每三个列表的 Python 总和

Question

I have the following problem.

I need to read x number of text files and get the sum of every third list. An example would be:

file 1

[1.0, 2.0, 3.0]
[3.1, 2.1, 2.1]
[3.4, 3.4, 4.4]

file 2

[1.0, 2.0, 3.0]
[3.1, 2.1, 2.1]
[3.4, 3.4, 4.4]

result:

[2.0, 4.0, 6.0]
[6.2, 4.2, 4.2]
[6.8, 6.8, 8.8]

The question is how to get the matching lists to sum?

I know I can use zip to get sum of lists and I get the files opened, and read by line and stored into lists, but I'm not sure how to target the right ones to sum?

If the case was only for 2, I could check with an if condition if the counter is even or odd, but how to handle it when there are 3 cases? Does Python have a solution built in there? The solution also has to be scalable to an x amount of files containing 3 lists. This also means I cant just make a list of cases.

Answer 1

result = list()
for l_index, num_list in enumerate(file1):
    result_list = list()
        for e_index, element in enumerate(num_list):
            result_list.append(file1[l_index][e_index] + file2[l_index][e_index])
result.append(result_list)

I just assumed every file is saved in a nested list, if this is not the case i don't really know how you should do it.

I'm not sure if 1. the code I wrote is correct (but you should get what i was doing) and 2. if i understood right what you were trying to do.

EDIT: Adjusted the code so now it does what it should

Answer 2

Would this work?

Given a nested list as in your case problem:

file1 = [[1.0,2.0,3.0],[3.1,2.1,2.1],[3.4,3.4,4.4]]
file2 = [[1.0,2.0,3.0],[3.1,2.1,2.1],[3.4,3.4,4.4]] 

simply concat the nested lists after n files. (order remains so we are good)

comb = file1+file2

Perform your operations in a function so its "modular"

def operate(data,n):#heres the function
    if n == 2: #naive way to initialize your output
        summary = [[],[]]
    elif n == 3:
        summary = [[],[],[]]
    for index,dat in enumerate(data): #getting index +the list
        f = [] #holder for the temp data. habit of overwriting stuff
        if index%n == 1: #modulo function lets you do for every n list. 
                         #Good experiment to extend this dynamically for n cases instead of just up to 3
            if len(summary[1]) == 0:
                summary[1] = dat
            else:
                for a,b in zip(summary[1],dat): #your zip 
                    f.append(a+b)
                summary[1] = f #since its a sum we just do it for each pair and replace 


        elif index%n == 2: 
            if len(summary[2])== 0:
                summary[2] = dat
            else:
                for a,b in zip(summary[2],dat):
                    f.append(a+b)
                summary[2] = f
        elif index%n == 0:
            if len(summary[0])== 0:
                summary[0] = dat
            else:
                for a,b in zip(summary[0],dat):
                    f.append(a+b)
                summary[0] = f



    return summary
file1 = [[1.0,2.0,3.0],[3.1,2.1,2.1],[3.4,3.4,4.4]]
file2 = [[1.0,2.0,3.0],[3.1,2.1,2.1],[3.4,3.4,4.4]] 

comb = file1+file2
t2 = operate(comb,2)
t3 = operate(comb,3)

print("for every second list sum: ",t2)
print("for every third list sum: ",t3)

Theoretically you can extend this for any set by either programming the cases dynamically but you get the gist of it I think.

Output:

for every second list sum:  [[7.5, 7.5, 9.5], [7.5, 7.5, 9.5]]
for every third list sum:  [[2.0, 4.0, 6.0], [6.2, 4.2, 4.2], [6.8, 6.8, 8.8]]

Do try to initialize the summary variable nicer. It will be part of your solution for extending the cases to orders >3