CODEWITHSUNDEEP
pythonfunctionnumberssum
O.rka
O.rka

Reputation: 30677

Sum consecutive numbers in a list. Python

i'm trying to sum consecutive numbers in a list while keeping the first one the same.

so in this case 5 would stay 5, 10 would be 10 + 5 (15), and 15 would be 15 + 10 + 5 (30)

x = [5,10,15]
y = []

for value in x:
   y.append(...)

print y

[5,15,30]

Upvotes: 8

Views: 17993

Answers (4)

hurrdrought
hurrdrought

Reputation: 492

With numpy.cumsum:

In[1]: import numpy as np
In[2]: x = [5,10,15]
In[3]: x = np.array(x)
In[4]: y = x.cumsum()
In[5]: y
Out[6]: array([ 5, 15, 30])

I am using Python 3.4

Upvotes: 4

will
will

Reputation: 10650

As in sum all the elements up to the element you're on?

x = [5,10,15]
y = [sum(x[:i+1]) for i in range(len(x))]

Upvotes: 1

Brent Washburne
Brent Washburne

Reputation: 13138

y = [sum(x[:i+1]) for i in range(len(x))]

Upvotes: 6

Gareth Latty
Gareth Latty

Reputation: 88977

You want itertools.accumulate() (added in Python 3.2). Nothing extra needed, already implemented for you.

In earlier versions of Python where this doesn't exist, you can use the pure python implementation given:

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

This will work perfectly with any iterable, lazily and efficiently. The itertools implementation is implemented at a lower level, and therefore even faster.

If you want it as a list, then naturally just use the list() built-in: list(accumulate(x)).

Upvotes: 13

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