CODEWITHSUNDEEP
arraysnumpy
diegus
diegus

Reputation: 1188

compare consecutive values of an array numpy

I have a vector A and I want to find all the "i" values of the array which are bigger than the neighboring values, i.e. right before (i-1) and right after (i+1). I wrote a simple code (surely with mistakes) that should print the array (y_maxtab) containing the "i" values. I try to run and It gets stacked. Could you help me on that? 

import numpy as np

y_maxtab = []


A=np.array([2.0,2.5,1.7,5,7,8,9,3,5,3,7,8])

i=1

while i <= len(A):
   if A[i] > A[i-1] and a[i] > A[i+1]:
         y_maxtab.append(A[i])
         i=i+1
print y_maxtab

Upvotes: 0

Views: 809

Answers (2)

wflynny
wflynny

Reputation: 18521

Well, your code as is has some problems.

  1. If the first if statement fails, you will never increment i, and will be stuck in an infinite loop.
  2. The second clause of the if statement has an undefined reference (a).
  3. The while statement will allow values of i which will cause IndexErrors with A[i] and A[i+1].

If you are committed to keeping the code in the same form, you could change it so that it looks like

i = 1
while i < len(A)-1:
    if A[i] > A[i-1] and a[i] > A[i+1]:
        y_maxtab.append(A[i])
    i=i+1
print y_maxtab
# [2.5, 9.0, 5.0]

However, a more pythonic way would be something like

for first, middle, last in zip(A[:], A[1:], A[2:]):
    if first < middle > last:
        y_maxtab.append(middle)
print y_maxtab
# [2.5, 9.0, 5.0]

Upvotes: 1

Eelco Hoogendoorn
Eelco Hoogendoorn

Reputation: 10759

Here is a solution that uses numpy directly. Far faster and cleaner than a python loop, and moreover, it does what you intend it to.

local_maximum = (A[:-2]<A[1:-1]) & (A[2:]<A[1:-1])
print A[1:-1][local_maximum]

Or slightly more verbose, but perhaps more readable:

left, mid, right = A[:-2], A[1:-1], A[2:]
local_maximum = (left<mid) & (right<mid)
print mid[local_maximum]

Upvotes: 1

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