列表的python平均值

Question

Example: A list as

x = [1, 5, 3, 6, 12, 55, 68, 24,]

I want a list of averages with increasing number of elements ie

y = [3, 3, 3.75, 5.4...]

the first element in y is average of first two elements, next is avg of 1st 3 elements, next is avg of 1st 4 elements. The actual list contains thousands of elements. Right now doing it by slicing and sum/length. But it takes a long time.

Answer 1

An O(n) solution would be like this

>>>x=[1,2,3,4,5,6,7,8]
>>> ss=x[0]
>>> y=[]
>>> for ind,j in enumerate(x[1:]):
...     ss+=j
...     y.append(ss*1.0/(ind+2))
... 
>>> y
[1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5]

Answer 2

Mathematical induction yields an algorithm:

1. For i == 0 , the average of x[0:0] == 0.0 ,
2. for i == 1 , the average of x[0:1] == x[0] .
3. And finally, for i > 1 the average of x[0:i] is (x[0:i - 1] + x[i]) / float(i) .

Now we have an incremental algorithm to generate our accumulating average :

x = [1, 5, 3, 6, 12, 55, 68, 24,]

def acc_avg(seq):
    acc_sum = 0.0
    for pos, item in enumerate(seq):
        acc_sum += item
        # don't forget the float(), otherwise we might be 
        # running integer division.
        yield acc_sum / float(pos + 1)

for z in y:
    # this calculates the averages on the fly
    print z   

# if we need a list, make one from the generator
y = list(acc_avg(x))