Pythonic way to access numpy array in reverse order

Question

I'm trying to re-write some code that looks like it was written by a FORTRAN programmer to make it more Pythonic/readable. Below is the code snippet of interest. The overall behavior of the code is to store the first three elements of Z into Zstart so long as the values are less than 1, and also store the last three values of Z into Zend as long as they are less than 1 as well.

import numpy as np
nbpoints = 3
Z = np.linspace(0,1.0,10)
Zstart = np.ones(nbpoints)
Zend = np.ones(nbpoints)
Zpts = np.size(Z)
for j in range(nbpoints):
    if Z[j] < Zstart[j]:
       Zstart[j] = Z[j]
    if Z[Zpts - 1 - j] < Zend[nbpoints - 1 - j]:
        Zend[nbpoints - 1 - j] = Z[Zpts - 1 - j]

The counter moving access of Zstart and Zend from both ends is tripping me up a bit. My current solution is the following.

import numpy as np
nbpoints = 3
Z = np.linspace(0,1.0,10)
Zstart = np.ones(nbpoints)
Zend = np.ones(nbpoints)
Zpts = np.size(Z)
for j in range(nbpoints):
    if Z[j] < Zstart[j]:
       Zstart[j] = Z[j]
    if Z[-(j+1)] < Zend[-(j+1)]:
        Zend[-(j+1)] = Z[-(j+1)]

Sample output from this code is:

Z = [ 0.0 0.11111111  0.22222222  0.33333333  0.44444444  0.55555556 
0.66666667  0.77777778  0.88888889  1.0 ]

Zstart = [ 0.0 0.11111111  0.22222222]
Zend = [ 0.77777778  0.88888889  1.0 ]

My solution feels like I'm still just re-writing poorly written code e.g. rearranging chairs on the deck of the Titanic. Is there a more Pythonic way to perform this operation?

@JulienSpronck I added sample output. The input is just the Z array. A one-dimensional numpy array that varies from 0 to 1. In the actual code it is passed in, but for the purposes of this snippet I just used the numpy linspace function to emulate input data. — wandadars
– wandadars, Commented Oct 12, 2017 at 15:46
@wandadars What happens if the starting values are > 1? Do you store only those that are <1, or none at all? — Brad Solomon
– Brad Solomon, Commented Oct 12, 2017 at 16:09
@BradSolomon In the context of the code that this snippet comes from, the values of Z are bound between 0 and 1. As for the initial values of Zstart and Zend, only the original author knows why they did that. — wandadars
– wandadars, Commented Oct 12, 2017 at 16:12
What? If the values in Z are guaranteed to be 0 < value < 1, then why are you testing that they are <1 in the first place? — Brad Solomon
– Brad Solomon, Commented Oct 12, 2017 at 16:14

Brad Solomon · Accepted Answer · 2017-10-12 16:16:08Z

1

You don't need to instantiate Zstart and Zend with np.ones. Just build them directly:

nbpoints = 3
Z = np.linspace(0,1.0,10)
Zstart = Z[:nbpoints][Z[:nbpoints] < 1]
Zend = Z[-nbpoints:][Z[-nbpoints:] < 1]

print(Zstart)
print(Zend)
# [ 0.          0.11111111  0.22222222]    
# [ 0.77777778  0.88888889]

Notice that Zend has only 2 elements here, because the final element in Z is not less than 1.

answered Oct 12, 2017 at 16:16

Brad Solomon

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triphook · Accepted Answer · 2017-10-12 15:59:05Z

1

This code gives the same results without a moving counter

nbpoints = 3
Z=np.linspace(0,1.,10.)               
Zstart = np.ones(nbpoints)            
Zend = np.ones(nbpoints)              

Zstart[Z[:nbpoints] < 1] = Z[:nbpoints][Z[:nbpoints] < 1]  
Zend[Z[-nbpoints:] < 1] = Z[-nbpoints:][Z[-nbpoints:] < 1]

edited Oct 12, 2017 at 15:59

answered Oct 12, 2017 at 15:42

triphook

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