time_interval = [4, 6, 12]
I want to sum up the numbers like [4, 4+6, 4+6+12] in order to get the list t = [4, 10, 22].
I tried the following:
t1 = time_interval[0]
t2 = time_interval[1] + t1
t3 = time_interval[2] + t2
print(t1, t2, t3) # -> 4 10 22
If you're doing much numerical work with arrays like this, I'd suggest numpy, which comes with a cumulative sum function cumsum:
import numpy as np
a = [4,6,12]
np.cumsum(a)
#array([4, 10, 22])
Numpy is often faster than pure python for this kind of thing, see in comparison to @Ashwini's accumu:
In [136]: timeit list(accumu(range(1000)))
10000 loops, best of 3: 161 us per loop
In [137]: timeit list(accumu(xrange(1000)))
10000 loops, best of 3: 147 us per loop
In [138]: timeit np.cumsum(np.arange(1000))
100000 loops, best of 3: 10.1 us per loop
But of course if it's the only place you'll use numpy, it might not be worth having a dependence on it.
np.cumsun case that starts with a list, to take into account the conversion time. –
Polik list I would not recommend numpy. –
Silverfish timeit, "if -n is not given, a suitable number of loops is calculated by trying successive powers of 10 until the total time is at least 0.2 seconds." If you expect it to make a difference, you can supply -n 1000 to make them all equivalent. –
Silverfish np.cumsum for np.array lists and itertools.accumulate otherwise. This answer is still the best one, but it's outdated. –
Lowgrade In Python 2 you can define your own generator function like this:
def accumu(lis):
total = 0
for x in lis:
total += x
yield total
In [4]: list(accumu([4,6,12]))
Out[4]: [4, 10, 22]
And in Python 3.2+ you can use itertools.accumulate():
In [1]: lis = [4,6,12]
In [2]: from itertools import accumulate
In [3]: list(accumulate(lis))
Out[3]: [4, 10, 22]
total = 0; partial_sums = [total := total + v for v in values]. I would still expect accumulate to be faster. –
Extenuate if (value := some_dict.get(key)) is not None: ... which avoids the notorious double lookup. Or also in list comprehensions with an if filter like [func_result for x in xs if satisfies_property(func_result := f(x))] which also avoids the notorious double evaluation of the function f(x). –
Avignon Try the
itertools.accumulate() function.
import itertools
list(itertools.accumulate([1,2,3,4,5]))
# [1, 3, 6, 10, 15]
operator.add as the default operation is addition anyway. –
Hogle I did a bench-mark of the top two answers with Python 3.4 and I found itertools.accumulate is faster than numpy.cumsum under many circumstances, often much faster. However, as you can see from the comments, this may not always be the case, and it's difficult to exhaustively explore all options. (Feel free to add a comment or edit this post if you have further benchmark results of interest.)
Some timings...
