Suppose i have list of numbers [3, 51, 34]
. I want to add to each element the sum of the previous elements and return a new list with these new values.
So here the
numpy.cumsum
might be good choice for something like this.
In [1]: import numpy as np
In [2]: a = [3,51,34]
In [3]: np.cumsum(a)
Out[3]: array([ 3, 54, 88])
A list comprehension approach, because that's always fun:
>>> data = [3, 51, 34]
>>> result = [n + sum(data[:i]) for i, n in enumerate(data)]
>>> result
[3, 54, 88]
generator based solution
In [25]: ll = [3,51,34]
In [26]: def acc(myiter):
....: it = iter(myiter)
....: total = it.next()
....: yield total
....: for element in it:
....: total = total + element
....: yield total
....:
In [27]: acc(ll)
Out[27]: <generator object acc at 0x1ec9e10>
In [28]: [x for x in acc(ll)]
Out[28]: [3, 54, 88]
as it will be the fastest one:
In [29]: %timeit [x for x in acc(ll)]
1000000 loops, best of 3: 1.26 µs per loop
In [30]: import numpy as np
In [31]: np.cumsum(ll)
Out[31]: array([ 3, 54, 88])
In [32]: %timeit np.cumsum(ll)
100000 loops, best of 3: 15.8 µs per loop
In [33]: %timeit reduce(lambda x, y: [y] if not x else x + [y + x[-1]], ll, None)
1000000 loops, best of 3: 1.87 µs per loop
In [34]: %timeit [n + sum(ll[:i]) for i, n in enumerate(ll)]
100000 loops, best of 3: 1.99 µs per loop
And a trivial, procedural solution, for completeness:
a = [3,51,34]
def cumsum(numbers):
accum = 0
result = []
for n in numbers:
accum += n
result.append(accum)
return result
print cumsum(a)
It prints [3, 54, 88]
a simple reduce:
nums = [3,51,34]
reduce(lambda x, y: [y] if not x else x + [y + x[-1]], nums, None)
# [3, 54, 88]