checking for truthiness on a numpy array but for 2 values

Question

I have a numpy array such as

import numpy as np
x = np.array(range(1, 10))

assuming that x is a 'timeseries' I am testing for truthiness on 2 conditions, if x t is larger than 5 and x t-1 is larger than 5 at the same time

is there a more pythonic way to write this test:

np.where((x[1:] > 5) & (x[0:-1] > 5), 1, 0)
array([0, 0, 0, 0, 0, 1, 1, 1])

I feel like calling x[1:] and x[0:-1] to get a lag value is kind of weird.

any better way?

thanks!

FYI: You don't need to use where(...). The expression (x[1:] > 5) & (x[0:-1] > 5) is a boolean array that you could probably use in place of the where(...) result, or, if you need integers, convert using the .astype(int) method: ((x[1:] > 5) & (x[0:-1] > 5)).astype(int). — Warren Weckesser
– Warren Weckesser, Commented Dec 16, 2016 at 19:53
I wouldn't call your expression "weird"; using shifted slices like that is pretty common in numpy code. There is some inefficiency, because you are repeating the same comparison len(x) - 1 times. For a small array, it might not matter, but if in your actual code x can be much larger, you could do something like: xgt5 = x > 5; result = xgt5[1:] & xgt5[:-1] — Warren Weckesser
– Warren Weckesser, Commented Dec 16, 2016 at 20:02
Yes my real arrays are much larger. I tested for speed, and your solution is about 2times faster. I can accept it as the answer if you want to post it below! — Steven G
– Steven G, Commented Dec 16, 2016 at 20:06

Warren Weckesser · Accepted Answer · 2016-12-16 20:10:02Z

1

I wouldn't call your expression "weird"; using shifted slices like that is pretty common in numpy code. There is some inefficiency, because you are repeating the same comparison len(x) - 1 times. For a small array, it might not matter, but if in your actual code x can be much larger, you could do something like:

xgt5 = x > 5
result = xgt5[1:] & xgt5[:-1]

answered Dec 16, 2016 at 20:10

Warren Weckesser

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