Scale each column of islands to their length in a 2D NumPy array… here is a solution to the problem.
Scale each column of islands to their length in a 2D NumPy array
For example, I have a numpy array like this:
array([[0, 0, 0, 1, 0, 1],
[0, 0, 0, 1, 0, 1],
[1, 1, 1, 1, 0, 1],
[1, 0, 0, 0, 1, 1],
[0, 0, 0, 0, 1, 0],
[1, 1, 0, 0, 0, 1]])
I want to find contiguous pixels with value 1 in each column and set these pixel values to the length obtained to get this output:
array([[0, 0, 0, 3, 0, 4],
[0, 0, 0, 3, 0, 4],
[2, 1, 1, 3, 0, 4],
[2, 0, 0, 0, 2, 4],
[0, 0, 0, 0, 2, 0],
[1, 1, 0, 0, 0, 1]])
Thank you for your help
Solution
Method #1
def scaleby_grouplen(ar):
a = ar==1
a1 = np.pad(a, ((1, 1), (0, 0)), 'constant')
a2 = a1.ravel('F')
idx = np.flatnonzero(a2[1:] != a2[:-1])
start, stop = idx[::2], idx[1::2]
id_ar = np.zeros(len(a2), dtype=int)
id_ar[start+1] = 1
idx_ar = id_ar.cumsum()-1
lens = stop - start
out = a*lens[idx_ar].reshape(-1,a.shape[0]+2). T[1:-1]
return out
Method #2
Alternatively, replace the cumsum
part with np.maximum.accumulate
def scaleby_grouplen_v2(ar):
a = ar==1
a1 = np.pad(a, ((1, 1), (0, 0)), 'constant')
a2 = a1.ravel('F')
idx = np.flatnonzero(a2[1:] != a2[:-1])
start, stop = idx[::2], idx[1::2]
id_ar = np.zeros(len(a2), dtype=int)
id_ar[start+1] = np.arange(len(start))
idx_ar = np.maximum.accumulate(id_ar)
lens = stop - start
out = a*lens[idx_ar].reshape(-1,a.shape[0]+2). T[1:-1]
return out
Method #3
Use np.repeat to repeat
the group length and therefore fill in the –
def scaleby_grouplen_v3(ar):
a = ar==1
a1 = np.pad(a, ((1, 1), (0, 0)), 'constant')
a2 = a1.ravel('F')
idx = np.flatnonzero(a2[1:] != a2[:-1])
lens = idx[1::2] - idx[::2]
out = ar.copy()
out. T[a.T] = np.repeat(lens, lens)
return out
sample run-
In [177]: a
Out[177]:
array([[0, 0, 0, 1, 0, 1],
[0, 0, 0, 1, 0, 1],
[1, 1, 1, 1, 0, 1],
[1, 0, 0, 0, 1, 1],
[0, 0, 0, 0, 1, 0],
[1, 1, 0, 0, 0, 1]])
In [178]: scaleby_grouplen(a)
Out[178]:
array([[0, 0, 0, 3, 0, 4],
[0, 0, 0, 3, 0, 4],
[2, 1, 1, 3, 0, 4],
[2, 0, 0, 0, 2, 4],
[0, 0, 0, 0, 2, 0],
[1, 1, 0, 0, 0, 1]])
Benchmarks
Other methods-
from numpy import array
from itertools import chain, groupby
# @timgeb's soln
def chain_groupby(a):
groups = (groupby(col, bool) for col in a.T)
unrolled = ([(one, list(sub)) for one, sub in grp] for grp in groups)
mult = ([[x*len(sub) for x in sub] if one else sub for one, sub in grp] for grp in unrolled)
chained = [list(chain(*sub)) for sub in mult]
result = array(chained). T
return result
Time –
In [280]: np.random.seed(0)
In [281]: a = np.random.randint(0,2,(1000,1000))
In [282]: %timeit chain_groupby(a)
1 loop, best of 3: 667 ms per loop
In [283]: %timeit scaleby_grouplen(a)
100 loops, best of 3: 17.7 ms per loop
In [284]: %timeit scaleby_grouplen_v2(a)
100 loops, best of 3: 17.1 ms per loop
In [331]: %timeit scaleby_grouplen_v3(a)
100 loops, best of 3: 18.6 ms per loop