Estimating the runtime of an R script
Background
Recently a person on Stack Overflow asked a question about how to estimate the runtime of an R script. She was attempting to produce correlation tests for 60 questions in a survey, using the corr.test() function from the psych package.
The answers from each question were coded as 5 point scales from 1 - 5. The input data frame included 8,219 observations. When she ran corr.test(), her computer did not complete the analysis after 2 hours. This led her to post a question on Stackoverflow.com, “Is there any method to estimate the R script running time?”
An initial answer
Although the question did not include enough data to be considered a reproducible example on Stack, one of the things that I appreciate about R is its ability to quickly generate simulated data. Three lines of code later, I have a simulated survey including 9,000 respondents of 60 questions with responses from 1 - 5.
# create 9000 rows of data w/ 60 columns
system.time(data <- as.data.frame(matrix(round(runif(9000*60,min = 1, max = 5)),
nrow = 9000)))
id <- 1:9000
data <- cbind(id,data)
With some additional code we can calculate timings on the psych::corr.test() function. Given that the question noted that corr.test() failed to produce a result in 2 hours on a laptop with 8Gb RAM and a 2.5Ghz two core processor, I used lapply() to process a vector of numbers of observations and process them with corr.test().
observations <- c(100,200,500,1000,2000)
theTimings <- lapply(observations,function(x){
system.time(r <- corr.test(data[id <= x,2:61],method = "kendall"))
})
theNames <- paste0("timings_",observations,"_obs")
names(theTimings) <- theNames
theTimings
Elapsed times for the analysis ranged from 0.46 seconds with 100 observations to 106.6 seconds with 2,000 observations, as illustrated below.
> theTimings
$timings_100_obs
user system elapsed
0.435 0.023 0.457
$timings_200_obs
user system elapsed
1.154 0.019 1.174
$timings_500_obs
user system elapsed
5.969 0.026 5.996
$timings_1000_obs
user system elapsed
24.260 0.045 24.454
$timings_2000_obs
user system elapsed
106.465 0.109 106.603
Generating Predictions
We can quickly fit a linear model to these timings and use it to predict the runtime for larger data sets. We create a data frame with the timing information, fit a model, and print the model summary to check the R^2 for goodness of fit. Since this is an exploratory data analysis, I didn’t bother writing code to extract the timings returned by lapply() into a vector.
time <- c(0.457,1.174,5.996,24.454,106.603)
timeData <- data.frame(observations,time)
fit <- lm(time ~ observations, data = timeData)
summary(fit)
The summary indicates that a linear model appears to be a good fit with the data, recognizing we used a small number of observations as input to the model.
> summary(fit)
Call:
lm(formula = time ~ observations, data = timeData)
Residuals:
1 2 3 4 5
9.808 4.906 -7.130 -16.769 9.186
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -14.970240 8.866838 -1.688 0.18993
observations 0.056193 0.008612 6.525 0.00731 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 13.38 on 3 degrees of freedom
Multiple R-squared: 0.9342, Adjusted R-squared: 0.9122
F-statistic: 42.57 on 1 and 3 DF, p-value: 0.007315
Next, we build another data frame with additional numbers of observations and use it to generate predicted timings via the stats::predict() function.
predictions <- data.frame(observations = c(3000,4000,5000,6000,7000,8000,9000))
data.frame(observations = predictions,predicted = predict(fit,predictions))
Given this model, the 9,000 observation data frame should take about 8.2 minutes to run corr.test() on the laptop I used for this analysis, a 2015 era Macbook Pro 15 with an Intel i7-4870HQ four core processor.
> data.frame(observations = predictions,predicted = predict(fit,predictions))
observations predicted
1 3000 153.6102
2 4000 209.8037
3 5000 265.9971
4 6000 322.1906
5 7000 378.3841
6 8000 434.5776
7 9000 490.7710
> 490 / 60
[1] 8.166667
>
The original question stated that corr.test() failed to complete within 2 hours on a 2 core 2.4Ghz processor, which leads us to the hypothesis that there is a non-linear effect in runtime that becomes prominent beyond 2,000 observations. We need to generate data on runs with more observations to determine whether we can discern an effect that makes the algorithm degrade to less than linear scalability.
Improving the Model
One way to capture a non-linear effect is to add a quadratic effect as an independent variable in the model. We collect data for 3,000, 4,000, and 5,000 observations in order to increase the degrees of freedom in the model, as well as to provide more data from which we might detect a quadratic effect.
