Pick First And Last Int Value From And List And Continue The Sequence Till The End Of List

Input: l1 = [1,2,3,5,7,9,10,11,12,15] Expected output: l2 = [[1,3],[5],[7],[9,12],[15]] In l2, the first nested list is the sequence from 1 to 3, so the first and last values are

Solution 1:

You could use a library, intspan, to get results like what you want.

from intspan import intspan

l1 = [1,2,3,5,7,9,10,11,12,15]
M = []

for tpl in intspan(l1).ranges():
    M.append(tpl)

print(M)

Prints:

[(1, 3), (5, 5), (7, 7), (9, 12), (15, 15)]

To get the same data structure as posted originally:

for a,b in intspan(l1).ranges():
    if a != b:
        M.append([a,b])
    else:
        M.append([a])


[[1, 3], [5], [7], [9, 12], [15]]

Solution 2:

Try this:

l=[]
cnt=0
a+=[-1] #added dummy value

for i in range(1,len(a)):

  if a[i] == a[i-1]+1:
    cnt+=1
  else:
    if cnt > 0  :
      l.append([a[i-1]-cnt,a[i-1]])   
    else:
      l.append([a[i-1]])  
    cnt=0

a=a[:-1] #remove the dummy value

Dummy value is added so that the iteration will loop through all the element of the original list

Solution 3:

This solution is not perfect but it works.

l1 = [1,2,3,5,7,9,10,11,12,15]
l2 = []

start_ind = Nonefor x inrange(len(l1)-1):
    """
    Lets asign a start index from where to make the slice
    """if start_ind == None:
        start_ind = x

    """
    Check the two nearest numbers
    """ifabs(l1[x] - l1[x+1]) != 1:
        """
        If the start index is the same as the end index:
            add first and last values
        Else:
            Only add the first value
        """if start_ind != x:
            l2.append([l1[start_ind], l1[x]])
        else:
            l2.append([l1[start_ind]])

        start_ind = None"""
This is for the last value if it hasn't been added
"""if l1[-1] != l2[-1][-1]:
    l2.append([l1[-1]])

print(l2)

Solution 4:

And another solution :-)

l1 = [1,2,3,5,7,9,10,11,12,15]
l2 = []
sequence = []
for i in range(1, len(l1)+1):
    sequence.append(l1[i-1])
    if i >= len(l1):
        l2.append(sequence)
        breakif l1[i-1] != l1[i] - 1:
        l2.append(sequence)
        sequence = []

print(l2)

Solution 5:

Your problem is the Gaps and Islands problem.

l1 = [1, 2, 3, 5, 7, 9, 43, 44, 10, 11, 12, 1, 1, 7, 9, 42, 170, 170, 170]
l1.sort()
prev_elem = l1[0]
dense_rank = 1
curr_group = prev_elem - dense_rank
islands = [[prev_elem]]
for elem in l1[1:]:
    if elem != prev_elem:
        dense_rank += 1
    group = elem - dense_rank
    if group != curr_group:
        if islands[-1][0] != prev_elem:
            islands[-1].append(prev_elem)
        islands.append([elem])
        curr_group = group
    prev_elem = elem
print(islands)

Output:

[[1, 3], [5], [7], [9, 12], [42, 44], [170]]