convert a Python 2D list to 3D list, if there are empty elements in 2D list

Please try the script below:

a =[[],[1,2],[0,1],[],[2,2],[],[2,13],[1,4],[6,1],[],[2,7]]
newlist=[]
temp=[]
for element in a:
    if element: # if element is not empty add to temp
        temp.append(element)
    elif temp: # if element is empty and temp is not empty add temp to newlist
        newlist.append(temp)
        temp=[]
if temp: # if temp is not empty add to newlist
    newlist.append(temp)        
print(newlist) 

Using itertools's groupby

from itertools import groupby

lst = []

for i, g in groupby(a, key=lambda x: len(x) >= 1):
    grp = list(g)
    if i and len(grp) >= 1:
        lst.append(grp)

lst:

[[[1, 2], [0, 1]], [[2, 2]], [[2, 13], [1, 4], [6, 1]], [[2, 7]]]

You can do this

def func(a):
    b = [[]]
    for i in (a): # iterate over a
        if i: # if ith element is not empty list
            b[-1].append(i) # add it to the latest list in b
        else: # otherwise append new list to b
            b.append([])
    return b

OR Similarly,

def func(a):
    b, c = [], []
    c = [b[-1].append(i) for i in ([[]] + a ) if (i or b.append([]))]
    return b

Both follow basically the same process,

2nd snippet wastes an extra variable

You can try this.

from collections import deque

a = [[1,2],[0,1],[],[2,2],[],[2,13],[1,4],[6,1],[],[2,7]]

d = deque(a)
if d[0] == []:
    d.popleft()
if len(d) > 0:
    if d[-1] == []:
        d.pop()
tmp = []; b = []
for x in d:
    if x == []:
        b.append(tmp)
        tmp = []
    else:
        tmp.append(x)
b.append(tmp)     
print(b)


 

outputs:

[[[1, 2], [0, 1]], [[2, 2]], [[2, 13], [1, 4], [6, 1]], [[2, 7]]]