如何从python列表中分离连续和非连续数字

Question

Input: lst = [101,102,103,104,120,131,132,133,134]

Expected output: consecutive - [101,102,103,104,131,132,133,134] non consecutive = [120]

def first_consecutive(lst):
    a = []
    for i,j in enumerate(lst,lst[0]):
        if i!=j:
            a.append(j)
    return a

Any suggestion?

Answer 1

You can just keep a list as you loop through the (sorted) numbers. When you encounter something that's consecutive, add it to the list. Otherwise put the values in the right place and redefine the list. Easier to show than explain:

l = [101,102,103,104, 120,131,132,133,134]

def partition_groups(l):
    out = [[], []]
    current = [l[0]]
    for n in l[1:]:
        if current[-1] + 1 == n:
            current.append(n)
        else:
            out[len(current) > 1].extend(current)
            current = [n]
    out[len(current) > 1].extend(current)
    return out
            
nonconsecutive, consecutive = partition_groups(l)      

print(nonconsecutive)
# [120]

print(consecutive)
# [101, 102, 103, 104, 131, 132, 133, 134]

Answer 2

This? (assumes list is already sorted, if not just call lst.sort() first)

(Simply compare each element to its previous (if any) and next (if any) neighbour.)

consec = []
non_consec = []
L = len(lst)
for i in range(L):
    if i>0 and lst[i-1]==lst[i]-1 or i<L-1 and lst[i+1]==lst[i]+1:
        consec.append(lst[i])
    else:
        non_consec.append(lst[i])

Answer 3

You could do something like this

def sort_list(list):
    sorted_list = []
    non_consecutive_list = []

    for i in range(len(list)):
        if list[i] + 1 in list or list[i] - 1 in list:
           sorted_list.append(list[i])
        
        else:
           non_consecutive_list.append(list[i])

    return sorted_list, non_consecutive_list

num_list = [101,102,103,104,120,131,132,133,134]

sorted_list, non_consecutive_list = sort_list(num_list)
print(sorted_list, non_consecutive_list)