Let's consider the following array:
x = np.array(["john", "john", "ellis", "lambert", "john"])
Is there a way to compare every element of the array to the previous and return a boolean array.
In the present example, the result would be [True,False,False,False].
Is there any function (similar as np.diff) to achieve that?
You can do this with indexing:
array[:-1] == array[1:]
The first item in the list cannot be compared with a previous value and should likely default to np.nan.
To maintain the same shape as the original array:
>>> np.concatenate([np.array([np.nan]), x[:-1] == x[1:]])
array([ nan, 1., 0., 0., 0.])
The inclusion of nan changes to type to floats.
Using indexing, you can do this with ease.
import numpy as np
x = np.array(["john", "john", "ellis", "lambert", "john"])
print x[1:] == x[:-1]
The statement x[1:] == x[:-1] works because the == operand returns Boolean values, and in this case presented in an array due to the types of the two compared items.
x[1:] represents all the elements in the array except item 0, the first item, while x[:-1] represents all the elements in the array except the last item.
x[1:] == x[:-1]?listfor this...