Max in a sliding window in a NumPy array

Question

I want to create an array that contains all max()es of a window navigating through a given numpy array. Sorry if this sounds confusing. I will give an example. Input data:

[ 6,4,8,7,1,4,3,5,7,2,4,6,2,1,3,5,6,3,4,7,1,9,4,3,2 ]

My output with a window width of 5 will be as follows:

[     8,8,8,7,7,7,7,7,7,6,6,6,6,6,6,7,7,9,9,9,9     ]

Each number must be a maximum subarray of width 5 of the input array:

[ 6,4,8,7,1,4,3,5,7,2,4,6,2,1,3,5,6,3,4,7,1,9,4,3,2 ]
  \       /                 \       /
   \     /                   \     /
    \   /                     \   /
     \ /                       \ /
[     8,8,8,7,7,7,7,7,7,6,6,6,6,6,6,7,7,9,9,9,9     ]

I did not find a function out of the box in numpy that would do this (but I would not be surprised if it were like this, I did not always think about the terms that numpy developers thought about). I thought about creating a shifted 2D version of my input:

[ [ 6,4,8,7,1,4,3,5,7,8,4,6,2,1,3,5,6,3,4,7,1 ]
  [ 4,8,7,1,4,3,5,7,8,4,6,2,1,3,5,6,3,4,7,1,9 ]
  [ 8,7,1,4,3,5,7,8,4,6,2,1,3,5,6,3,4,7,1,9,4 ]
  [ 7,1,4,3,5,7,8,4,6,2,1,3,5,6,3,4,7,1,9,4,3 ]
  [ 1,4,3,5,7,8,4,6,2,1,3,5,6,3,4,7,1,9,4,3,2 ] ]

np.max(input, 0) . , , ( > 1000000 > 100000 ). .

np.convolve() - , .

, ?

Answer 1

Pandas , DataFrames, :

import pandas as pd

lst = [6,4,8,7,1,4,3,5,7,8,4,6,2,1,3,5,6,3,4,7,1,9,4,3,2]
lst1 = pd.Series(lst).rolling(5).max().dropna().tolist()

# [8.0, 8.0, 8.0, 7.0, 7.0, 8.0, 8.0, 8.0, 8.0, 8.0, 6.0, 6.0, 6.0, 6.0, 6.0, 7.0, 7.0, 9.0, 9.0, 9.0, 9.0]

lst1 int:

[int(x) for x in lst1]

# [8, 8, 8, 7, 7, 8, 8, 8, 8, 8, 6, 6, 6, 6, 6, 7, 7, 9, 9, 9, 9]

Answer 2

№1: 1D max Scipy -

from scipy.ndimage.filters import maximum_filter1d

def max_filter1d_valid(a, W):
    hW = (W-1)//2 # Half window size
    return maximum_filter1d(a,size=W)[hW:-hW]

№ 2: strides: strided_app, 2D , -

def max_filter1d_valid_strided(a, W):
    return strided_app(a, W, S=1).max(axis=1)

-

In [55]: a = np.random.randint(0,10,(10000))

# @Abdou solution using pandas rolling
In [56]: %timeit pd.Series(a).rolling(5).max().dropna().tolist()
1000 loops, best of 3: 999 µs per loop

In [57]: %timeit max_filter1d_valid(a, W=5)
    ...: %timeit max_filter1d_valid_strided(a, W=5)
    ...: 
10000 loops, best of 3: 90.5 µs per loop
10000 loops, best of 3: 87.9 µs per loop

Answer 3

, , , 10- imput 8, , , 2.

, , , , , :

import numpy as np
a=np.array([ 6,4,8,7,1,4,3,5,7,8,4,6,2,1,3,5,6,3,4,7,1,9,4,3,2 ])
window=5
for i in range(0,len(a)-window,1): 
    b[i] = np.amax(a[i:i+window])

, , 2D- imput, , ​​, , imput, , .

Answer 4

Pandas . , ( Python) . ( ). , , - ; , , ; . Python .

, , , . , , , . 1M , rollMax'ed 100k:

         old (slow HW)           new (better HW)
scipy:   0.9.0:  21.2987391949   0.13.3:  11.5804400444
pandas:  0.7.0:  13.5896410942   0.18.1:   0.0551438331604
numpy:   1.6.1:   1.17417216301  1.8.2:    0.537392139435

numpy ringbuffer:

def rollingMax(a, window):
  def eachValue():
    w = a[:window].copy()
    m = w.max()
    yield m
    i = 0
    j = window
    while j < len(a):
      oldValue = w[i]
      newValue = w[i] = a[j]
      if newValue > m:
        m = newValue
      elif oldValue == m:
        m = w.max()
      yield m
      i = (i + 1) % window
      j += 1
  return np.array(list(eachValue()))

, . (, -np.arange(10000000)), (, , ).

, - .