如何在python中获得列表中每10个数字的平均值

如何从列表中获得每10个数字的平均值

下面是示例代码。对每一步重复你的数字列表，例如，在你的案例中为10。计算步骤之间元素的平均值

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]

---

无需创建子列表即可计算平均值：

>>> 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

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]

具有子列表的替代实现：

>>> 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

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]

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注意：这是Python3代码，其中int/int是float。在Python2中，int/int是int。

解释代码应该做什么。输入和预期输出是什么。现在代码中发生了什么使您相信它不起作用。最后，请阅读如何组合一个更好的答案，并相应地编辑您的问题。平均每10个数字意味着平均1-10，然后是11-20，依此类推？每10个数字组合产生3000万个平均值！是的，莫努丁。我可以做average=sumlist[0:10]/10，但对于比这个更大的列表，这不是很有效。太棒了！！非常感谢你。