AGreatPigeon - 7 months ago 70

Python Question

On attempting to produce Automatic Peak Detection in Noisy Periodic and Quasi-Periodic Signals, by Felix Scholkmann, Jens Boss and Martin Wolf in Python, I've hit a stumbling block in the implementation.

Upon attempting to optimise, I've noticed that the nested for loops are creating a bottleneck in processing time (taking 115394 ms on average to complete).

Is there a more efficient means of constructing the nested for loop?

N.B:

The parameter, signal, is a list of co-ordinates to which the algorithm will process which is of the form

-48701.0

-20914.0

-1757.0

-49278.0

-106781.0

-88139.0

-13587.0

28071.0

11880.0

-13375.0

-18056.0

-15248.0

-12476.0

-9832.0

-26365.0

-65734.0

-81657.0

-41566.0

6382.0

872.0

-30666.0

-20261.0

17543.0

6278.0

...

The list is 32768 lines long.

The function returns the indexes of the peaks detected to which is processed in another function.

`def ampd(signal):`

s_time = range(1, len(signal)+1)

[fitPolynomial, fitError] = np.polyfit(s_time, signal, 1)

fitSignal = np.polyval([fitPolynomial, fitError], s_time)

dtrSignal = signal - fitSignal

N = len(dtrSignal)

L = math.ceil(N/2.0)-1

creation_start = time.time()

np.random.seed(1969)

LSM = np.random.uniform(0, 2, size=(L, N))

creation_elapsedTime = time.time() - creation_start

print('LSM created in %s ms' % int(creation_elapsedTime * 1000))

loop_start = time.time()

for k in range(1, L):

for i in range(k+2, N-k+1):

if signal[i-1]>signal[i-k-1] and signal[i-1]>signal[i+k-1]:

LSM[k,i] = 0

loop_elapsedTime = time.time() - loop_start

print('Loop completed in %s ms' % int(loop_elapsedTime * 1000))

G = np.sum(LSM, axis=1)

l = min(enumerate(G), key=itemgetter(1))[0]

MLSM = LSM[0:l]

S = np.std(MLSM, ddof=1)

found_indices = np.where(MLSM == ((S-1) == 0))

del LSM

del MLSM

return found_indices[1]

Answer

Here is a solution which uses only one loop

```
for k in range(1, L):
mat=1-((signal[k+1:N-k]>signal[1:N-2*k]) & (signal[k+1:N-k]>signal[2*k+1:N]))
LSM[k,k+2:N-k+1]*=mat
```

it's faster and seems do give the same solutions. You compare slices (as suggested by Ami Tavory), combine the comparisons with a `&`

, which gives a `True/False`

array; with `1-`

operation, you transform it to zeros and ones, the zeros corresponding to where the conditions are met. And lastly you multiply the row by the result.

Source (Stackoverflow)