如何获取列表的第n个元素

Question

I have the following output from my code:

['test1', ':', '(', '83.25,', '78.32)', '*33.90*', '2.29', '2.14', '0.82', '65.95', 'test2', ':', '(', '101.80,', '95.10)', '*31.73*', '12.05', '0.60', '0.96', '-26.46', 'test3', ':', '(', '49.84,', '42.29)', '*33.19*', '6.54', '1.24', '0.50', '67.42']

I want the 6th, 7th, 16th, 17th, 26th and 27th element of the list and so on for any length If possible putting them in two separate lists. So eventually I would like an output that's just two lists like the following:

list1 = [2.29, 12.05, 6.54]
list2 = [2.14, 0.60, 1.24]

Does anyone know how to get this result?

Answer 1

If your list is called data you can do the following.

data_starting_at_position_6 = data[6::10]
data_starting_at_position_7 = data[7::10]

You start at the correct position and use 10 as the step.

At the moment you have strings in the list, so data[6::10] gives you ['2.29', '12.05', '6.54'] . If you would like to have floats you can use list(map(float, data[6::10])) and get [2.29, 12.05, 6.54] .

Answer 2

Use the mod of index!

6_list = []
7_list = []
for i in xrange(len(your_list) + 1):
    if i%10 == 6:
        6_list.append(your_list[i])
    elif i%10 == 7:
        7_list.append(your_list[i])