So, i simply want to make this faster:
for x in range(matrix.shape[0]):
for y in range(matrix.shape[1]):
if matrix[x][y] == 2 or matrix[x][y] == 3 or matrix[x][y] == 4 or matrix[x][y] == 5 or matrix[x][y] == 6:
if x not in heights:
heights.append(x)
Simply iterate over a 2x2 matrix (usually round 18x18 or 22x22) and check it's x. But its kinda slow, i wonder which is the fastest way to do this.
Thank you very much!
For a numpy based approach, you can do:
np.flatnonzero(((a>=2) & (a<=6)).any(1))
# array([1, 2, 6], dtype=int64)
Where:
a = np.random.randint(0,30,(7,7))
print(a)
array([[25, 27, 28, 21, 18, 7, 26],
[ 2, 18, 21, 13, 27, 26, 2],
[23, 27, 18, 7, 4, 6, 13],
[25, 20, 19, 15, 8, 22, 0],
[27, 23, 18, 22, 25, 17, 15],
[19, 12, 12, 9, 29, 23, 21],
[16, 27, 22, 23, 8, 3, 11]])
Timings on a larger array:
a = np.random.randint(0,30, (1000,1000))
%%timeit
heights=[]
for x in range(a.shape[0]):
for y in range(a.shape[1]):
if a[x][y] == 2 or a[x][y] == 3 or a[x][y] == 4 or a[x][y] == 5 or a[x][y] == 6:
if x not in heights:
heights.append(x)
# 3.17 s ± 59.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
yatu = np.flatnonzero(((a>=2) & (a<=6)).any(1))
# 965 µs ± 11.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
np.allclose(yatu, heights)
# true
Vectorizing with numpy yields to roughly a 3200x speedup
heights was [1,2,6] for this example array a @LaioZattfindHeightsTime = time.time() heights = numpy.flatnonzero(((matrix>=2) & (matrix<=6)).any(1)) heights = heights[::-1] print("findHeightsTime : " + str(time.time() - findHeightsTime)) Maybe im doing something wrong? Thanks again =)It looks like you want to find if 2, 3, 4, 5 or 6 appear in the matrix.
You can use np.isin() to create a matrix of true/false values, then use that as an indexer:
>>> arr = np.array([1,2,3,4,4,0]).reshape(2,3)
>>> arr[np.isin(arr, [2,3,4,5,6])]
array([2, 3, 4, 4])
Optionally, turn that into a plain Python set() for faster in lookups and no duplicates.
To get the positions in the array where those numbers appear, use argwhere:
>>> np.argwhere(np.isin(arr, [2,3,4,5,6]))
array([[0, 1],
[0, 2],
[1, 0],
[1, 1]])
