如何从前一个整数中减去数组中的每个整数并求和

Question

I have a numpy array that has the size (122,) and is made up of integers

I want to use the following formula:

I tried doing

value=sum((I[:-1]-I[1:])**2))

This line itself does not create an error, but whenever I try to use this value in the bigger formula it gives an invalid syntax error.

I want to use this value to plug into a bigger formula

I would like an answer of how to plug my array into this formula For instance, the code for the second term in the bigger formula, I have written as follows:

calc = np.linalg.norm((RR_intervals[:-1])-(RR_intervals[1:]))                                                                     

p=(n-1)*sqrt(2)
o=(1/p)*calc
t=o**2

If I try and combine this with the first term I get an invalid syntax error

Full interpreter message: runfile('/home/user_1/p01.py', wdir='/home/user_1') Traceback (most recent call last):

File "/home/user_1/.local/lib/python3.6/site-packages/IPython/core/interactiveshell.py", line 3291, in run_code exec(code_obj, self.user_global_ns, self.user_ns)

File "", line 1, in runfile('/home/user_1/p01.py', wdir='/home/user_1')

File "/usr/lib/python3/dist-packages/spyder/utils/site/sitecustomize.py", line 705, in runfile execfile(filename, namespace)

File "/usr/lib/python3/dist-packages/spyder/utils/site/sitecustomize.py", line 102, in execfile exec(compile(f.read(), filename, 'exec'), namespace)

File "/home/user_1/p01.py", line 91 dispersion = (sqrt(mt)) ^ SyntaxError: invalid syntax

Answer 1

You can do using list :

a = list(numpy_array)
sum([(i-j)**2 for i,j in zip(a,a[1:])])

Answer 2

这应该这样做：

result = sum((ar[i] - ar[i + 1])**2 for i in range(len(ar) - 1))

Answer 3

Using numpy:

(((a-np.roll(a,1))**2)[1:]).sum()

Note: The [1:] is necessary because roll reinserts the element at the beginning of the array.