CODEWITHSUNDEEP
rstatistics
user1830307
user1830307

Reputation:

Calculating AIC (in R or any other software)

Thank you for opening this message!

I am trying to fit the log-log plot of the cumulative distribution of a network to one of three models: Exponential (EXP) (P(k)~e^(-ak)), Exponentially truncated power law (TRU) (P(k)~k^(a-1)e^(k/kc)), and Power law (POWER) (P(k)~k^-a). I know this is a low-information test, but I am simply trying to determine which of the three models is the best fit (or possibly the least terrible fit!)

I realize that TRU has 2 degrees of freedom, while EXP and POWER have only 1 degree of freedom.

I have the following data in R:

logCPK<-c(0, -0.0014, -0.0038, -0.0056, -0.007, -0.013, -0.0165, -0.0222, -0.0261, -0.0354, -0.0478, -0.0581, -0.0714, -0.0846, -0.1068, -0.1205, -0.1403, -0.1638, -0.1837, -0.2091, -0.2265, -0.2517, -0.2766, -0.3121, -0.3381, -0.3744, -0.4079, -0.4485, -0.4869, -0.5326, -0.5881, -0.6383, -0.6959, -0.7586, -0.8199, -0.8853, -0.9624, -1.0352, -1.1341, -1.2294, -1.3521, -1.4982, -1.6824, -1.892, -2.182)

dataPOW<-c(0.3387, 0.2175, 0.1197, 0.0441, -0.0198, -0.0777, -0.1262, -0.1717, -0.2108, -0.2482, -0.281, -0.3113, -0.341, -0.3673, -0.3921, -0.4167, -0.4387, -0.4607, -0.4806, -0.4996, -0.5187, -0.536, -0.5535, -0.5695, -0.5856, -0.6004, -0.6154, -0.6292, -0.6426, -0.6561, -0.6686, -0.6808, -0.6932, -0.7046, -0.7158, -0.7272, -0.7383, -0.7486, -0.7586, -0.7689, -0.7785, -0.7883, -0.7974, -0.8064, -0.8155)

dataEXP<-c(0.1981, 0.168, 0.1364, 0.1063, 0.0762, 0.0445, 0.0144, -0.0172, -0.0473, -0.0789, -0.109, -0.1391, -0.1707, -0.2008, -0.2309, -0.2625, -0.2926, -0.3243, -0.3544, -0.3845, -0.4161, -0.4462, -0.4778, -0.5079, -0.5395, -0.5696, -0.6012, -0.6313, -0.6614, -0.6931, -0.7232, -0.7533, -0.7849, -0.815, -0.8451, -0.8767, -0.9083, -0.9384, -0.9685, -1.0001, -1.0302, -1.0619, -1.092, -1.1221, -1.1537)

dataTRU<-c(-0.1867, -0.0857, -0.0193, 0.0204, 0.0443, 0.057, 0.0602, 0.0562, 0.0467, 0.0319, 0.0139, -0.0073, -0.0327, -0.0592, -0.088, -0.1202, -0.1525, -0.1881, -0.2234, -0.26, -0.2995, -0.3383, -0.38, -0.4205, -0.464, -0.5062, -0.5512, -0.5947, -0.6388, -0.6858, -0.731, -0.7767, -0.8253, -0.8719, -0.919, -0.9689, -1.0191, -1.0674, -1.116, -1.1673, -1.2165, -1.2685, -1.3183, -1.3684, -1.4212)

So my question is, how can I get the AIC values for the three models?

I have read, but not fully understand, the documentation for AIC{}, which has a general syntax of: AIC(object, ..., k = 2).

Not understanding what I am doing, I have tried commands in R, like:

AIC(logCPK~dataPOW) Error in UseMethod("logLik") : no applicable method for 'logLik' applied to an object of class "formula"

If you have any advise on how to calculate the AIC with the type of data I listed above in R, or any other software, I would be so, so, thankful!!

Thank you...

Upvotes: 1

Views: 3167

Answers (1)

Marc in the box
Marc in the box

Reputation: 12005

You should be able to give the name of the fitted model object in AIC, e.g.

fit <- lm(logCPK~dataPOW)
AIC(fit)

Upvotes: 1

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