For more verbose algorithms, determining the time complexity (i.e. BigO) is a pain. My solution has been to time the execution of the algorithm with parameters n and k, and come up with a function (time function) that varies with n and k.
My data looks something like the below:
n k executionTime
500 1 0.02
500 2 0.03
500 3 0.05
500 ... ...
500 10 0.18
1000 1 0.08
... ... ...
10000 1 9.8
... ... ...
10000 10 74.57
I've been using the lm() function in the stats R package. I don't know how to interpret the output of the multiple regression, to determine a final Big-O. This is my main question: how do you translate the output of a multiple variable regression, to a final ruling on the best Big-O time complexity rating?
Here's the output of the lm():
Residuals:
Min 1Q Median 3Q Max
-14.943 -5.325 -1.916 3.681 31.475
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -2.130e+01 1.591e+00 -13.39 <2e-16 ***
n 4.080e-03 1.953e-04 20.89 <2e-16 ***
k 2.361e+00 1.960e-01 12.05 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 7.962 on 197 degrees of freedom
Multiple R-squared: 0.747, Adjusted R-squared: 0.7444
F-statistic: 290.8 on 2 and 197 DF, p-value: < 2.2e-16
Here's the output of log(y) ~ log(n) + log(k):
Residuals:
Min 1Q Median 3Q Max
-0.4445 -0.1136 -0.0253 0.1370 0.5007
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -16.80405 0.13749 -122.22 <2e-16 ***
log(n) 2.02321 0.01609 125.72 <2e-16 ***
log(k) 1.01216 0.01833 55.22 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.1803 on 197 degrees of freedom
Multiple R-squared: 0.9897, Adjusted R-squared: 0.9896
F-statistic: 9428 on 2 and 197 DF, p-value: < 2.2e-16
Here's the output of the principle components, showing both n and k are contributing to the spread of the multivariate model:
PC1(This is n) PC2 (this is k) PC3 (noise?)
Standard deviation 1.3654 1.0000 0.36840
Proportion of Variance 0.6214 0.3333 0.04524
Cumulative Proportion 0.6214 0.9548 1.00000
Related
I have 4 data sets which is CO2 emission , GDP per capita, GDP per capita square and Number of tourist arrival. I am trying to run a model to observe the number of tourist arrival impact on Co2 emission in order to derive the Tourism induced Environmental Kuznets Curve. Below is the code and summary results .
Without log
Yt<-Data$`CO2 emissions`
X1t<-Data$`GDP per capita`
X2t<-Data$`GDP per caita square`
X3t<-Data$`Number of Tourist arrival`
model<-lm(Yt~X1t+X2t+X3t)
summary(model)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.238e-02 7.395e-03 1.674 0.100187
X1t -2.581e-05 6.710e-05 -0.385 0.702139
X2t 1.728e-07 4.572e-08 3.780 0.000413 ***
X3t 1.928e-07 3.501e-08 5.507 1.2e-06 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.02252 on 51 degrees of freedom
Multiple R-squared: 0.9475, Adjusted R-squared: 0.9444
F-statistic: 306.5 on 3 and 51 DF, p-value: < 2.2e-16
With log
LYt<-(log(Yt))
LX1t<-(log(X1t))
LX2t<-(log(X2t))
LX3t<-(log(X3t))
model1<-lm(LYt~LX1t+LX2t+LX3t)
summary(model1)
Coefficients: (1 not defined because of singularities)
Estimate Std. Error t value Pr(>|t|)
(Intercept) -9.38623 0.46040 -20.387 < 2e-16 ***
LX1t 0.83679 0.09834 8.509 2.01e-11 ***
LX2t NA NA NA NA
LX3t 0.17802 0.06888 2.585 0.0126 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.2863 on 52 degrees of freedom
Multiple R-squared: 0.9074, Adjusted R-squared: 0.9038
F-statistic: 254.7 on 2 and 52 DF, p-value: < 2.2e-16
It is pretty evident that GDP per capita and GDP per capita square are perfectly collinear. However, why does the regression coefficients show missing values (NA) only in the case of log transformed model?