For short lists accumulate is about 4 times faster:
from timeit import timeit
def sum1(l):
from itertools import accumulate
return list(accumulate(l))
def sum2(l):
from numpy import cumsum
return list(cumsum(l))
l = [1, 2, 3, 4, 5]
timeit(lambda: sum1(l), number=100000)
# 0.4243644131347537
timeit(lambda: sum2(l), number=100000)
# 1.7077815784141421
For longer lists accumulate is about 3 times faster:
l = [1, 2, 3, 4, 5]*1000
timeit(lambda: sum1(l), number=100000)
# 19.174508565105498
timeit(lambda: sum2(l), number=100000)
# 61.871223849244416
If the numpy array is not cast to list, accumulate is still about 2 times faster:
from timeit import timeit
def sum1(l):
from itertools import accumulate
return list(accumulate(l))
def sum2(l):
from numpy import cumsum
return cumsum(l)
l = [1, 2, 3, 4, 5]*1000
print(timeit(lambda: sum1(l), number=100000))
# 19.18597290944308
print(timeit(lambda: sum2(l), number=100000))
# 37.759664884768426
If you put the imports outside of the two functions and still return a numpy array, accumulate is still nearly 2 times faster:
from timeit import timeit
from itertools import accumulate
from numpy import cumsum
def sum1(l):
return list(accumulate(l))
def sum2(l):
return cumsum(l)
l = [1, 2, 3, 4, 5]*1000
timeit(lambda: sum1(l), number=100000)
# 19.042188624851406
timeit(lambda: sum2(l), number=100000)
# 35.17324400227517
list of five items, especially if you are unwilling to accept an array in return. If the list in question is really so short, then their running time would be inconsequential---dependencies and legibility would surely dominate. But wide usage of a list of uniform numerical data type of significant length would be silly; for that, a numpy array would be appropriate, and usually faster. –
Silverfish numpy to be faster, unless I've overlooked something? –
Mispleading sum2 function is probably in converting l into an array. Try timing a = np.array(l) and np.cumsum(a) separately. Then try a = np.tile(np.arange(1, 6), 1000) vs l = [1,2,3,4,5]*1000. In a program conducting other numerical processes (like the creation or loading of l in the first place) your working data would probably be in an array already, and the creation would be a constant cost. –
Silverfish Behold:
a = [4, 6, 12]
reduce(lambda c, x: c + [c[-1] + x], a, [0])[1:]
Will output (as expected):
[4, 10, 22]
c + [c[-1] + x] over and over adds up to a total runtime quadratic in the input length. –
Syverson Assignment expressions from PEP 572 (new in Python 3.8) offer yet another way to solve this:
time_interval = [4, 6, 12]
total_time = 0
cum_time = [total_time := total_time + t for t in time_interval]
You can calculate the cumulative sum list in linear time with a simple for loop:
def csum(lst):
s = lst.copy()
for i in range(1, len(s)):
s[i] += s[i-1]
return s
time_interval = [4, 6, 12]
print(csum(time_interval)) # [4, 10, 22]
The standard library's itertools.accumulate may be a faster alternative (since it's implemented in C):
from itertools import accumulate
time_interval = [4, 6, 12]
print(list(accumulate(time_interval))) # [4, 10, 22]
Since python 3.8 it's possible to use Assignment expressions, so things like this became easier to implement
nums = list(range(1, 10))
print(f'array: {nums}')
v = 0
cumsum = [v := v + n for n in nums]
print(f'cumsum: {cumsum}')
produces
array: [1, 2, 3, 4, 5, 6, 7, 8, 9]
cumsum: [1, 3, 6, 10, 15, 21, 28, 36, 45]
The same technique can be applied to find the cum product, mean, etc.
p = 1
cumprod = [p := p * n for n in nums]
print(f'cumprod: {cumprod}')
s = 0
c = 0
cumavg = [(s := s + n) / (c := c + 1) for n in nums]
print(f'cumavg: {cumavg}')
results in
cumprod: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880]
cumavg: [1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0]
If You want a pythonic way without numpy working in 2.7 this would be my way of doing it
l = [1,2,3,4]
_d={-1:0}
cumsum=[_d.setdefault(idx, _d[idx-1]+item) for idx,item in enumerate(l)]
now let's try it and test it against all other implementations
import timeit, sys
L=list(range(10000))
if sys.version_info >= (3, 0):
reduce = functools.reduce
xrange = range
def sum1(l):
cumsum=[]
total = 0
for v in l:
total += v
cumsum.append(total)
return cumsum
def sum2(l):
import numpy as np
return list(np.cumsum(l))
def sum3(l):
return [sum(l[:i+1]) for i in xrange(len(l))]
def sum4(l):
return reduce(lambda c, x: c + [c[-1] + x], l, [0])[1:]
def this_implementation(l):
_d={-1:0}
return [_d.setdefault(idx, _d[idx-1]+item) for idx,item in enumerate(l)]
# sanity check
sum1(L)==sum2(L)==sum3(L)==sum4(L)==this_implementation(L)
>>> True
# PERFORMANCE TEST
timeit.timeit('sum1(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.001018061637878418
timeit.timeit('sum2(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.000829620361328125
timeit.timeit('sum3(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.4606760001182556
timeit.timeit('sum4(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.18932826995849608
timeit.timeit('this_implementation(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.002348129749298096
There could be many answers for this depending on the length of the list and the performance. One very simple way which I can think without thinking of the performance is this:
a = [1, 2, 3, 4]
a = [sum(a[0:x]) for x in range(1, len(a)+1)]
print(a)
[1, 3, 6, 10]
This is by using list comprehension and this may work fairly well it is just that here I am adding over the subarray many times, you could possibly improvise on this and make it simple!
