Combine strings. Count how many indices (from list) are in original strings. Python

Question

junctions = [2,9,15,20]

seq_1 = 'sauron'
seq_2 = 'corrupted'
seq_3 = 'numenor'
combined = 'sauroncorruptednumenor' #seq_1 + seq_2 + seq_3

count_1 = 1
count_2 = 1
count_3 = 2

I have a list of 3 strings (seq_1-3). I combine them to create 1 long string (combined) I have a list of indices (junctions). I have 3 different counters set to zero for each string (count_1-3)

What I am trying to do is find the position of each junction [2,9,15,20] in the combined sequence . . . if it is from seq_1 --> count_1 += 1, if it is from seq_2 --> count_2 += 1, from seq_3 --> count_3 += 1

example

junctions = [2,9,15,20]
count_1 = 0
count_2 = 0
count_3 = 0
combined = 'sauroncorruptednumenor'
seq_1 = 'sauron' #index 2 would be on 'u' in combined but originally from seq_1 so count_1 = count_1 + 1 
seq_2 = 'corrupted' #index 9 would be on 'r' in combined so count_2 += 1
seq_3 = 'numenor' #index 15 would be 'n' in combined so count_3 += 1, and 20 would be 'o' so count_3 += 1

let me know if i need to clarify any differently

Answer 1

You can use collections.Counter and bisect.bisect_left here:

>>> from collections import Counter
>>> import bisect
>>> junctions = [2,9,15,20]
>>> seq_1 = 'sauron'
>>> seq_2 = 'corrupted'
>>> seq_3 = 'numenor'
>>> lis  = [seq_1, seq_2, seq_3]

Create a list containing the indexes at which at each seq_ ends:

>>> start = -1
>>> break_points = []
for item in lis:
    start += len(item) 
    break_points.append(start)
...     
>>> break_points
[5, 14, 21]

Now we can simply loop over junctions and find each junction's position in the break_points list using bisect.bisect_left function.

>>> Counter(bisect.bisect_left(break_points, jun)+1  for jun in junctions)
Counter({3: 2, 1: 1, 2: 1})

Better output using collections.defaultdict :

>>> from collections import defaultdict
>>> dic = defaultdict(int)
for junc in junctions:
    ind = bisect.bisect_left(break_points, junc) +1
    dic['count_'+str(ind)] += 1
...     
>>> dic
defaultdict(<type 'int'>,
{'count_3': 2,
 'count_2': 1,
 'count_1': 1})

#accessing these counts
>>> dic['count_3']
2

Answer 2

You could try something basic like

L_1 = len(seq_1)
L_2 = len(seq_2)
L_3 = len(seq_3)

junctions = [2, 9, 15, 20]
c_1, c_2, c_3 = (0, 0, 0)

for j in junctions:
    if j < L_1:
        c_1 += 1
    elif j < L_1 + L_2:
        c_2 += 1
    elif j < L_1 + L_2 + L_3:
        c_3 += 1
    else:
        Raise error

Answer 3

Could use collections.Counter , and repeat and chain from itertools, eg:

from itertools import chain, repeat
from operator import itemgetter
from collections import Counter

junctions = [2,9,15,20]
seq_1 = 'sauron'
seq_2 = 'corrupted'
seq_3 = 'numenor'

indices = list(chain.from_iterable(repeat(i, len(j)) for i, j in enumerate([seq_1, seq_2, seq_3], start=1)))
print Counter(itemgetter(*junctions)(indices))
# Counter({3: 2, 1: 1, 2: 1})