Python中绝对差异的平均值

I am trying to write a programme in Python 3which calculates the mean of the absolute differences between successive values.

我正在尝试用Python 3编写一个程序，它计算连续值之间绝对差值的平均值。

4 个解决方案

#1

EDIT: As the code was removed from the question, updating answer, moving the issues with code to bottom.

编辑：当代码从问题中删除，更新答案，将代码问题移到底部。

As given in comments you can use enumerate() to get the index as well as element from the array and then use that to calculate the mean. Example -

如注释中所示，您可以使用enumerate（）来获取索引以及数组中的元素，然后使用它来计算平均值。示例 -

>>> def absolute_difference(v):
...     sum_diff = 0
...     for i,x in enumerate(v):
...         if i+1 < len(v):
...             sum_diff += abs(v[i+1] - x)
...     r = sum_diff/len(v)
...     return r
...
>>> absolute_difference([4.0, 4.2, 4.1, 4.4, 4.3])
0.1400000000000004

Lots of issues in the code (that you seem to have removed) -

代码中的很多问题（您似乎已删除） -

Why are you converting your absolute difference to float ? float mathematics is not accurate , as you can see from the sum of difference in your code - 0.20000000000000018 . In your case you do not need to convert them to float.

你为什么把你的绝对差异转换为浮动？浮动数学是不准确的，你可以从代码中的差异总和看出 - 0.20000000000000018。在您的情况下，您不需要将它们转换为浮动。
The main issue of 0.0 for r occurs because you are using // to divide, // truncates the division to the nearest integer, so diving 7.0 by something grater than that using // would always result in 0.0 . Example -

出现0.0的主要问题是因为你使用//来划分，//将除法截断为最接近的整数，所以以比使用//更大的方式跳过7.0将总是得到0.0。示例 -
```
>>> 7.0 // 8.0
0.0
>>> 7.0/8.0
0.875
```
For your case, you should divide using / .

对于您的情况，您应该使用/来划分。
You are taking the mean in each iteration of the loop, though that is not an issue , it may not be completely needed. If you do not want to take the mean in each iteration of the loop, you should indent it outside the loop.

你在循环的每次迭代中取平均值，虽然这不是问题，但可能并不完全需要。如果你不想在循环的每次迭代中取平均值，你应该在循环之外缩进它。

#2

You are using // which means integer division in python 3

你正在使用//这意味着python 3中的整数除法

That is

那是

i.e)

即）

2/4 =0.5

2//4=0

Just replace the // with / when calculating r

只需在计算r时用//替换//

#3

Here is another approach:

这是另一种方法：

def absolute_difference(values):
    last = values[0]
    total = 0

    for value in values[1:]:
        total += abs(value - last)
        last = value

    return total/(len(values)-1)

print('{:.5f}'.format(absolute_difference([4.0, 4.2, 4.1, 4.4, 4.3])))

Giving the answer: 0.17500

给出答案：0.17500

#4

Also, to prevent None from appearing at the end, you must have return at the end of your definition. This happens if you make another variable "equal to" (=) your definition. This was shown in the other posts, but I'm stating this just to highlight things out.

另外，为了防止None出现在最后，您必须在定义的末尾返回。如果您将另一个变量“等于”（=）定义，则会发生这种情况。这在其他帖子中有所体现，但我说这只是为了强调一些事情。

#1