Cheers to your endeavor!
O(n²), so it only makes sense to use it for tiny lists. –
Hogle First, you want a running list of subsequences:
subseqs = (seq[:i] for i in range(1, len(seq)+1))
Then you just call sum on each subsequence:
sums = [sum(subseq) for subseq in subseqs]
(This isn't the most efficient way to do it, because you're adding all of the prefixes repeatedly. But that probably won't matter for most use cases, and it's easier to understand if you don't have to think of the running totals.)
If you're using Python 3.2 or newer, you can use itertools.accumulate to do it for you:
sums = itertools.accumulate(seq)
And if you're using 3.1 or earlier, you can just copy the "equivalent to" source straight out of the docs (except for changing next(it) to it.next() for 2.5 and earlier).
range than to hack around it by doing [1:] at the end, or to ignore it.) –
Heikeheil [4,6,12] since, as he wrote in the question, he already knows what that is! –
Jackquelinejackrabbit values = [4, 6, 12]
total = 0
sums = []
for v in values:
total = total + v
sums.append(total)
print 'Values: ', values
print 'Sums: ', sums
Running this code gives
Values: [4, 6, 12]
Sums: [4, 10, 22]
Try this:
result = []
acc = 0
for i in time_interval:
acc += i
result.append(acc)
lst = [4, 6, 12]
[sum(lst[:i+1]) for i in xrange(len(lst))]
If you are looking for a more efficient solution (bigger lists?) a generator could be a good call (or just use numpy if you really care about performance).
def gen(lst):
acu = 0
for num in lst:
yield num + acu
acu += num
print list(gen([4, 6, 12]))
l = [1,-1,3]
cum_list = l
def sum_list(input_list):
index = 1
for i in input_list[1:]:
cum_list[index] = i + input_list[index-1]
index = index + 1
return cum_list
print(sum_list(l))
In Python3, To find the cumulative sum of a list where the ith element
is the sum of the first i+1 elements from the original list, you may do:
a = [4 , 6 , 12]
b = []
for i in range(0,len(a)):
b.append(sum(a[:i+1]))
print(b)
OR you may use list comprehension:
b = [sum(a[:x+1]) for x in range(0,len(a))]
Output
[4,10,22]
In [42]: a = [4, 6, 12]
In [43]: [sum(a[:i+1]) for i in xrange(len(a))]
Out[43]: [4, 10, 22]
This is slighlty faster than the generator method above by @Ashwini for small lists
In [48]: %timeit list(accumu([4,6,12]))
100000 loops, best of 3: 2.63 us per loop
In [49]: %timeit [sum(a[:i+1]) for i in xrange(len(a))]
100000 loops, best of 3: 2.46 us per loop
For larger lists, the generator is the way to go for sure. . .
In [50]: a = range(1000)
In [51]: %timeit [sum(a[:i+1]) for i in xrange(len(a))]
100 loops, best of 3: 6.04 ms per loop
In [52]: %timeit list(accumu(a))
10000 loops, best of 3: 162 us per loop
Somewhat hacky, but seems to work:
def cumulative_sum(l):
y = [0]
def inc(n):
y[0] += n
return y[0]
return [inc(x) for x in l]
I did think that the inner function would be able to modify the y declared in the outer lexical scope, but that didn't work, so we play some nasty hacks with structure modification instead. It is probably more elegant to use a generator.
