this is probably really easy to do but I am looking to calculate the length of consecutive positive occurrences in a list in python. For example, I have a and I am looking to return b:
a=[0,0,1,1,1,1,0,0,1,0,1,1,1,0]
b=[0,0,4,4,4,4,0,0,1,0,3,3,3,0]
I note a similar question on Counting consecutive positive value in Python array but this only returns consecutive counts but not the length of the belonging group.
Thanks
This is similar to a run length encoding problem, so I've borrowed some ideas from that Rosetta code page:
import itertools
a=[0,0,1,1,1,1,0,0,1,0,1,1,1,0]
b = []
for item, group in itertools.groupby(a):
size = len(list(group))
for i in range(size):
if item == 0:
b.append(0)
else:
b.append(size)
b
Out[8]: [0, 0, 4, 4, 4, 4, 0, 0, 1, 0, 3, 3, 3, 0]
At last after so many tries came up with these two lines.
In [9]: from itertools import groupby
In [10]: lst=[list(g) for k,g in groupby(a)]
In [21]: [x*len(_lst) if x>=0 else x for _lst in lst for x in _lst]
Out[21]: [0, 0, 4, 4, 4, 4, 0, 0, 1, 0, 3, 3, 3, 0]
Here's one approach.
The basic premise is that when in a consecutive run of positive values, it will remember all the indices of these positive values. As soon as it hits a zero, it will backtrack and replace all the positive values with the length of their run.
a=[0,0,1,1,1,1,0,0,1,0,1,1,1,0]
glob = []
last = None
for idx, i in enumerate(a):
if i>0:
glob.append(idx)
if i==0 and last != i:
for j in glob:
a[j] = len(glob)
glob = []
# > [0, 0, 4, 4, 4, 4, 0, 0, 1, 0, 3, 3, 3, 0]