I am a beginner in R and statistics in general. I am trying to build a regression model with 4 variables (one of them is nominal data, with 3 alt categories). I think i managed to build a model with my raw data. But I wanted to standardize my data-set and the lm essentially perfect fit: summary may be unreliable message.
This is the summary of the model with raw data;
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -2.0713 5.9131 -0.350 0.727
Vdownload 8.6046 0.5286 16.279 < 2e-16 ***
DownloadDegisim 2.8854 0.6822 4.229 4.25e-05 ***
Vupload -4.2877 0.5418 -7.914 7.32e-13 ***
Saglayici2 -8.2084 0.6043 -13.583 < 2e-16 ***
Saglayici3 -9.8869 0.5944 -16.634 < 2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 2.885 on 138 degrees of freedom
Multiple R-squared: 0.8993, Adjusted R-squared: 0.8956
F-statistic: 246.5 on 5 and 138 DF, p-value: < 2.2e-16
I wrote these codes to standardize my data
memnuniyet_scaled <-scale(Vdownload, center = TRUE, scale = TRUE)
Vdownload_scaled <-scale(Vdownload, center = TRUE, scale = TRUE)
Vupload_scaled <-scale(Vupload, center = TRUE, scale = TRUE)
DownloadD_scaled <- scale(DownloadDegisim, center = TRUE, scale = TRUE)
result<-lm(memnuniyet_scaled~Vdownload_scaled+DownloadD_scaled+Vupload_scaled+Saglayıcı)
summary(result)
And this is the summary of my standardized data
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5.079e-17 5.493e-17 9.250e-01 0.357
Vdownload_scaled 1.000e+00 6.667e-17 1.500e+16 <2e-16 ***
DownloadD_scaled -4.591e-17 8.189e-17 -5.610e-01 0.576
Vupload_scaled 9.476e-18 6.337e-17 1.500e-01 0.881
Saglayici2 -6.523e-17 7.854e-17 -8.300e-01 0.408
Saglayici3 -8.669e-17 7.725e-17 -1.122e+00 0.264
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 3.75e-16 on 138 degrees of freedom
Multiple R-squared: 1, Adjusted R-squared: 1
F-statistic: 2.034e+32 on 5 and 138 DF, p-value: < 2.2e-16
I do know that R value should not have changed with standardization and have no idea what I did wrong.
I ran a model explaining the weight of some plant as a function of time and trying to incorporate the treatment effect.
mod <- lm(weight ~time + treatment)
The model looks like this:
with model summary being:
Call:
lm(formula = weight ~ time + treatment, data = df)
Residuals:
Min 1Q Median 3Q Max
-21.952 -7.674 0.770 6.851 21.514
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -37.5790 3.2897 -11.423 < 2e-16 ***
time 4.7478 0.2541 18.688 < 2e-16 ***
treatmentB 8.2000 2.4545 3.341 0.00113 **
treatmentC 5.4633 2.4545 2.226 0.02797 *
treatmentD 20.3533 2.4545 8.292 2.36e-13 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 9.506 on 115 degrees of freedom
Multiple R-squared: 0.7862, Adjusted R-squared: 0.7788
F-statistic: 105.7 on 4 and 115 DF, p-value: < 2.2e-16
ANOVA table
Analysis of Variance Table
Response: weight
Df Sum Sq Mean Sq F value Pr(>F)
time 1 31558.1 31558.1 349.227 < 2.2e-16 ***
treatment 3 6661.9 2220.6 24.574 2.328e-12 ***
Residuals 115 10392.0 90.4
I want to test the H0 that intercept1=intercept2=intercept3=intercept4. Is this done by simply interpreting the t-value and p-value for the intercept ( I guess not because this is the baseline (treatment A in this case))? I'm a bit puzzled by this as not much attention is paid on difference in intercept on most sources i looked up.
