Python列表重新排序，还记得原始顺序吗？

Question

I'm working on a Bayesian probability project, in which I need to adjust probabilities based on new information. I have yet to find an efficient way to do this. What I'm trying to do is start with an equal probability list for distinct scenarios. Ex. There are 6 people: E, T, M, Q, L, and Z, and their initial respective probabilities of being chosen are represented in

myList=[.1667, .1667, .1667, .1667, .1667, .1667]

New information surfaces that people in the first third alphabetically have a collective 70% chance of being chosen. A new list is made, sorted alphabetically by name (E, L, M, Q, T, Z), that just includes the new information. (.7/.333=2.33, .3/.667=.45)

    newList=[2.33, 2.33, .45, .45, .45, .45)

I need a way to order the newList the same as myList so I can multiply the right values in list comprehension, and reach the adjust probabilities. Having a single consistent order is important because the process will be repeated several times, each with different criteria (vowels, closest to P, etc), and in a list with about 1000 items. Each newList could instead be a newDictionary, and then once the adjustment criteria are created they could be ordered into a list, but transforming multiple dictionaries seems inefficient. Is it? Is there a simple way to do this I'm entirely missing?

Thanks!

Answer 1

For what it's worth, the best thing you can do for the speed of your methods in Python is to use numpy instead of the standard types (you'll thus be using pre-compiled C code to perform arithmetic operations). This will lead to a dramatic speed increase. Numpy arrays have fixed orderings anyway, and syntax is more directly applicable to mathematical operations. You just need to consider how to express the operations as matrix operations. Eg your example:

myList = np.ones(6) / 6.
newInfo = np.array( [.7/2, .7/2, .3/4, .3/4, .3/4, .3/4] )
result = myList * newInfo

Since both vectors have unit sum there's no need to normalise (I'm not sure what you were doing in your example, I confess, so if there's a subtlety I've missed let me know), but if you do need to it's trivial:

result /= np.sum(result)

Answer 2

Try storing your info as a list of tuples:

bayesList = [('E', 0.1667), ('M', 0.1667), ...]

your list comprehension can be along the lines of

newBayes = [(person, prob * normalizeFactor) for person, prob in bayesList]

where you've normalizeFactor was calculated before setting up your list comprehension