I have read a lot about the pain of replicate the easy robust option from STATA to R to use robust standard errors. I replicated following approaches: StackExchange and Economic Theory Blog. They work but the problem I face is, if I want to print my results using the stargazer function (this prints the .tex code for Latex files).
Here is the illustration to my problem:
reg1 <-lm(rev~id + source + listed + country , data=data2_rev)
stargazer(reg1)
This prints the R output as .tex code (non-robust SE) If i want to use robust SE, i can do it with the sandwich package as follow:
vcov <- vcovHC(reg1, "HC1")
if I now use stargazer(vcov) only the output of the vcovHC function is printed and not the regression output itself.
With the package lmtest() it is possible to print at least the estimator, but not the observations, R2, adj. R2, Residual, Residual St.Error and the F-Statistics.
lmtest::coeftest(reg1, vcov. = sandwich::vcovHC(reg1, type = 'HC1'))
This gives the following output:
t test of coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -2.54923 6.85521 -0.3719 0.710611
id 0.39634 0.12376 3.2026 0.001722 **
source 1.48164 4.20183 0.3526 0.724960
country -4.00398 4.00256 -1.0004 0.319041
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
How can I add or get an output with the following parameters as well?
Residual standard error: 17.43 on 127 degrees of freedom
Multiple R-squared: 0.09676, Adjusted R-squared: 0.07543
F-statistic: 4.535 on 3 and 127 DF, p-value: 0.00469
Did anybody face the same problem and can help me out?
How can I use robust standard errors in the lm function and apply the stargazer function?
You already calculated robust standard errors, and there's an easy way to include it in the stargazeroutput:
library("sandwich")
library("plm")
library("stargazer")
data("Produc", package = "plm")
# Regression
model <- plm(log(gsp) ~ log(pcap) + log(pc) + log(emp) + unemp,
data = Produc,
index = c("state","year"),
method="pooling")
# Adjust standard errors
cov1 <- vcovHC(model, type = "HC1")
robust_se <- sqrt(diag(cov1))
# Stargazer output (with and without RSE)
stargazer(model, model, type = "text",
se = list(NULL, robust_se))
Solution found here: https://www.jakeruss.com/cheatsheets/stargazer/#robust-standard-errors-replicating-statas-robust-option
Update I'm not so much into F-Tests. People are discussing those issues, e.g. https://stats.stackexchange.com/questions/93787/f-test-formula-under-robust-standard-error
When you follow http://www3.grips.ac.jp/~yamanota/Lecture_Note_9_Heteroskedasticity
"A heteroskedasticity-robust t statistic can be obtained by dividing an OSL estimator by its robust standard error (for zero null hypotheses). The usual F-statistic, however, is invalid. Instead, we need to use the heteroskedasticity-robust Wald statistic."
and use a Wald statistic here?
This is a fairly simple solution using coeftest:
reg1 <-lm(rev~id + source + listed + country , data=data2_rev)
cl_robust <- coeftest(reg1, vcov = vcovCL, type = "HC1", cluster = ~
country)
se_robust <- cl_robust[, 2]
stargazer(reg1, reg1, cl_robust, se = list(NULL, se_robust, NULL))
Note that I only included cl_robust in the output as a verification that the results are identical.
Related
I'm trying to replicate values from pine script cci() function in golang. I've found this lib https://github.com/markcheno/go-talib/blob/master/talib.go#L1821
but it gives totally different values than cci function does
pseudo code how do I use the lib
cci := talib.Cci(latest14CandlesHighArray, latest14CandlesLowArray, latest14CandlesCloseArray, 14)
The lib gives me the following data
Timestamp: 2021-05-22 18:59:27.675, Symbol: BTCUSDT, Interval: 5m, Open: 38193.78000000, Close: 38122.16000000, High: 38283.55000000, Low: 38067.92000000, StartTime: 2021-05-22 18:55:00.000, EndTime: 2021-05-22 18:59:59.999, Sma: 38091.41020000, Cci0: -16.63898084, Cci1: -53.92565811,
While current cci values on TradingView are: cci0 - -136, cci1 - -49
could anyone guide what do I miss?