Without having to use Numpy, you can loop directly over the array and accumulate the sum along the way. For example:
a=range(10)
i=1
while((i>0) & (i<10)):
a[i]=a[i-1]+a[i]
i=i+1
print a
Results in:
[0, 1, 3, 6, 10, 15, 21, 28, 36, 45]
A pure python oneliner for cumulative sum:
cumsum = lambda X: X[:1] + cumsum([X[0]+X[1]] + X[2:]) if X[1:] else X
This is a recursive version inspired by recursive cumulative sums. Some explanations:
X[:1] is a list containing the previous element and is almost the same as [X[0]] (which would complain for empty lists).cumsum call in the second term processes the current element [1] and remaining list whose length will be reduced by one.if X[1:] is shorter for if len(X)>1.Test:
cumsum([4,6,12])
#[4, 10, 22]
cumsum([])
#[]
And simular for cumulative product:
cumprod = lambda X: X[:1] + cumprod([X[0]*X[1]] + X[2:]) if X[1:] else X
Test:
cumprod([4,6,12])
#[4, 24, 288]
Here's another fun solution. This takes advantage of the locals() dict of a comprehension, i.e. local variables generated inside the list comprehension scope:
>>> [locals().setdefault(i, (elem + locals().get(i-1, 0))) for i, elem
in enumerate(time_interval)]
[4, 10, 22]
Here's what the locals() looks for each iteration:
>>> [[locals().setdefault(i, (elem + locals().get(i-1, 0))), locals().copy()][1]
for i, elem in enumerate(time_interval)]
[{'.0': <enumerate at 0x21f21f7fc80>, 'i': 0, 'elem': 4, 0: 4},
{'.0': <enumerate at 0x21f21f7fc80>, 'i': 1, 'elem': 6, 0: 4, 1: 10},
{'.0': <enumerate at 0x21f21f7fc80>, 'i': 2, 'elem': 12, 0: 4, 1: 10, 2: 22}]
Performance is not terrible for small lists:
>>> %timeit list(accumulate([4, 6, 12]))
387 ns ± 7.53 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
>>> %timeit np.cumsum([4, 6, 12])
5.31 µs ± 67.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
>>> %timeit [locals().setdefault(i, (e + locals().get(i-1,0))) for i,e in enumerate(time_interval)]
1.57 µs ± 12 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
And obviously falls flat for larger lists.
>>> l = list(range(1_000_000))
>>> %timeit list(accumulate(l))
95.1 ms ± 5.22 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
>>> %timeit np.cumsum(l)
79.3 ms ± 1.07 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
>>> %timeit np.cumsum(l).tolist()
120 ms ± 1.23 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
>>> %timeit [locals().setdefault(i, (e + locals().get(i-1, 0))) for i, e in enumerate(l)]
660 ms ± 5.14 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Even though the method is ugly and not practical, it sure is fun.
I think the below code is the easiest:
a=[1,1,2,1,2]
b=[a[0]]+[sum(a[0:i]) for i in range(2,len(a)+1)]
def cumulative_sum(list):
l = []
for i in range(len(list)):
new_l = sum(list[:i+1])
l.append(new_l)
return l
time_interval = [4, 6, 12]
print(cumulative_sum(time_interval)
Maybe a more beginner-friendly solution.
So you need to make a list of cumulative sums. You can do it by using for loop and .append() method
time_interval = [4, 6, 12]
cumulative_sum = []
new_sum = 0
for i in time_interval:
new_sum += i
cumulative_sum.append(new_sum)
print(cumulative_sum)
or, using numpy module
import numpy
time_interval = [4, 6, 12]
c_sum = numpy.cumsum(time_interval)
print(c_sum.tolist())
This would be Haskell-style:
def wrand(vtlg):
def helpf(lalt,lneu):
if not lalt==[]:
return helpf(lalt[1::],[lalt[0]+lneu[0]]+lneu)
else:
lneu.reverse()
return lneu[1:]
return helpf(vtlg,[0])
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