Python List accumulation

Question

So the following code reads through two text files both containing 25 ints on 25 lines and puts them in two respective lists:

num_DIR = '/path/to/num.txt'
den_DIR = '/path/to/den.txt'

def makeList(DIR):
    list = []
    for line in open(DIR).readlines():
            list.append(line.strip())
    return list

num_list = makeList(num_DIR)
den_list = makeList(den_DIR)

Outputting:

num_list = ['76539', '100441', '108637', '108874', '103580', '91869', '78458', '61955', '46100', '32701', '21111', '13577', '7747', '4455', '2309', '1192', '554', '264', '134', '63', '28', '15', '12', '7', '5']

den_list = ['621266', '496647', '436229', '394595', '353249', '305882', '253983', '199455', '147380', '102872', '67255', '41934', '24506', '13778', '7179', '3646', '1778', '816', '436', '217', '114', '74', '49', '44', '26']

How would I go about making each value in both lists the added sum of all the value after it, as in an accumulation list?

You don't appear to have made any effort in that direction; this isn't a code-writing service. For one thing, you should use integers not strings. — jonrsharpe, Jun 18 '14 at 19:09
You really want each value to be the sum of all those _after_ it, not before, as in `['1', '2', '3']` -> `['6', '5', '3']`? — martineau, Jun 18 '14 at 19:12

timgeb · Answer 1 · 2014-06-18T19:19:31.853

4

You can also use sum and a list comprehension if performance is not critical:

>>> num_list=[1,2,3]
>>> [sum(num_list[:y]) for y in range(1, len(num_list) + 1)]
[1, 3, 6]

edited Jun 18 '14 at 19:19

answered Jun 18 '14 at 19:08

timgeb

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score 4 · Accepted Answer · answered Jun 18 '14 at 19:24

4

You seem to want the sum of all values to the end of the list after each element

summed=[]
den_list = map(int,den_list)
for i,j in enumerate(den_list[:-1]):
    summed += j,sum(den_list[i+1:])

If you don't want to keep the original values in the list

summed=[sum(num_list[i+1:])for i,j in enumerate(num_list[:-1])]

answered Jun 18 '14 at 19:24

Padraic Cunningham

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The summed list comprehension is exactly what I needed! – Sean Jun 18 '14 at 19:53

score 3 · Answer 3 · answered Jun 18 '14 at 19:09

If you are running python 3.2 or later, batteries are included in itertools.accumulate.

If you're running a version prior to that, the pure-python equivalent provided in the docs works equivalently (albeit not at C speed):

def accumulate(iterable, func=operator.add):
    'Return running totals'
    # accumulate([1,2,3,4,5]) --> 1 3 6 10 15
    # accumulate([1,2,3,4,5], operator.mul) --> 1 2 6 24 120
    it = iter(iterable)
    total = next(it)
    yield total
    for element in it:
        total = func(total, element)
        yield total

demo:

from itertools import accumulate

list(accumulate(map(int,num_list)))
Out[3]: 
[76539,
 176980,
 285617,
 394491,
 498071,
 589940,
 668398,
 730353,
 776453,
 809154,
 830265,
 843842,
 #snip...,
 860627
]

Note that you'll need to turn all those strings into ints, as I did with map.

Python List accumulation

3 Answers3