Thank you
P.S. cci0 - current candle cci, cci1 - previous candle cci
PineScript has really great reference when looking for functions, usually even supplying the pine code to recreate it.
https://www.tradingview.com/pine-script-reference/v4/#fun_cci
The code wasn't provided for cci, but a step-by-step explanation was.
Here is how I managed to recreate the cci function using Pine, following the steps in the reference:
// This source code is subject to the terms of the Mozilla Public License 2.0 at https://mozilla.org/MPL/2.0/
// © bajaco
//#version=4
study("CCI Breakdown", overlay=false, precision=16)
cci_breakdown(src, p) =>
// The CCI (commodity channel index) is calculated as the
// 1. difference between the typical price of a commodity and its simple moving average,
// divided by the
// 2. mean absolute deviation of the typical price.
// 3. The index is scaled by an inverse factor of 0.015
// to provide more readable numbers
// 1. diff
ma = sma(src,p)
diff = src - ma
// 2. mad
s = 0.0
for i = 0 to p - 1
s := s + abs(src[i] - ma)
mad = s / p
// 3. Scaling
mcci = diff/mad / 0.015
mcci
plot(cci(close, 100))
plot(cci_breakdown(close,100))
I didn't know what mean absolute deviation meant, but at least in their implementation it appears to be taking the difference from the mean for each value in the range, but NOT changing the mean value as you go back.
I don't know Go but that's the logic.
I'm trying to learn more about pyhf and my understanding of what the goals are might be limited. I would love to fit my HEP data outside of ROOT, but I could be imposing expectations on pyhf which are not what the authors intended for it's use.
I'd like to write myself a hello-world example, but I might just not know what I'm doing. My misunderstanding could also be gaps in my statistical knowledge.
With that preface, let me explain what I'm trying to explore.
I have some observed set of events for which I calculate some observable and make a binned histogram of that data. I hypothesize that there are two contributing physics processes, which I call signal and background. I generate some Monte Carlo samples for these processes and the theorized total number of events is close to, but not exactly what I observe.
I would like to:
Fit the data to this two process hypothesis
Get from the fit the optimal values for the number of events for each process
Get the uncertainties on these fitted values
If appropriate, calculate an upper limit on the number of signal events.
My starter code is below, where all I'm doing is an ML fit but I'm not sure where to go. I know it's not set up to do what I want, but I'm getting lost in the examples I find on RTD. I'm sure it's me, this is not a criticism of the documentation.
import pyhf
import numpy as np
import matplotlib.pyplot as plt
nbins = 15
# Generate a background and signal MC sample`
MC_signal_events = np.random.normal(5,1.0,200)
MC_background_events = 10*np.random.random(1000)
signal_data = np.histogram(MC_signal_events,bins=nbins)[0]
bkg_data = np.histogram(MC_background_events,bins=nbins)[0]
# Generate an observed dataset with a slightly different
# number of events
signal_events = np.random.normal(5,1.0,180)
background_events = 10*np.random.random(1050)
observed_events = np.array(signal_events.tolist() + background_events.tolist())
observed_sample = np.histogram(observed_events,bins=nbins)[0]
# Plot these samples, if you like
plt.figure(figsize=(12,4))
plt.subplot(1,3,1)
plt.hist(observed_events,bins=nbins,label='Observations')
plt.legend()
plt.subplot(1,3,2)
plt.hist(MC_signal_events,bins=nbins,label='MC signal')
plt.legend()
plt.subplot(1,3,3)
plt.hist(MC_background_events,bins=nbins,label='MC background')
plt.legend()
# Use a very naive estimate of the background
# uncertainties
bkg_uncerts = np.sqrt(bkg_data)
print("Defining the PDF.......")
pdf = pyhf.simplemodels.hepdata_like(signal_data=signal_data.tolist(), \
bkg_data=bkg_data.tolist(), \
bkg_uncerts=bkg_uncerts.tolist())
print("Fit.......")
data = pyhf.tensorlib.astensor(observed_sample.tolist() + pdf.config.auxdata)
bestfit_pars, twice_nll = pyhf.infer.mle.fit(data, pdf, return_fitted_val=True)
print(bestfit_pars)
print(twice_nll)
plt.show()
Note: this answer is based on pyhf v0.5.2.
