List comprehension for running total

Question

I want to get a running total from a list of numbers.

For demo purposes, I start with a sequential list of numbers using range

a = range         


        

Answer 1

This can be implemented in 2 lines in Python.

Using a default parameter eliminates the need to maintain an aux variable outside, and then we just do a map to the list.

def accumulate(x, l=[0]): l[0] += x; return l[0];
map(accumulate, range(20))

Answer 2

I wanted to do the same thing to generate cumulative frequencies that I could use bisect_left over - this is the way I've generated the list;

[ sum( a[:x] ) for x in range( 1, len(a)+1 ) ]

Answer 3

Another one-liner, in linear time and space.

def runningSum(a):
    return reduce(lambda l, x: l.append(l[-1]+x) or l if l else [x], a, None)

I'm stressing linear space here, because most of the one-liners I saw in the other proposed answers --- those based on the pattern list + [sum] or using chain iterators --- generate O(n) lists or generators and stress the garbage collector so much that they perform very poorly, in comparison to this.

Answer 4

This is inefficient as it does it every time from beginning but possible it is:

a = range(20)
runtot=[sum(a[:i+1]) for i,item in enumerate(a)]
for line in zip(a,runtot):
    print line

Answer 5

Here's a linear time solution one liner:

list(reduce(lambda (c,s), a: (chain(c,[s+a]), s+a), l,(iter([]),0))[0])

Example:

l = range(10)
list(reduce(lambda (c,s), a: (chain(c,[s+a]), s+a), l,(iter([]),0))[0])
>>> [0, 1, 3, 6, 10, 15, 21, 28, 36, 45]

In short, the reduce goes over the list accumulating sum and constructing an list. The final x[0] returns the list, x[1] would be the running total value.

Answer 6

If you can use numpy, it has a built-in function named cumsum that does this.

import numpy
tot = numpy.cumsum(a)  # returns a numpy.ndarray
tot = list(tot)        # if you prefer a list