汇总列表中每隔四个元素

Question

I am new to python, I have very long list (around 100 elements in list). I would like to add every 4th element of this list and store result as sum1 , sum2 , sum3 , or sum4 . I used the following code :

while y in range(0,104):
    if y%4==0:
    sum=last_indexes[y]+sum
    y=y+4

Questions:

1. If the lists increases the compilation time increases. Is there any function in python by which I can fasten this process?
2. Also if the list is not a multiple of 4 and still I want to store it in sum1,sum2,sum3,sum4 then what modification should be done.

Thanks for your time and consideration

Answer 1

No, you cannot make code run (not compile) at the same speed if the input is larger.

The more Pythonic way to handle this problem, which should work even if the length of the list is not a multiple of 4 is to use list slicing:

sum1 = sum(last_indices[::4])
sum2 = sum(last_indices[1::4])
sum3 = sum(last_indices[2::4])
sum4 = sum(last_indices[3::4])

The slice notation of a:b:c says "Start at index a , go to index b , in increments of c ". When a is omitted, it is assumed to be 0 . when b is omitted, it is assumed to be the end of the list.

Answer 2

For a fast and clean solution, use itertools.islice() to extract the data elements at fixed intervals:

from itertools import islice

multiple = 4
for start in range(multiple):
    print sum(islice(data_array, start, None, multiple))

For large lists, this is fast because:

• it doesn't do expensive modulo computations (islice has an internal counter)
• it doesn't build temporary lists like regular slices do (islice uses iterators)
• both sum() and islice() run at C speed (that is the whole point of the itertools module)

Answer 3

Any time you have a series of variables named like varX where X is a number, you should probably use a list, rather than multiple variables (exceptions may exist for very small numbers of variables, like two or maybe three). In your situation, you should probably use a list with four integers, rather than four separate sumX variables. As a side benefit, this will allow your code to be very simple!

sums = [0]*4                  # initialize list of sums
for i, v in enumerate(lst):   # iterate over indexes and values from the source list
    sums[i%4] += v            # add each value to the appropriate sum

Answer 4

You could you a "for loop" for that. For loops are very commonly used for list management. One that would solve your problem could be like this:

for i in range(len(your_list)-1):
    if i % 4 == 0:
        sum = your_list[i] + sum

Answer 5

Use the following code:

>>> lst = [5, 7, 3, 5, 4, 2, 7, 9, 5, 6, 1, 3, 2, 7, 5, 7, 5, 2, 9, 8, 9, 5, 3, 6, 2]
>>> len(lst)
25
>>> sum1 = sum2 = sum3 = sum4 = 0
>>> for k in range(len(lst)):
...     i = k % 4
...     j = lst[k]    
...     if i == 1:
...             sum1+=j
...     elif i == 2:
...             sum2+=j
...     elif i == 3:
...             sum3+=j
...     elif i == 0:
...             sum4+=j
... 
>>> sum1
29
>>> sum2
28
>>> sum3
38
>>> sum4
32
>>> 

Note: I have truncated lst to a length of 25 for demonstration purposes

We first assign lst to be our desired list. We then make sum1 , sum2 , sum3 , and sum4 integers with a value of 0. We loop over the range of the list, and using modulo, assign i to be the remainder of k/4 .

>>> 5 % 4
1
>>> 10 % 4
2
>>> 23 % 4
3
>>> 16 % 4
0

According to the value of i , we add i to a specific integer. We then print it out.

Answer 6

If you want really fast one. This is the fastest way I think. This will raise Exception ,but there is no problem in the result. If you don't want to finish with Exception, you can use try...except block like myfunc in measurement code below.

it = iter(data_array)
sum1 = 0
sum2 = 0
sum3 = 0
sum4 = 0
while(1):
    sum1 += it.next()
    sum2 += it.next()
    sum3 += it.next()
    sum4 += it.next()

Measurement code:

#my answer
def myfunc(data_array):
    it = iter(data_array)
    sum1 = 0
    sum2 = 0
    sum3 = 0
    sum4 = 0
    while(1):
        try:
            sum1 += it.next()
            sum2 += it.next()
            sum3 += it.next()
            sum4 += it.next()
        except:
            break

#@Blckknght answer
def myfunc2(data_array):
    sums = [0]*4
    for i, v in enumerate(data_array):
        sums[i%4] += v

#@aj8uppal answer
def myfunc3(data_array):
    sum1 = 0
    sum2 = 0
    sum3 = 0
    sum4 = 0
    for k in range(len(data_array)):
        i = k % 4
        j = data_array[k]
        if i == 1:
                sum1+=j
        elif i == 2:
                sum2+=j
        elif i == 3:
                sum3+=j
        elif i == 0:
                sum4+=j

#@Raymond Hettinger answer
from itertools import islice
def myfunc4(data_array):
    sum1 = sum(islice(data_array, 0, None, 4))
    sum2 = sum(islice(data_array, 1, None, 4))
    sum3 = sum(islice(data_array, 2, None, 4))
    sum4 = sum(islice(data_array, 3, None, 4))

#@merlin2011 answer
def myfunc5(data_array):
    sum1 = sum(data_array[::4])
    sum2 = sum(data_array[1::4])
    sum3 = sum(data_array[2::4])
    sum4 = sum(data_array[3::4])

Measurement:

>>> data_array = range(100)

>>> %timeit myfunc(data_array)
100000 loops, best of 3: 12.5 us per loop

>>> %timeit myfunc2(data_array)
100000 loops, best of 3: 14.6 us per loop

>>> %timeit myfunc3(data_array)
100000 loops, best of 3: 16.7 us per loop

>>> %timeit myfunc4(data_array)
10000 loops, best of 3: 33.3 us per loop

>>> %timeit myfunc5(data_array)
10000 loops, best of 3: 56 us per loop