Alright, so it looks like you've managed to figure most of the big pieces for sure. However, there's two different ways to do this depending on how you prefer to set things up. In both cases, I assume you want an unconstrained fit and you want to...
fit your signal+background model to observed data
fit your background model to observed data
First, let's discuss uncertainties briefly. At the moment, we default to numpy for the tensor background and scipy for the optimizer. See documentation:
numpy backend
scipy optimizer
However, one unfortunate drawback right now with the scipy optimizer is that it cannot return the uncertainties. What you need to do anywhere in your code before the fit (although we generally recommend as early as possible) is to use the minuit optimizer instead:
pyhf.set_backend('numpy', 'minuit')
This will get you the nice features of being able to get the correlation matrix, the uncertainties on the fitted parameters, and the hessian -- amongst other things. We're working to make this consistent for scipy as well, but this is not ready right now.
All optimizations go through our optimizer API which you can currently view through the mixin here in our documentation. Specifically, the signature is
minimize(
objective,
data,
pdf,
init_pars,
par_bounds,
fixed_vals=None,
return_fitted_val=False,
return_result_obj=False,
do_grad=None,
do_stitch=False,
**kwargs)
There are a lot of options here. Let's just focus on the fact that one of the keyword arguments we can pass through is return_uncertainties which will change the bestfit parameters by adding a column for the fitted parameter uncertainty which you want.
1. Signal+Background
In this case, we want to just use the default model
result, twice_nll = pyhf.infer.mle.fit(
data,
pdf,
return_uncertainties=True,
return_fitted_val=True
)
bestfit_pars, errors = result.T
2. Background-Only
In this case, we need to turn off the signal. The way we do this is by setting the parameter of interest (POI) fixed to 0.0. Then we can get the fitted parameters for the background-only model in a similar way, but using fixed_poi_fit instead of an unconstrained fit:
result, twice_nll = pyhf.infer.mle.fixed_poi_fit(
0.0,
data,
pdf,
return_uncertainties=True,
return_fitted_val=True
)
bestfit_pars, errors = result.T
Note that this is quite simply a quick way of doing the following unconstrained fit
bkg_params = pdf.config.suggested_init()
fixed_params = pdf.config.suggested_fixed()
bkg_params[pdf.config.poi_index] = 0.0
fixed_params[pdf.config.poi_index] = True
result, twice_nll = pyhf.infer.mle.fit(
data,
pdf,
init_pars=bkg_params,
fixed_params=fixed_params,
return_uncertainties=True,
return_fitted_val=True
)
bestfit_pars, errors = result.T
Hopefully that clarifies things up more!
Giordon's solution should answer all of your question, but I thought I'd also write out the code to basically address everything we can.
I also take the liberty of changing some of your values a bit so that the signal isn't so strong that the observed CLs value isn't far off to the right of the Brazil band (the results aren't wrong obviously, but it probably makes more sense to be talking about using the discovery test statistic at that point then setting limits. :))
Environment
For this example I'm going to setup a clean Python 3 virtual environment and then install the dependencies (here we're going to be using pyhf v0.5.2)
$ python3 -m venv "${HOME}/.venvs/question"
$ . "${HOME}/.venvs/question/bin/activate"
(question) $ cat requirements.txt
pyhf[minuit,contrib]~=0.5.2
black
(question) $ python -m pip install -r requirements.txt
Code
While we can't easily get the best fit value for both the number of signal events as well as the background events we definitely can do inference to get the best fit value for the signal strength.
