Question

Very quick question, can’t find an answer with these keywords. What is a better way of doing the following?

t = linspace(0,1000,300)    
x0 = generic_function(t)

x1 = x0[x0>0.8]
t1 = t[t>t[len(x0)-len(x1)-1]]

The operation I’m using @t1 strikes me as very un-pythonic and inefficient. Any pointers?

Answer 1

IIUC, you can simply reuse the cut array. For example:

>>> from numpy import arange, sin
>>> t = arange(5)
>>> t
array([0, 1, 2, 3, 4])
>>> y = sin(t)
>>> y
array([ 0.        ,  0.84147098,  0.90929743,  0.14112001, -0.7568025 ])

As you’ve already done, you can make a bool array:

>>> y > 0.8
array([False,  True,  True, False, False], dtype=bool)

and then you can use this to filter both t and y:

>>> t[y > 0.8]
array([1, 2])
>>> y[y > 0.8]
array([ 0.84147098,  0.90929743])

No use of len or assumptions about monotonicity involved.