numpy.diff with parameter n=2 produces strange result-CodePudding

I'm having a hard time understanding the behaviour of np.diff when n>1

The documentation gives the following example :

x = np.array([1, 2, 4, 7, 0])
np.diff(x)
array([ 1,  2,  3, -7])
np.diff(x, n=2)
array([  1,   1, -10])

It seems from the first example that we are substracting each number by the previous one (x[i 1]-x[i]) and all results make sense.

The second time the function is called, with n=2, it seems that we're doing x[i 2]-x[i 1]-x[i] and the two first numbers (1 and 1) in the resulting array make sense but I am surprised the last number is not -11 (0 -7 -4) but -10.

Looking in the documentation I found this explaination

The first difference is given by out[i] = a[i 1] - a[i] along the given axis, higher differences are calculated by using diff recursively.

I fail to understand this 'recursively' so I'd be glad if someone had a clearer explanation !

CodePudding user response：

np.diff(x, n=2) is the same as np.diff(np.diff(x)) (that's what "recursively" means in this case).

CodePudding user response：

"Recursively" in this case simply means it's performing the same operation multiple times, each time on the array resulting from the previous step. So:

x = np.array([1, 2, 4, 7, 0])
output = np.diff(x)

produces

output = [2-1, 4-2, 7-4, 0-7] = [1, 2, 3, -7]

If you use n=2, it simply does the same thing 2 times:

output = np.diff(x, n=2)
# first step, you won't see this result
output = [2-1, 4-2, 7-4, 0-7] = [1, 2, 3, -7]
# and then again (this will be your actual output)
output = [2-1, 3-2, -7-3] = [1, 1, -10]