Pythonic way to generate a list of list

Question

What would be the most pythonic way of achieving the transformation from the following input:

 input = [('a', 1), ('a', 10), ('b', 244), ('c', 31) , ('c',45)]

to the desired output:

output = [[('a',1),('a',10)],[('c',31),('c',45)]]

where I have grouped in lists the tuples which have the same first element.

Feeling that Python has a strong potential (I'm new with it) in writing complicated things on one line, I have decided to use comprehensive lists. My initial try is something like:

output = [x for x in input if [k[0] for f in input].count(x[0])>1] 

giving me a nice list of all my "pseudo" duplicates :

output = [('a',1),('a',10),('c',31),('c',45)]

which I further process to obtain my result.

My question is: is there a way to achieve this result in one line using comprehensive lists instead of two (ugly) steps?

Answer 1

Use groupby from itertools and list comprehension. This will give you a simple one liner:

from itertools import groupby

filter(lambda x: len(x)>1, [list(g) for i,g in groupby(input, key=lambda x: x[0])])
[[('a', 1), ('a', 10)], [('c', 31), ('c', 45)]]

Answer 2

Using a 1-liner list comprehension:

>>> L=[('a', 1), ('a', 10), ('b', 244), ('c', 31) , ('c',45)]
>>> [list(filter(lambda x:x[0]==i, L)) for i in set(map(lambda x:x[0], L)) if len(list(filter(lambda x:x[0]==i, L)))>1]
[[('a', 1), ('a', 10)], [('c', 31), ('c', 45)]]

Answer 3

Use itertools.groupby . My solution is not one-liner, but more readable.

import itertools

lists_in = [('a', 1), ('a', 10), ('b', 244), ('c', 31) , ('c',45)]

lists_out = list()
for name, group in itertools.groupby(lists_in, key=lambda x:x[0]):
    l = list(group)
    if len(l) == 2:
        lists_out.extend(l)

print(lists_out)
# Output
[('a', 1), ('a', 10), ('c', 31), ('c', 45)]

Answer 4

There is nothing wrong with the following:

input = [('a', 1), ('a', 10), ('b', 244), ('c', 31) , ('c',45)]

d = {}
for i in input:
  if i[0] in d:
    d[i[0]].append(i)
  else:
    d[i[0]] = [i]

print([d[k] for k in d if len(d[k]) > 1])

Don't forget, you have to mantain a balance between readability and cleverness.

Later edit : I actually gathered the other solutions from other answers and measured time execution (200000 uniformly distributed tuples with 'a'-'z' first element), see below:

# 0.048532 s
def foo(input):
  d = {}
  for i in input:
    if i[0] in d:
      d[i[0]].append(i)
    else:
      d[i[0]] = [i]

  return len(([d[k] for k in d if len(d[k]) > 1]))

# 1.9594 s
def foo2(input):
  [list(filter(lambda x:x[0]==i, input)) for i in set(map(lambda x:x[0], input)) if len(list(filter(lambda x:x[0]==i, input)))>1]

# 0.209639 s
def foo3(input):
  [filter(lambda x: len(x)>1, [list(g) for i,g in itertools.groupby(input, key=lambda x: x[0])])]

# 0.188625
def foo4(input):
  lists = list()
  for name, group in itertools.groupby(input, key=lambda x: x[0]):
    l = list(group)
    if len(l) == 2:
      lists.extend(l)

# didn't even finish, >120 s
def foo5(input_list):
  [[x for x in input_list if x[0]==a] for a in {x[0] for x in input_list if [k[0] for k in input].count(x[0])>1}]

So yes, more clever one-line solutions, but slower and harder to read are not really the "most pythonic".

Answer 5

Here is one solution:

>>> input_list = [('a', 1), ('a', 10), ('b', 244), ('c', 31) , ('c',45)]
>>> [[x for x in input_list if x[0]==a] for a in {x[0] for x in input_list if [k[0] for k in input].count(x[0])>1}]

will print

>>> [[('a', 1), ('a', 10)], [('c', 31), ('c', 45)]]