The following chunk of code (which is long only because of the visualization) should address all of the points of your question.
# answer.py
import numpy as np
import pyhf
import matplotlib.pyplot as plt
import pyhf.contrib.viz.brazil
# Goals:
# - Fit the model to the observed data
# - Infer the best fit signal strength given the model
# - Get the uncertainties on the best fit signal strength
# - Calculate an 95% CL upper limit on the signal strength
def plot_hist(ax, bins, data, bottom=0, color=None, label=None):
bin_width = bins[1] - bins[0]
bin_leftedges = bins[:-1]
bin_centers = [edge + bin_width / 2.0 for edge in bin_leftedges]
ax.bar(
bin_centers, data, bin_width, bottom=bottom, alpha=0.5, color=color, label=label
)
def plot_data(ax, bins, data, label="Data"):
bin_width = bins[1] - bins[0]
bin_leftedges = bins[:-1]
bin_centers = [edge + bin_width / 2.0 for edge in bin_leftedges]
ax.scatter(bin_centers, data, color="black", label=label)
def invert_interval(test_mus, hypo_tests, test_size=0.05):
# This will be taken care of in v0.5.3
cls_obs = np.array([test[0] for test in hypo_tests]).flatten()
cls_exp = [
np.array([test[1][idx] for test in hypo_tests]).flatten() for idx in range(5)
]
crossing_test_stats = {"exp": [], "obs": None}
for cls_exp_sigma in cls_exp:
crossing_test_stats["exp"].append(
np.interp(
test_size, list(reversed(cls_exp_sigma)), list(reversed(test_mus))
)
)
crossing_test_stats["obs"] = np.interp(
test_size, list(reversed(cls_obs)), list(reversed(test_mus))
)
return crossing_test_stats
def main():
np.random.seed(0)
pyhf.set_backend("numpy", "minuit")
observable_range = [0.0, 10.0]
bin_width = 0.5
_bins = np.arange(observable_range[0], observable_range[1] + bin_width, bin_width)
n_bkg = 2000
n_signal = int(np.sqrt(n_bkg))
# Generate simulation
bkg_simulation = 10 * np.random.random(n_bkg)
signal_simulation = np.random.normal(5, 1.0, n_signal)
bkg_sample, _ = np.histogram(bkg_simulation, bins=_bins)
signal_sample, _ = np.histogram(signal_simulation, bins=_bins)
# Generate observations
signal_events = np.random.normal(5, 1.0, int(n_signal * 0.8))
bkg_events = 10 * np.random.random(int(n_bkg + np.sqrt(n_bkg)))
observed_events = np.array(signal_events.tolist() + bkg_events.tolist())
observed_sample, _ = np.histogram(observed_events, bins=_bins)
# Visualize the simulation and observations
fig, ax = plt.subplots()
fig.set_size_inches(7, 5)
plot_hist(ax, _bins, bkg_sample, label="Background")
plot_hist(ax, _bins, signal_sample, bottom=bkg_sample, label="Signal")
plot_data(ax, _bins, observed_sample)
ax.legend(loc="best")
ax.set_ylim(top=np.max(observed_sample) * 1.4)
ax.set_xlabel("Observable")
ax.set_ylabel("Count")
fig.savefig("components.png")
# Build the model
bkg_uncerts = np.sqrt(bkg_sample)
model = pyhf.simplemodels.hepdata_like(
signal_data=signal_sample.tolist(),
bkg_data=bkg_sample.tolist(),
bkg_uncerts=bkg_uncerts.tolist(),
)
data = pyhf.tensorlib.astensor(observed_sample.tolist() + model.config.auxdata)
# Perform inference
fit_result = pyhf.infer.mle.fit(data, model, return_uncertainties=True)
bestfit_pars, par_uncerts = fit_result.T
print(
f"best fit parameters:\
\n * signal strength: {bestfit_pars[0]} +/- {par_uncerts[0]}\
\n * nuisance parameters: {bestfit_pars[1:]}\
\n * nuisance parameter uncertainties: {par_uncerts[1:]}"
)
