I have a list of tuples, each containing two elements. The first element of few sublists is common. I want to compare the first element of these sublists and append the second element in one lists. Here is my list:

myList=[(1,2),(1,3),(1,4),(1,5),(2,6),(2,7),(2,8),(3,9),(3,10)]

I would like to make a list of lists out of it which looks something like this:`

NewList=[(2,3,4,5),(6,7,8),(9,10)]

I hope if there is any efficient way.

You can use an OrderedDict to group the elements by the first subelement of each tuple:

myList=[(1,2),(1,3),(1,4),(1,5),(2,6),(2,7),(2,8),(3,9),(3,10)]from collections import OrderedDictod  = OrderedDict()for a,b in myList:od.setdefault(a,[]).append(b)print(list(od.values()))
[[2, 3, 4, 5], [6, 7, 8], [9, 10]]

If you really want tuples:

print(list(map(tuple,od.values())))
[(2, 3, 4, 5), (6, 7, 8), (9, 10)]

If you did not care about the order the elements appeared and just wanted the most efficient way to group you could use a collections.defaultdict:

from collections import defaultdictod  = defaultdict(list)for a,b in myList:od[a].append(b)print(list(od.values()))

Lastly, if your data is in order as per your input example i.e sorted you could simply use itertools.groupby to group by the first subelement from each tuple and extract the second element from the grouped tuples:

from itertools import groupby
from operator import itemgetter
print([tuple(t[1] for t in v) for k,v in groupby(myList,key=itemgetter(0))])

Output:

[(2, 3, 4, 5), (6, 7, 8), (9, 10)]

Again the groupby will only work if your data is sorted by at least the first element.

Some timings on a reasonable sized list:

In [33]: myList = [(randint(1,10000),randint(1,10000)) for _ in range(100000)]In [34]: myList.sort()In [35]: timeit ([tuple(t[1] for t in v) for k,v in groupby(myList,key=itemgetter(0))])
10 loops, best of 3: 44.5 ms per loopIn [36]: %%timeit                                                               od = defaultdict(list)
for a,b in myList:od[a].append(b)....: 
10 loops, best of 3: 33.8 ms per loopIn [37]: %%timeit
dictionary = OrderedDict()
for x, y in myList:if x not in dictionary:dictionary[x] = [] # new empty listdictionary[x].append(y)....: 
10 loops, best of 3: 63.3 ms per loopIn [38]: %%timeit   
od = OrderedDict()
for a,b in myList:od.setdefault(a,[]).append(b)....: 
10 loops, best of 3: 80.3 ms per loop

If order matters and the data is sorted, go with the groupby, it will get even closer to the defaultdict approach if it is necessary to map all the elements to tuple in the defaultdict.

If the data is not sorted or you don't care about any order, you won't find a faster way to group than using the defaultdict approach.