比较Python中连续元组列表的第一个元素

Question

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.

Answer 1

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 OrderedDict

od  = 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 defaultdict

od  = 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 ie 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 loop

In [36]: %%timeit                                                               od = defaultdict(list)
for a,b in myList:
    od[a].append(b)
   ....: 
10 loops, best of 3: 33.8 ms per loop

In [37]: %%timeit
dictionary = OrderedDict()
for x, y in myList:
     if x not in dictionary:
        dictionary[x] = [] # new empty list
    dictionary[x].append(y)
   ....: 
10 loops, best of 3: 63.3 ms per loop

In [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.

Answer 2

This feels like a task for a dictionary (if you don't know dictionaries yet, look them up on python.org). This is a very verbose example, so it's not what I'd write in everyday coding, but it's better to be verbose than unclear:

dictionary = collections.OrderedDict()
for x, y in myList:
    if not dictionary.has_key(x):
        dictionary[x] = [] # new empty list
    # append y to that list
    dictionary[x].append(y)

Answer 3

Having thought about this, the most efficient approach is probably this one-liner (assuming dictionary is an empty dict , ie dictionary = {} or dictionary = OrderedDict() like in Padraic' excellent answer ):

for x,y in myList: dictionary.setdefault(x,[]).append(y)

I'm not saying this is the easiest to read approach, but I like it :)

EDIT Ha! Benchmarking proved me wrong; the setdefault approach is slower than the if not dictionary.has_key(x): dictionary[x]=[] approach:

>>> timeit.timeit("for x,y in myList:\n    if not dictionary.has_key(x):\n        dictionary[x]=[]\n    dictionary[x].append(y)", "from collections import OrderedDict\nmyList=[(1,2),(1,3),(
1,4),(1,5),(2,6),(2,7),(2,8),(3,9),(3,10)]\ndictionary=OrderedDict()")
2.2573769092559814
>>> timeit.timeit("for x,y in myList: dictionary.setdefault(x,[]).append(y)", "from collections import OrderedDict\nmyList=[(1,2),(1,3),(1,4),(1,5),(2,6),(2,7),(2,8),(3,9),(3,10)]\ndictiona
ry=OrderedDict()")
3.3534231185913086

Of course, Padraic was still right: his defaultdict approach uses but 0.82s on my machine, so it's faster by a factor of 3.

Also, as Padraic pointed out: dict.has_key(x) has been deprecated, and one should use x in dict instead; however, I couldn't measure a speed difference.

Answer 4

The following should work:

import itertools

myList = [(1,2),(1,3),(1,4),(1,5),(2,6),(2,7),(2,8),(3,9),(3,10)]
print [tuple(x[1] for x in g) for k, g in itertools.groupby(myList, key=lambda x: x[0])]

Which displays:

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