# Perform hypothesis test scan
_start = 0.0
_stop = 5
_step = 0.1
poi_tests = np.arange(_start, _stop + _step, _step)
print("\nPerforming hypothesis tests\n")
hypo_tests = [
pyhf.infer.hypotest(
mu_test,
data,
model,
return_expected_set=True,
return_test_statistics=True,
qtilde=True,
)
for mu_test in poi_tests
]
# Upper limits on signal strength
results = invert_interval(poi_tests, hypo_tests)
print(f"Observed Limit on µ: {results['obs']:.2f}")
print("-----")
for idx, n_sigma in enumerate(np.arange(-2, 3)):
print(
"Expected {}Limit on µ: {:.3f}".format(
" " if n_sigma == 0 else "({} σ) ".format(n_sigma),
results["exp"][idx],
)
)
# Visualize the "Brazil band"
fig, ax = plt.subplots()
fig.set_size_inches(7, 5)
ax.set_title("Hypothesis Tests")
ax.set_ylabel(r"$\mathrm{CL}_{s}$")
ax.set_xlabel(r"$\mu$")
pyhf.contrib.viz.brazil.plot_results(ax, poi_tests, hypo_tests)
fig.savefig("brazil_band.png")
if __name__ == "__main__":
main()
which when run gives
(question) $ python answer.py
best fit parameters:
* signal strength: 1.5884737977889158 +/- 0.7803435235862329
* nuisance parameters: [0.99020988 1.06040191 0.90488207 1.03531383 1.09093327 1.00942088
1.07789316 1.01125627 1.06202964 0.95780043 0.94990993 1.04893286
1.0560711 0.9758487 0.93692481 1.04683181 1.05785515 0.92381263
0.93812855 0.96751869]
* nuisance parameter uncertainties: [0.06966439 0.07632218 0.0611428 0.07230328 0.07872258 0.06899675
0.07472849 0.07403246 0.07613661 0.08606657 0.08002775 0.08655314
0.07564512 0.07308117 0.06743479 0.07383134 0.07460864 0.06632003
0.06683251 0.06270965]
Performing hypothesis tests
/home/stackoverflow/.venvs/question/lib/python3.7/site-packages/pyhf/infer/calculators.py:229: RuntimeWarning: invalid value encountered in double_scalars
teststat = (qmu - qmu_A) / (2 * self.sqrtqmuA_v)
Observed Limit on µ: 2.89
-----
Expected (-2 σ) Limit on µ: 0.829
Expected (-1 σ) Limit on µ: 1.110
Expected Limit on µ: 1.542
Expected (1 σ) Limit on µ: 2.147
Expected (2 σ) Limit on µ: 2.882
Let us know if you have any further questions!
I'm trying to train a dataset with 357 features using Isolation Forest sklearn implementation. I can successfully train and get results when the max features variable is set to 1.0 (the default value).
However when max features is set to 2, it gives the following error:
ValueError: Number of features of the model must match the input.
Model n_features is 2 and input n_features is 357
It also gives the same error when the feature count is 1 (int) and not 1.0 (float).
How I understood was that when the feature count is 2 (int), two features should be considered in creating each tree. Is this wrong? How can I change the max features parameter?
The code is as follows:
from sklearn.ensemble.iforest import IsolationForest
def isolation_forest_imp(dataset):
estimators = 10
samples = 100
features = 2
contamination = 0.1
bootstrap = False
random_state = None
verbosity = 0
estimator = IsolationForest(n_estimators=estimators, max_samples=samples, contamination=contamination,
max_features=features,
bootstrap=boostrap, random_state=random_state, verbose=verbosity)
model = estimator.fit(dataset)
In the documentation it states:
max_features : int or float, optional (default=1.0)
The number of features to draw from X to train each base estimator.
- If int, then draw `max_features` features.
