如何獲得Python中列表中每10個數字的平均值

Question

list [] 
list contains [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 
19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30] 

如何從列表中獲得每10個數字的平均值？

Answer 1

下面是示例代碼。遍歷您的案例中每個step即10的數字列表。計算步驟之間元素的average。

>>> my_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30] 
>>> step = 10 
>>> for i, _ in enumerate(my_list[::step]): 
...  sub_list = my_list[i*10:] if (i+1)*10 > len(my_list) else my_list[i*10:(i+1)*10] # Condition if the len(my_list) % step != 0 
...  print sum(sub_list)/float(len(sub_list)) # Dividing by float' to get decimal value as average (Not needed in Python 3) 
... 
5.5 
15.5 
25.5 

Answer 2

的平均值可被計算，而不創建子列表：

testdata = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 
     19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30] 

LEN = 10 

def avg(data): 
    datasum = cnt = 0 
    for num in data: 
     datasum += num 
     cnt += 1 
     if cnt == LEN: 
      yield datasum/LEN 
      datasum = cnt = 0 
    if cnt: 
     yield datasum/cnt 

print(list(avg(testdata))) 
# [5.5, 15.5, 25.5] 

---

另一個實現與子列表：

def avg(data): 
    for i in range((len(data) + LEN - 1) // LEN): 
     sublist = data[i*LEN:(i+1)*LEN] 
     yield sum(sublist)/len(sublist) 

注意：這是Python3代碼其中int/int是浮動。在Python2中int/int是int。