order must be 'n'o'n'-'n'egative but got 'n'
Package:
numpy
18118

Exception Class:
ValueError
Raise code
""" np.diff(x)
array([1, 1], dtype='timedelta64[D]')
"""
if n == 0:
return a
if n < 0:
raise ValueError(
"order must be non-negative but got " + repr(n))
a = asanyarray(a)
nd = a.ndim
if nd == 0:
raise ValueError("diff requires input that is at least one dimensional")
axis = normalize_axis_index(axis, nd)
com
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Links to the raise (1)
https://github.com/numpy/numpy/blob/5f3c3181b5d8db0e430e5f605cc45c4392f04934/numpy/lib/function_base.py#L1252Ways to fix
The method numpy.diff should be given a non negative order argument.
Code to reproduce the exception:
import numpy as np
x = np.array([1, 2, 4, 7, 0])
np.diff(x, n=-2)
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-24-0c42bcf56fc6> in <module>() 1 import numpy as np 2 x = np.array([1, 2, 4, 7, 0]) ----> 3 np.diff(x, n=-2)
<__array_function__ internals> in diff(*args, **kwargs)
/usr/local/lib/python3.7/dist-packages/numpy/lib/function_base.py in diff(a, n, axis, prepend, append) 1251 if n < 0: 1252 raise ValueError( -> 1253 "order must be non-negative but got " + repr(n)) 1254 1255 a = asanyarray(a)
ValueError: order must be non-negative but got -2
Fixed version of the code:
import numpy as np
x = np.array([1, 2, 4, 7, 0])
np.diff(x, n=2)
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