Most efficient way to split python list into sublists using key while keeping order

Question

The first poster is here. If I have a list containing a sublist with different information and I want to group all the subbooks with the same key value into a new sublist (for example, a unique-ify list based on key values), which is the most efficient way to do this ?

Example:

A = [[1,2,3], [1,3,7], [2,1,3], [8,9,6], [3,7,9], [2,3,8], [1,2,4]]

And I want to group all the sublists that have the same element in the index '0' into one new sublist, preferably keeping the original order:

B = [ [[1,2,3], [1,3,7], [1,2,4]], [[2,1,3],[2,3,8]], [[8,9,6]], [[3,7,9]] ] 

I have solved this problem many times, but I would like to know if there is a more efficient way to do this. Yes, I can use the numpy package for this reason, but I'm interested in a list solution.

Typically, I would first have a uniquely-ify element in the first index in each sublist (using the set method, if I'm not worried about efficiency or order), and then iterate over the entire list, combining the signatures into a new sublist

def getUniqueList(list):
    seen = []
    for e in list:
        if e not in seen:
        seen.append(e)
    return seen

def uniquify(list):
    # Get all items from the first column
    new_list = [item[0] for item in list]
    return getUniqueList(new_list)

def rearrangeList(A, A_0):
    B = []
    for i in range(len(A_0)):
        B.append([])
    for i in range(len(A)):
        index = A_0.index(A[i][0])
        B[index].append(A[i])
    return B

A = [[1,2,3], [1,3,7], [2,1,3], [8,9,6], [3,7,9], [2,3,8], [1,2,4]]
A_0 = uniquify(A) # Contains all unique keys in index 0 of list A
B = rearrangeList(A, A_0)
print(B)

This should (I have not tested this, but this is how I encoded it):

B = [[[1,2,3], [1,3,7], [1,2,4]], [[2,1,3], [2,3,8]], [[8 , 9.6]], [[3,7,9]]]

I wanted to know if 1) the best way to do this in terms of efficiency, and 2) is there a way to do this in smaller steps (not necessarily effective, just curious). Hope this question is posed well, but let me know if it needs editing!

Answer 1

, . , OrderedDict :

>>> from collections import OrderedDict
>>> groups = OrderedDict()
>>> A = [[1,2,3], [1,3,7], [2,1,3], [8,9,6], [3,7,9], [2,3,8], [1,2,4]]
>>> key = lambda l: l[0]
>>> for sub in A:
...   groups.setdefault(key(sub),[]).append(sub)
... 
>>> groups
OrderedDict([(1, [[1, 2, 3], [1, 3, 7], [1, 2, 4]]), (2, [[2, 1, 3], [2, 3, 8]]), (8, [[8, 9, 6]]), (3, [[3, 7, 9]])])
>>> B = list(groups.values())
>>> B
[[[1, 2, 3], [1, 3, 7], [1, 2, 4]], [[2, 1, 3], [2, 3, 8]], [[8, 9, 6]], [[3, 7, 9]]]

Answer 2

.

, A B, , , , :

from collections import defaultdict
B=[]
reduce(lambda a,n: a[n[0]].append(n) or a,A,defaultdict(lambda: B.append([]) or B[-1]))

defaultdict, , :

from collections import defaultdict
B=reduce(lambda a,n: a[n[0]].append(n) or a,A,defaultdict(list)).values()

, , .

Answer 3

( ) :

A = [[1,2,3], [1,3,7], [2,1,3], [8,9,6], [3,7,9], [2,3,8], [1,2,4]]
A = sorted(A, key=lambda x: x[0]) #sort just by the first index to preserve order
B = [[A[0]]]

for list in A[1:]:              # add to the last list or create new one
    if B[-1][-1][0]==list[0]:   #
        B[-1].append(list)      #
    else:                       #
        B.append([list])        #

B
#output: 
[[[1, 2, 3], [1, 3, 7], [1, 2, 4]], [[2, 1, 3], [2, 3, 8]], [[3, 7, 9]], [[8, 9, 6]]]