
www.psych.umn.edu/research/tricam
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Loop through each subject
Huh, not sure I understand your feedback. There is only 1 file that is 1
large dataframe. I want to execute the commands on each subject in the
dataframe and ouput their respective files with the subject # and file type
appended (10_green.txt).
How would I do this loop index?
- Show quoted text -
On Tue, Jun 28, 2011 at 1:27 PM, David Winsemius dwinsem...@comcast.netwrote:
On Jun 28, 2011, at 1:44 PM, Edward Patzelt wrote:
R help -
I am attempting to write a script that has multiple subjects in 1 data
file.
Each subject has multiple rows with columns as variables. Here is my
code,
I am having problem executing it on each unique subject id (dat$Subject).
One problem that I see is that you are calling all of your files the same
thing (i.e. overwriting earlier results. Why aren't you using the loop index
in the naming process?
(And aeppnd is a logical argument in write.table.)
?write.table
--
David.
getwd()
setwd(/Users/edwardpatzelt/**Desktop/Neuroimaging/MERIT/**SRRT/merge)
dat - read.table(test2.txt, header = TRUE, na.strings = NA,
stringsAsFactors = FALSE, sep = \t)
for(i in 1:length(dat))
{
for (i in 1:)dat[(unique(dat$Subject)),)**]
{
colg - dat[grep(Green, dat$CueProbe),]
colg - data.frame(colg$SRRTCue.**OnsetTime/1000,
(colg$SRRTFix2.OnsetTime-
colg$SRRTCue.OnsetTime)/1000, (ifelse((colg$SRRTProbe.ACC == 1 |
colg$Probe==
+), 1, 0)))
colr - dat[grep(Red, dat$CueProbe),]
colr - data.frame(colr$SRRTCue.**OnsetTime/1000,
(colr$SRRTFix2.OnsetTime-
colr$SRRTCue.OnsetTime)/1000, (ifelse((colr$SRRTProbe.ACC == 1 |
colr$Probe==
+), 1, 0)))
write.table(colg, file = paste(dat$Subject[[1]], sep = \t, append =
green.txt), col.names = FALSE, row.names = FALSE)
write.table(colr, file = paste(dat$Subject[[1]], sep = \t, append =
red.txt), col.names = FALSE, row.names = FALSE)
}
}
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota Psychology/Psychiatry
VA Medical Center
Office: S355 Elliot Hall - Twin Cities Campus
Phone: 612-626-0072 Email: patze...@umn.edu
Please consider the environment before printing this email
www.psych.umn.edu/research/**tricamhttp://www.psych.umn.edu/research/tricam
[[alternative HTML version deleted]]
__**
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/**
posting-guide.html http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
--
Edward H. Patzelt
Research Assistant TRiCAM Lab
University of Minnesota Psychology/Psychiatry
VA Medical Center
Office: S355 Elliot Hall - Twin Cities Campus
Phone: 612-626-0072 Email: patze...@umn.edu
Please consider the environment before printing this email
www.psych.umn.edu/research/tricam
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] nlme Output
Everyone - What do the NaN's mean here? Is this analysis a problem? Linear mixed-effects model fit by maximum likelihood Data: tmp.dat AIC BIClogLik 1611.251 1638.363 -797.6253 Random effects: Formula: ~1 | group_id (Intercept) Residual StdDev: 0.0003077668 9.236715 Fixed effects: AvgTrials ~ time + factor(group_id) + time * factor(group_id) Value Std.Error DF t-value p-value (Intercept)18.159722 3.576664 213 5.077279 0. time4.192708 1.655674 213 2.532327 0.0121 factor(group_id)2 -6.929563 5.235700 0 -1.323522 NaN factor(group_id)3 -1.654554 4.189575 0 -0.394922 NaN time:factor(group_id)2 1.729911 2.423658 213 0.713760 0.4762 time:factor(group_id)3 -2.555111 1.939396 213 -1.317478 0.1891 Correlation: (Intr) time fc(_)2 fc(_)3 t:(_)2 time -0.926 factor(group_id)2 -0.683 0.632 factor(group_id)3 -0.854 0.790 0.583 time:factor(group_id)2 0.632 -0.683 -0.926 -0.540 time:factor(group_id)3 0.790 -0.854 -0.540 -0.926 0.583 Standardized Within-Group Residuals: Min Q1Med Q3Max -1.8842754 -0.6979785 -0.3370998 0.5666704 3.0943948 Number of Observations: 219 Number of Groups: 3 Warning message: In pt(q, df, lower.tail, log.p) : NaNs produced [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] nlme Output
The design is a repeated measures with 3 instances. There are 3 groups: Controls, Heavy Cocaine Users, Light Cocaine Users. I reshaped the data so that there was one variable for the 3 instances called AvgTrials. Time is the indicator of each instance. Here is the model call: mod5 - lme(AvgTrials ~ time + factor(group_id) + time*factor(group_id), random = ~ 1 | group_id, data = tmp.dat, method = ML, na.action = na.omit) What else would you need here? -Edward On Mon, Sep 6, 2010 at 8:52 AM, ONKELINX, Thierry thierry.onkel...@inbo.bewrote: Dear Edward, You have no degrees of freedom left to estimate those p-values. Your design does not allows for the model your implemented. We need a brief summary of your design in order to help you further. HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek team Biometrie Kwaliteitszorg Gaverstraat 4 9500 Geraardsbergen Belgium Research Institute for Nature and Forest team Biometrics Quality Assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens Edward Patzelt Verzonden: maandag 6 september 2010 15:43 Aan: r-help@r-project.org Onderwerp: [R] nlme Output Everyone - What do the NaN's mean here? Is this analysis a problem? Linear mixed-effects model fit by maximum likelihood Data: tmp.dat AIC BIClogLik 1611.251 1638.363 -797.6253 Random effects: Formula: ~1 | group_id (Intercept) Residual StdDev: 0.0003077668 9.236715 Fixed effects: AvgTrials ~ time + factor(group_id) + time * factor(group_id) Value Std.Error DF t-value p-value (Intercept)18.159722 3.576664 213 5.077279 0. time4.192708 1.655674 213 2.532327 0.0121 factor(group_id)2 -6.929563 5.235700 0 -1.323522 NaN factor(group_id)3 -1.654554 4.189575 0 -0.394922 NaN time:factor(group_id)2 1.729911 2.423658 213 0.713760 0.4762 time:factor(group_id)3 -2.555111 1.939396 213 -1.317478 0.1891 Correlation: (Intr) time fc(_)2 fc(_)3 t:(_)2 time -0.926 factor(group_id)2 -0.683 0.632 factor(group_id)3 -0.854 0.790 0.583 time:factor(group_id)2 0.632 -0.683 -0.926 -0.540 time:factor(group_id)3 0.790 -0.854 -0.540 -0.926 0.583 Standardized Within-Group Residuals: Min Q1Med Q3Max -1.8842754 -0.6979785 -0.3370998 0.5666704 3.0943948 Number of Observations: 219 Number of Groups: 3 Warning message: In pt(q, df, lower.tail, log.p) : NaNs produced [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Druk dit bericht a.u.b. niet onnodig af. Please do not print this message unnecessarily. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. -- Edward H. Patzelt Research Assistant TRiCAM Lab University of Minnesota Psychology/Psychiatry VA Medical Center Office: N437 Elliot Hall - Twin Cities Campus Phone: 612-624-3892 Email: patze...@umn.edu Please consider the environment before printing this email www.psych.umn.edu/research/tricam [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
