python - numpy how find local minimum in neighborhood on 1darray -


i've got list of sorted samples. they're sorted sample time, each sample taken 1 second after previous one. i'd find minimum value in neighborhood of specified size.

for example, given neighborhood size of 2 , following sample size:

samples = [ 5, 12.3, 12.3, 7, 2, 6, 9, 10, 5, 9, 17, 2 ]

i'd expect following output: [5, 2, 5, 2] best way achieve in numpy / scipy

edited: explained reasoning behind min values:

  1. 5 - 2 number window next [12.3 12.3]. 5 smaller
  2. 2 - left [12.3, 7] right [6 9]. 2 min
  3. 5 - left [9 10] right [9 17]. 5 min

notice 9 isn't min there's 2 window left , right smaller value (2)

use scipy's argrelextrema:

>>> import numpy np >>> scipy.signal import argrelextrema >>> data = np.array([ 5, 12.3, 12.3, 7, 2, 6, 9, 10, 5, 9, 17, 2 ]) >>> radius = 2 # number of elements left , right compare >>> argrelextrema(data, np.less, order=radius) (array([4, 8]),)

which suggest numbers @ position 4 , 8 (2 , 5) smallest ones in within 2 size neighbourhood. numbers @ boundaries (5 , 2) not detected since argrelextrema supports clip or wrap boundary conditions. question, guess interested in them too. detect them, easy add reflect boundary conditions first:

>>> new_data = np.pad(data, radius, mode='reflect') >>> new_data array([ 12.3,  12.3,   5. ,  12.3,  12.3,   7. ,   2. ,   6. ,   9. ,         10. ,   5. ,   9. ,  17. ,   2. ,  17. ,   9. ])

with data corresponding boundary conditions, can apply previus extrema detector:

>>> arg_minimas = argrelextrema(new_data, np.less, order=radius)[0] - radius >>> arg_minimas array([ 0,  4,  8, 11])

which returns positions local extrema (minimum in case since np.less) happens in sliding window of radius=2.

note -radius fix +radius index after wrapping array reflect boundary conditions np.pad.

edit: if insterested in values , not in positions, straight forward:

>>>  data[arg_minimas] array([ 5.,  2.,  5.,  2.])

-

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