> theTimings
$timings_3000_obs
user system elapsed
259.444 0.329 260.149
$timings_4000_obs
user system elapsed
458.993 0.412 460.085
$timings_5000_obs
user system elapsed
730.178 0.839 731.915
>
Next, we run linear models with and without the quadratic effect, generate predictions, and compare the results. The summary() for the quadratic model is very interesting.
observations <- c(100,200,500,1000,2000,3000,4000,5000)
obs_squared <- observations^2
time <- c(0.457,1.174,5.996,24.454,106.603,260.149,460.085,731.951)
timeData <- data.frame(observations,obs_squared,time)
fitLinear <- lm(time ~ observations, data = timeData)
fitQuadratic <- lm(time ~ observations + obs_squared, data = timeData)
summary(fitQuadratic)
> summary(fitQuadratic)
Call:
lm(formula = time ~ observations + obs_squared, data = timeData)
Residuals:
1 2 3 4 5 6 7 8
-0.2651 0.2384 0.7455 -0.2363 -2.8974 4.5976 -2.7581 0.5753
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.121e+00 1.871e+00 0.599 0.5752
observations -7.051e-03 2.199e-03 -3.207 0.0238 *
obs_squared 3.062e-05 4.418e-07 69.307 1.18e-08 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 2.764 on 5 degrees of freedom
Multiple R-squared: 0.9999, Adjusted R-squared: 0.9999
F-statistic: 3.341e+04 on 2 and 5 DF, p-value: 4.841e-11
Not only has R^2 improved to .9999 with the quadratic term in the model, both the linear and quadratic terms are significantly different from zero at alpha = 0.05. Interestingly, with a quadratic term in the model, the linear effect is negative.
Comparing prediction results
Our penultimate step is to generate predictions for both models, combine them into a data frame and print the results.
predLinear = predict(fitLinear,predictions)
predQuadratic <- predict(fitQuadratic,predictions)
data.frame(observations = predictions$observations,
obs_squared = predictions$obs_squared,
predLinear,
predQuadratic)
observations obs_squared predLinear predQuadratic
1 3000 9.0e+06 342.6230 255.5514
2 4000 1.6e+07 482.8809 462.8431
3 5000 2.5e+07 623.1388 731.3757
4 6000 3.6e+07 763.3967 1061.1490
5 7000 4.9e+07 903.6546 1452.1632
6 8000 6.4e+07 1043.9125 1904.4181
7 9000 8.1e+07 1184.1704 2417.9139
Conclusions
First, as we added data at larger numbers of observations in the corr.test(), the linear prediction at 9,000 observations more than doubled from 491 seconds to 1,184 seconds. As expected, adding data to the model helped improve its accuracy.
Second, the time prediction of the quadratic model was more than twice the runtime as the linear model, 2,417.9 seconds.
And the drumroll, please…
I ran the 9,000 observation data frame through the test, and it took 40 minutes to complete. The runtime was 5X the duration of the original linear prediction from runs up to 2,000 observations, and slightly more than 2X the prediction from runs up to 4,000 observations.
> # validate model
> system.time(r <- corr.test(data[,2:61],method = "kendall"))
user system elapsed
2398.572 2.990 2404.175
> 2404.175 / 60
[1] 40.06958
While the both versions of the linear version of the model were more accurate than the IHME COVID-19 predicted fatalities model that originally predicted as many as 2.2 million fatalities in the United States, it’s too inaccurate to help one decide whether to go brew a cup of coffee or go for a 10K run while waiting for corr.test() to complete its work.
In contrast, the quadratic model is stunningly accurate in its prediction, where the predicted runtime of 2,418 seconds was within 0.6% of the actual value.
Why didn’t the original poster’s analysis complete in 2 hours?
In the back and forth comments posted on Stack as I developed the models, a question was raised about the relevance of multiple CPU cores in runtime performance when the corr.test() function uses a single thread to process the data.
My tests, as well as performance analyses I have done with R functions that support multithreading (e.g. Improving Performance of caret::train() with Random Forest) indicate that in practice CPUs with similar speed ratings but fewer cores are slower than those with more cores.
In this specific situation where we analyzed the performance of core.test(), I ran a second series of tests on an HP Spectre x-360 with an Intel i7-U6500 CPU that also runs at 2.5Ghz. Its processing time degrades faster than that of the Intel i7-4870HQ CPU (also at 2.5Ghz), as illustrated by the following table.
| observations | time x360-13 | time Macbook15 | pct difference |
|---|---|---|---|
| 100 | 0.56 | 0.457 | 22.53829 |
| 200 | 1.55 | 1.174 | 32.02726 |
| 500 | 8.36 | 5.996 | 39.42628 |
| 1000 | 35.33 | 24.454 | 44.47534 |
| 2000 | 151.92 | 106.603 | 42.51006 |
| 3000 | 357.45 | 260.149 | 37.40203 |
| 4000 | 646.56 | 460.085 | 40.53055 |
As we can see from the table, the i7-U6500 is 22.5% slower than the i7-4870HQ at 100 observations, and this deficit grows as the number of observations including in the timing simulations increases to 4,000.
© 2020 - 2024 Leonard M. Greski - copying with attribution permitted