Finding Patterns in a Numpy Array

Question

I am trying to find patterns in a numpy array, called values. I\'d like to return the starting index position of the pattern. I know I

Answer 1

If the input is random Ed Smith solution is faster. But if you has a few set of available values this hash-solution can help:

"""
Can be replaced with any revertable hash
"""
def my_hash(rem, h, add):
    return rem^h^add

"""
Imput
"""
values = np.array([0,1,2,1,2,4,5,6,1,2,1])
searchval = [1,2]


"""
Prepare
"""
sh = 0
vh = 0
ls = len(searchval)
lv = len(values)

for i in range(0, len(searchval)):
    vh = my_hash(0, vh, values[i])
    sh = my_hash(0, sh, searchval[i])

"""
Find matches
"""
for i in range(0, lv-ls):
    if sh == vh:
        eq = True
        for j in range(0, ls):
            if values[i+j] != searchval[j]:
                eq = False
                break
        if eq:
            print i
    vh = my_hash(values[i], vh, values[i+ls])

Answer 2

Here's a straight forward approach to using where. Start with a logical expression that finds the matches:

In [670]: values = np.array([0,1,2,1,2,4,5,6,1,2,1])
     ...: searchval = [1,2]
     ...: 
In [671]: (values[:-1]==searchval[0]) & (values[1:]==searchval[1])
Out[671]: array([False,  True, False,  True, False, False, False, False,  True, False], dtype=bool)
In [672]: np.where(_)
Out[672]: (array([1, 3, 8], dtype=int32),)

That could be generalized into a loop that operates on multiple searchval. Getting the slice range correct will take some fiddling. The roll suggested in another answer might be easier, but I suspect a bit slower.

As long as searchval is small compared to values this general approach should be efficient. There is a np.in1d that does this sort of match, but with a or test. So it isn't applicable. But it too uses this iterative approach is the searchval list is small enough.

Generalized slicing

In [716]: values
Out[716]: array([0, 1, 2, 1, 2, 4, 5, 6, 1, 2, 1])
In [717]: searchvals=[1,2,1]
In [718]: idx = [np.s_[i:m-n+1+i] for i in range(n)]
In [719]: idx
Out[719]: [slice(0, 9, None), slice(1, 10, None), slice(2, 11, None)]
In [720]: [values[idx[i]] == searchvals[i] for i in range(n)]
Out[720]: 
[array([False,  True, False,  True, False, False, False, False,  True], dtype=bool),
 array([False,  True, False,  True, False, False, False, False,  True], dtype=bool),
 array([False,  True, False, False, False, False,  True, False,  True], dtype=bool)]
In [721]: np.all(_, axis=0)
Out[721]: array([False,  True, False, False, False, False, False, False,  True], dtype=bool)
In [722]: np.where(_)
Out[722]: (array([1, 8], dtype=int32),)

I used the intermediate np.s_ to look at the slices and make sure they look reasonable.

as_strided

An advanced trick would be to use as_strided to construct the 'rolled' array and perform a 2d == test on that. as_strided is neat but tricky. To use it correctly you have to understand strides, and get the shape correct.

In [740]: m,n = len(values), len(searchvals)
In [741]: values.shape
Out[741]: (11,)
In [742]: values.strides
Out[742]: (4,)
In [743]: 
In [743]: M = as_strided(values, shape=(n,m-n+1),strides=(4,4))
In [744]: M
Out[744]: 
array([[0, 1, 2, 1, 2, 4, 5, 6, 1],
       [1, 2, 1, 2, 4, 5, 6, 1, 2],
       [2, 1, 2, 4, 5, 6, 1, 2, 1]])
In [745]: M == np.array(searchvals)[:,None]
Out[745]: 
array([[False,  True, False,  True, False, False, False, False,  True],
       [False,  True, False,  True, False, False, False, False,  True],
       [False,  True, False, False, False, False,  True, False,  True]], dtype=bool)
In [746]: np.where(np.all(_,axis=0))
Out[746]: (array([1, 8], dtype=int32),)

Answer 3

I think this does the job:

np.where((values == 1) & (np.roll(values,-1) == 2))[0]

Answer 4

Couldn't you simply use np.where (assuming this is the optimal way to find an element) and then only check pattens which satisfy the first condition.

import numpy as np
values = np.array([0,1,2,1,2,4,5,6,1,2,1])
searchval = [1,2]
N = len(searchval)
possibles = np.where(values == searchval[0])[0]

solns = []
for p in possibles:
    check = values[p:p+N]
    if np.all(check == searchval):
        solns.append(p)

print(solns)

Answer 5

Compact straitforward solution will be a "legal" variant of as_strided solution. Others have mentioned np.roll. But here is an universal solution with the only circle (132 µs).

seq = np.array([0,1,2,1,2,4,5,6,1,2,1])
patt = np.array([1,2])

Seq = np.vstack(np.roll(seq, shift) for shift in -np.arange(len(patt))).T
where(all(Seq == patt, axis=1))[0]

The another option for sequences with small integers will be converting to string. It is faster per near 6 times (20 µs). For small positive integers only!

import re

def to_string(arr):
    return ''.join(map(chr, arr))

array([m.start() for m in re.finditer(to_string(patt), to_string(seq))])