- If float, then draw `max_features * X.shape[1]` features.
So, 2 should mean take two features and 1.0 should mean take all of the features, 0.5 take half and so on, from what I understand.
I think this could be a bug, since, taking a look in IsolationForest's fit:
# Isolation Forest inherits from BaseBagging
# and when _fit is called, BaseBagging takes care of the features correctly
super(IsolationForest, self)._fit(X, y, max_samples,
max_depth=max_depth,
sample_weight=sample_weight)
# however, when after _fit the decision_function is called using X - the whole sample - not taking into account the max_features
self.threshold_ = -sp.stats.scoreatpercentile(
-self.decision_function(X), 100. * (1. - self.contamination))
then:
# when the decision function _validate_X_predict is called, with X unmodified,
# it calls the base estimator's (dt) _validate_X_predict with the whole X
X = self.estimators_[0]._validate_X_predict(X, check_input=True)
...
# from tree.py:
def _validate_X_predict(self, X, check_input):
"""Validate X whenever one tries to predict, apply, predict_proba"""
if self.tree_ is None:
raise NotFittedError("Estimator not fitted, "
"call `fit` before exploiting the model.")
if check_input:
X = check_array(X, dtype=DTYPE, accept_sparse="csr")
if issparse(X) and (X.indices.dtype != np.intc or
X.indptr.dtype != np.intc):
raise ValueError("No support for np.int64 index based "
"sparse matrices")
# so, this check fails because X is the original X, not with the max_features applied
n_features = X.shape[1]
if self.n_features_ != n_features:
raise ValueError("Number of features of the model must "
"match the input. Model n_features is %s and "
"input n_features is %s "
% (self.n_features_, n_features))
return X
So, I am not sure on how you can handle this. Maybe figure out the percentage that leads to just the two features you need - even though I am not sure it'll work as expected.
Note: I am using scikit-learn v.0.18
Edit: as #Vivek Kumar commented this is an issue and upgrading to 0.20 should do the trick.
I am currently runnuing training in matlab on a matrix of logspecrum samples I am constantly dealing with underflow problems.I understood that I need to work with log's in order to deal with underflowing.
I am still strugling with uderflow though , when i calculate the mean (mue) bucause it is negetive i cant work with logs so i need the real values that underflow.
These are equasions i am working with:
In MATLAB code i calulate log_tau in oreder avoid underflow but when calulating mue i need exp(log(tau)) which goes to zero.
I am attaching relevent MATLAB code
**in the code i called the variable alpha is tau ...
for i = 1 : 50
log_c = Logsum(log_alpha,1) - log(N);
c = exp(log_c);
mue = DataMat*alpha./(repmat(exp(Logsum(log_alpha,1)),FrameSize,1));
log_abs_mue = log(abs(mue));
log_SigmaSqr = log((DataMat.^2)*alpha) - repmat(Logsum(log_alpha,1),FrameSize,1) - 2*log_abs_mue;
SigmaSqr = exp(log_SigmaSqr);
for j=1:N
rep_DataMat(:,:,j) = repmat(DataMat(:,j),1,M);
log_gamma(j,:) = log_c - 0.5*(FrameSize*log(2*pi)+sum(log_SigmaSqr)) + sum((rep_DataMat(:,:,j) - mue).^2./(2*SigmaSqr));
end
log_alpha = log_gamma - repmat(Logsum(log_gamma,2),1,M);
alpha = exp(log_alpha);
end
c = exp(log_c);
SigmaSqr = exp(log_SigmaSqr);
does any one see how i can avoid this? or what needs to be fixed in code?
What i did was add this line to the MATLAB code:
mue(isnan(mue))=0; %fix 0/0 problem
and this one:
SigmaSqr(SigmaSqr==0)=1;%fix if mue_k = x_k
not sure if this is the best solution but is seems to work...
any have a better idea?
I'm working with a large data frame called exp (file here) in R. In the interests of performance, it was suggested that I check out the idata.frame() function from plyr. But I think I'm using it wrong.
