1

I have a list that stores lists with 5 elements. I want to create 5 new lists that store elements of each indexes. I have the following code but it seems not smart way.

>>> stats
[['1', '0', '36', '36', '3'], ['10', '0', '41', '77', '5'], ['1', '0', '631', '631', '63'], ['1', '0', '98', '98', '9'], ['9', '0', '52', '81', '6'], ['2', '0', '111', '167', '13'], ['1', '0', '98', '98', '9'], ['1', '0', '92', '92', '9'], ['2', '0', '241', '287', '26'], ['1', '0', '210', '210', '21'], ['2', '0', '336', '358', '34'], ['2', '0', '49', '57', '5'], ['5', '0', '52', '148', '7'], ['2', '0', '46', '76', '6'], ['3', '0', '33', '50', '4'], ['7', '0', '47', '70', '6'], ['1', '0', '94', '94', '9'], ['1', '0', '65', '65', '6'], ['1', '0', '66', '66', '6'], ['1', '0', '429', '429', '42'], ['1', '0', '337', '337', '33'], ['12', '0', '49', '126', '6'], ['1', '0', '47', '47', '4'], ['1', '0', '63', '63', '6'], ['1', '0', '79', '79', '7'], ['2', '0', '96', '100', '9'], ['1', '0', '36', '36', '3'], ['1', '0', '69', '69', '6'], ['6', '0', '44', '67', '5'], ['3', '0', '269', '385', '31'], ['2', '0', '78', '115', '9'], ['2', '0', '49', '52', '5'], ['3', '0', '26', '134', '9'], ['2', '0', '255', '561', '40'], ['1', '0', '75', '75', '7'], ['1', '0', '59', '59', '5'], ['2', '0', '59', '64', '6'], ['1', '0', '86', '86', '8'], ['1', '0', '63', '63', '6'], ['2', '0', '79', '100', '8'], ['4', '0', '825', '888', '86'], ['1', '0', '82', '82', '8'], ['3', '0', '65', '94', '7'], ['1', '0', '88', '88', '8'], ['1', '0', '344', '344', '34'], ['1', '0', '286', '286', '28'], ['1', '0', '73', '73', '7'], ['3', '0', '42', '69', '5'], ['1', '0', '151', '151', '15'], ['1', '0', '286', '286', '28'], ['2', '0', '47', '59', '5'], ['9', '0', '15', '41', '2'], ['2', '0', '343', '355', '34'], ['1', '0', '305', '305', '30'], ['1', '0', '238', '238', '23'], ['2', '0', '974', '2101', '153'], ['2', '0', '138', '142', '14'], ['7', '0', '45', '70', '5'], ['1', '0', '39', '39', '3']]
>>>
>>> num_requests,num_failures,min_response_time,max_response_time,avg_response_time = [], [], [], [], []
>>>
>>> for l in stats:
...  num_requests.append(l[0])
...  num_failures.append(l[1])
...  min_response_time.append(l[2])
...  max_response_time.append(l[3])
...  avg_response_time.append(l[4])
...
>>> num_requests
['1', '10', '1', '1', '9', '2', '1', '1', '2', '1', '2', '2', '5', '2', '3', '7', '1', '1', '1', '1', '1', '12', '1', '1', '1', '2', '1', '1', '6', '3', '2', '2', '3', '2', '1', '1', '2', '1', '1', '2', '4', '1', '3', '1', '1', '1', '1', '3', '1', '1', '2', '9', '2', '1', '1', '2', '2', '7', '1']

It could be stored in one list which stores 5 sublist.

1 Answer 1

3

Solution

Just use zip with *:

(num_requests, num_failures, min_response_time, max_response_time, 
 avg_response_time) = zip(*stats)

This gives you tuples. Convert to lists if you need lists:

(num_requests, num_failures, min_response_time, max_response_time, 
 avg_response_time) = (list(x) for x in zip(*stats))

Details

A shorter example:

>>> data = [[1, 2, 3], [10, 20, 30], [100, 200, 300]]
>>> a, b, c = zip(*data)
>>> a
(1, 10, 100)
>>> b
(2, 20, 200)
>>> c
(3, 30, 300)

This is equivalent to:

a, b, c = zip(data[0], data[1], data[2])

but works for any number of sublists.

The left side uses tuple unpacking. For example, this:

x, y, z = (10, 20, 30)

assigns 10 to x, 20 to y, and 30 to z.

Performance

Measure how fast it is.

Version with append:

%%timeit
num_requests,num_failures,min_response_time,max_response_time,avg_response_time = [], [], [], [], []

for l in stats:
    num_requests.append(l[0])
    num_failures.append(l[1])
    min_response_time.append(l[2])
    max_response_time.append(l[3])
    avg_response_time.append(l[4])

10000 loops, best of 3: 51 µs per loop

Version with zip:

%%timeit
(num_requests, num_failures, min_response_time, max_response_time, 
     avg_response_time) = zip(*stats)

100000 loops, best of 3: 8.58 µs per loop

It is about five times faster.

It takes a bit longer when you convert the tuples to lists:

%%timeit
(num_requests, num_failures, min_response_time, max_response_time, 
 avg_response_time) = (list(x) for x in zip(*stats))

100000 loops, best of 3: 13.3 µs per loop

Still, about four times faster.

Sign up to request clarification or add additional context in comments.

1 Comment

thank you it works great. can you evaluate about the performance for a long list?

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.