CODEWITHSUNDEEP
pythonperformancenumpymatrix
Leon Berkers
Leon Berkers

Reputation: 51

Numpy conversion of column values in to row values

I take 3 values of a column (third) and put these values into a row on 3 new columns. And merge the new and old columns into a new matrix A

Input timeseries in col nr3 values in col nr 1 and 2

[x x 1]
[x x 2]
[x x 3]

output : matrix A

[x x 1 0 0 0]
[x x 2 0 0 0]
[x x 3 1 2 3]
[x x 4 2 3 4]

So for brevity, first the code generates the matrix 6 rows / 3 col. The last column I want to use to fill 3 extra columns and merge it into a new matrix A. This matrix A was prefilled with 2 rows to offset the starting position.

I have implemented this idea in the code below and it takes a really long time to process large data sets. How to improve the speed of this conversion 

import  numpy as np

matrix = np.arange(18).reshape((6, 3))

nr=3 
A = np.zeros((nr-1,nr))

for x in range( matrix.shape[0]-nr+1):
    newrow =  (np.transpose( matrix[x:x+nr,2:3] ))
    A = np.vstack([A , newrow])

total= np.column_stack((matrix,A))
print (total)

Upvotes: 3

Views: 118

Answers (1)

Divakar
Divakar

Reputation: 221524

Here's an approach using broadcasting to get those sliding windowed elements and then just some stacking to get A -

col2 = matrix[:,2]
nrows = col2.size-nr+1
out = np.zeros((nr-1+nrows,nr))
col2_2D = np.take(col2,np.arange(nrows)[:,None] + np.arange(nr))
out[nr-1:] = col2_2D

Here's an efficient alternative using NumPy strides to get col2_2D -

n = col2.strides[0]
col2_2D = np.lib.stride_tricks.as_strided(col2, shape=(nrows,nr), strides=(n,n))

It would be even better to initialize an output array of zeros of the size as total and then assign values into it with col2_2D and finally with input array matrix.

Runtime test

Approaches as functions -

def org_app1(matrix,nr):    
    A = np.zeros((nr-1,nr))
    for x in range( matrix.shape[0]-nr+1):
        newrow =  (np.transpose( matrix[x:x+nr,2:3] ))
        A = np.vstack([A , newrow])
    return A

def vect_app1(matrix,nr):    
    col2 = matrix[:,2]
    nrows = col2.size-nr+1
    out = np.zeros((nr-1+nrows,nr))
    col2_2D = np.take(col2,np.arange(nrows)[:,None] + np.arange(nr))
    out[nr-1:] = col2_2D
    return out

def vect_app2(matrix,nr):    
    col2 = matrix[:,2]
    nrows = col2.size-nr+1
    out = np.zeros((nr-1+nrows,nr))
    n = col2.strides[0]
    col2_2D = np.lib.stride_tricks.as_strided(col2, \
                        shape=(nrows,nr), strides=(n,n))
    out[nr-1:] = col2_2D
    return out

Timings and verification -

In [18]: # Setup input array and params
    ...: matrix = np.arange(1800).reshape((60, 30))
    ...: nr=3
    ...: 

In [19]: np.allclose(org_app1(matrix,nr),vect_app1(matrix,nr))
Out[19]: True

In [20]: np.allclose(org_app1(matrix,nr),vect_app2(matrix,nr))
Out[20]: True

In [21]: %timeit org_app1(matrix,nr)
1000 loops, best of 3: 646 µs per loop

In [22]: %timeit vect_app1(matrix,nr)
10000 loops, best of 3: 20.6 µs per loop

In [23]: %timeit vect_app2(matrix,nr)
10000 loops, best of 3: 21.5 µs per loop

In [28]: # Setup input array and params
    ...: matrix = np.arange(7200).reshape((120, 60))
    ...: nr=30
    ...: 

In [29]: %timeit org_app1(matrix,nr)
1000 loops, best of 3: 1.19 ms per loop

In [30]: %timeit vect_app1(matrix,nr)
10000 loops, best of 3: 45 µs per loop

In [31]: %timeit vect_app2(matrix,nr)
10000 loops, best of 3: 27.2 µs per loop

Upvotes: 2

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