My original call, slow but it works:
df.median<-ddply(exp,
.(groupname,starttime,fPhase,fCycle),
numcolwise(median),
na.rm=TRUE)
With idata.frame, Error: is.data.frame(df) is not TRUE
library(plyr)
df.median<-ddply(idata.frame(exp),
.(groupname,starttime,fPhase,fCycle),
numcolwise(median),
na.rm=TRUE)
So, I thought, perhaps it is my data. So I tried the baseball dataset. The idata.frame example works fine: dlply(idata.frame(baseball), "id", nrow) But if I try something similar to my desired call using baseball, it doesn't work:
bb.median<-ddply(idata.frame(baseball),
.(id,year,team),
numcolwise(median),
na.rm=TRUE)
>Error: is.data.frame(df) is not TRUE
Perhaps my error is in how I'm specifying the groupings? Anyone know how to make my example work?
ETA:
I also tried:
groupVars <- c("groupname","starttime","fPhase","fCycle")
voi<-c('inadist','smldist','lardist')
i<-idata.frame(exp)
ag.median <- aggregate(i[,voi], i[,groupVars], median)
Error in i[, voi] : object of type 'environment' is not subsettable
which uses a faster way of getting the medians, but gives a different error. I don't think I understand how to use idata.frame at all.
Given you are working with 'big' data and looking for perfomance, this seems a perfect fit for data.table.
Specifically the lapply(.SD,FUN) and .SDcols arguments with by
Setup the data.table
library(data.table)
DT <- as.data.table(exp)
iexp <- idata.frame(exp)
Which columns are numeric
numeric_columns <- names(which(unlist(lapply(DT, is.numeric))))
dt.median <- DT[, lapply(.SD, median), by = list(groupname, starttime, fPhase,
fCycle), .SDcols = numeric_columns]
some benchmarking
library(rbenchmark)
benchmark(data.table = DT[, lapply(.SD, median), by = list(groupname, starttime,
fPhase, fCycle), .SDcols = numeric_columns],
plyr = ddply(exp, .(groupname, starttime, fPhase, fCycle), numcolwise(median), na.rm = TRUE),
idataframe = ddply(exp, .(groupname, starttime, fPhase, fCycle), function(x) data.frame(inadist = median(x$inadist),
smldist = median(x$smldist), lardist = median(x$lardist), inadur = median(x$inadur),
smldur = median(x$smldur), lardur = median(x$lardur), emptyct = median(x$emptyct),
entct = median(x$entct), inact = median(x$inact), smlct = median(x$smlct),
larct = median(x$larct), na.rm = TRUE)),
aggregate = aggregate(exp[, numeric_columns],
exp[, c("groupname", "starttime", "fPhase", "fCycle")],
median),
replications = 5)
## test replications elapsed relative user.self
## 4 aggregate 5 5.42 1.789 5.30
## 1 data.table 5 3.03 1.000 3.03
## 3 idataframe 5 11.81 3.898 11.77
## 2 plyr 5 9.47 3.125 9.45
Strange behaviour, but even in the docs it says that idata.frame is experimental. You probably found a bug. Perhaps you could rewrite the check at the top of ddply that tests is.data.frame().
In any case, this cuts about 20% off the time (on my system):
system.time(df.median<-ddply(exp, .(groupname,starttime,fPhase,fCycle), function(x) data.frame(
inadist=median(x$inadist),
smldist=median(x$smldist),
lardist=median(x$lardist),
inadur=median(x$inadur),
smldur=median(x$smldur),
lardur=median(x$lardur),
emptyct=median(x$emptyct),
entct=median(x$entct),
inact=median(x$inact),
smlct=median(x$smlct),
larct=median(x$larct),
na.rm=TRUE))
)
Shane asked you in another post if you could cache the results of your script. I don't really have an idea of your workflow, but it may be best to setup a chron to run this and store the results, daily/hourly whatever.