如何计算python列表中的组合数量

Question

I need some help to count the amount of combinations in a list array in python.

I need to count the amount of possible combinations between three letters in all of the elements and then find the most repeated one. eg, ABC, CDA, CCA, etc...

I have created a for loop to look in each element of the list, then I have another loop to check each combo of three letters and add it to a new list. I am not sure about how to count the amount of times a combination is repeated, and then to find the mode, I think I might use the max() function.

this is part of the code I have, but it does not work as I am expecting, because it is just adding each item of the list into an independent list.

lst = ["ABCDABCD", "ABDCABD", "ACCACABB", "BACDABC"] 

for combo in lst:
   for i in range (0,3):      
      combolst = []
      combolst.append(lst[i].split())
      print(combolst)

I am new to coding so that's why I'm here. Thanks!

Answer 1

(Assuming my math memory isn't garbage) So okay, we are interested in combinations. Your code simply splits the list and creates a new one (as you said). Then we would use the combination formula : n!/(z!(nz)!). Where:

• n is the number of elements, in this case the length of our string in question
• z would be how many objects we wish to choose

Thus you would get:

for combo in lst:
    n = math.factorial(len(combo))
    r = math.factorial(3)
    nMinR = math.factorial((len(combo) - 3))
    result = n/(r*nMinR)
    print(result)

This is for combination, if we want permutations (where order does matter)

for combo in lst:
        n = math.factorial(len(combo))
        nMinR = math.factorial((len(combo) - 3))
        result = n/(nMinR)
        print(result)

I hope I understood your question correctly. Here is some reading about combinations vs permutations ( https://medium.com/i-math/combinations-permutations-fa7ac680f0ac ). Keep in mind, the above code will only print out how many possible combinations or permutations are possible; it won't actually try to construct the possible values