Entropy is related to the number of possible microstates according to S = k_Bln(\Omega), where S is the entropy of the system, k B is Boltzmann’s constant, and Ω is the number of microstates (e.g. This statement holds true if the perfect crystal has only one state with minimum energy. Specifically, the entropy of a pure crystalline substance at absolute zero temperature is zero.Īt zero temperature the system must be in a state with the minimum thermal energy. The entropy of a system at absolute zero is typically zero, and in all cases is determined only by the number of different ground states it has. The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches zero. ferromagneticthe basic mechanism by which certain materials form permanent magnets, or are attracted to magnets.paramagneticattracted to the poles of a magnet.third law of thermodynamicsa law which states that the entropy of a perfect crystal at absolute zero is exactly equal to zero.For the entropy at absolute zero to be zero, the magnetic moments of a perfectly ordered crystal must themselves be perfectly ordered.Mathematically, the absolute entropy of any system at zero temperature is the natural log of the number of ground states times Boltzmann’s constant kB.At zero temperature the system must be in a state with the minimum thermal energy.Timeit results: # for loop to print out results of timeitįor approach,timeit_results in zip(, ): Setup='''labels= from _main_ import entropy4''', Setup='''labels= from _main_ import entrop圓''',ĭ = timeit.repeat(stmt='''entropy4(labels)''', Setup='''labels= from _main_ import entropy2''',Ĭ = timeit.repeat(stmt='''entrop圓(labels)''', Setup='''labels= from _main_ import entropy1''',ī = timeit.repeat(stmt='''entropy2(labels)''', Timeit operations: repeat_number = 1000000Ī = timeit.repeat(stmt='''entropy1(labels)''', Return -(norm_counts * np.log(norm_counts)/np.log(base)).sum() Return -(vc * np.log(vc)/np.log(base)).sum() Vc = pd.Series(labels).value_counts(normalize=True, sort=False) """ Computes entropy of label distribution. Value,counts = np.unique(labels, return_counts=True) This question is specifically asking about the "Fastest" way but I only see times on one answer so I'll post a comparison of using scipy and numpy to the original poster's entropy2 answer with slight alterations.įour different approaches: scipy/numpy, numpy/math, pandas/numpy, numpy import numpy as np Gupta answer is good but could be condensed.
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