I am running a simple regression (OLS)
> lm_1 <- lm(Dependent_variable_1 ~ Independent_variable_1, data = data_1)
> summary(lm_1)
Call:
lm(formula = Dependent_variable_1 ~ Independent_variable_1,
data = data_1)
Residuals:
Min 1Q Median 3Q Max
-143187 -34084 -4990 37524 136293
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 330853 13016 25.418 < 2e-16 ***
`GDP YoY% - Base` 3164631 689599 4.589 0.000118 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 66160 on 24 degrees of freedom
(4 observations deleted due to missingness)
Multiple R-squared: 0.4674, Adjusted R-squared: 0.4452
F-statistic: 21.06 on 1 and 24 DF, p-value: 0.0001181
The autocorrelation and heteroskedasticity tests follow:
> dwtest(lm_1,alternative="two.sided")
Durbin-Watson test
data: lm_1
DW = 0.93914, p-value = 0.001591
alternative hypothesis: true autocorrelation is not 0
> bptest(lm_1)
studentized Breusch-Pagan test
data: lm_1
BP = 9.261, df = 1, p-value = 0.002341
then I run a robust regression for autocorrelation and heteroskedasticity (HAC - Newey-West):
> coeftest(lm_1, vocv=NeweyWest(lm_1,lag=2, prewhite=FALSE))
t test of coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 330853 13016 25.4185 < 2.2e-16 ***
Independent_variable_1 3164631 689599 4.5891 0.0001181 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
and I get the same results for coefficients / standard errors.
Is this normal? Is this due to the small sample size?
Normally, me and you(assuming you're not a bot) are easily able to identify whether a predictor is categorical or quantitative. Like, for example, gender is obviously categorical. Your last vote can be classified categorically.
Basically, we can identify categorical predictors easily. But what happens when we input some data in R, and it's lm function makes dummy variables for a predictor? How does it do that?
Somewhat related Question on StackOverflow.
Search R factor function. Here is a small demo, first model uses number of cylinder as a numerical valuable. Second model uses it as a categorical variable.
> summary(lm(mpg~cyl,mtcars))
Call:
lm(formula = mpg ~ cyl, data = mtcars)
Residuals:
Min 1Q Median 3Q Max
-4.9814 -2.1185 0.2217 1.0717 7.5186
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 37.8846 2.0738 18.27 < 2e-16 ***
cyl -2.8758 0.3224 -8.92 6.11e-10 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 3.206 on 30 degrees of freedom
Multiple R-squared: 0.7262, Adjusted R-squared: 0.7171
F-statistic: 79.56 on 1 and 30 DF, p-value: 6.113e-10
> summary(lm(mpg~factor(cyl),mtcars))
Call:
lm(formula = mpg ~ factor(cyl), data = mtcars)
Residuals:
Min 1Q Median 3Q Max
-5.2636 -1.8357 0.0286 1.3893 7.2364
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 26.6636 0.9718 27.437 < 2e-16 ***
factor(cyl)6 -6.9208 1.5583 -4.441 0.000119 ***
factor(cyl)8 -11.5636 1.2986 -8.905 8.57e-10 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 3.223 on 29 degrees of freedom
Multiple R-squared: 0.7325, Adjusted R-squared: 0.714
F-statistic: 39.7 on 2 and 29 DF, p-value: 4.979e-09
Hxd1011 adressed the more difficult case, when a categorical variable is stored as a number and therefore R understands by default that it is a numerical value - and if this is not the desired behaviour we must use factor function.
Your example with predictor ShelveLoc in dataset Carseats is easier because it's a text (character) variable, and therefore it can only be a categorical variable.
> head(Carseats$ShelveLoc)
[1] Bad Good Medium Medium Bad Bad
Levels: Bad Good Medium
R decides that thing from the features type. You can check that by using the str(dataset).If the feature is factor type then it will create dummies for that feature.