Generate all subsets of size k (containing k elements) in Python

I have a set of values and would like to create list of all subsets containing 2 elements.

For example, a source set ([1,2,3]) has the following 2-element subsets:

set([1,2]), set([1,3]), set([2,3])

Is there a way to do this in python?

edited Mar 19 '16 at 14:19

omerbp

2,49732038

asked Sep 11 '11 at 12:30

John Manak

7,4152562105

add a comment |

I have a set of values and would like to create list of all subsets containing 2 elements.

For example, a source set ([1,2,3]) has the following 2-element subsets:

set([1,2]), set([1,3]), set([2,3])

Is there a way to do this in python?

edited Mar 19 '16 at 14:19

omerbp

2,49732038

asked Sep 11 '11 at 12:30

John Manak

7,4152562105

add a comment |

I have a set of values and would like to create list of all subsets containing 2 elements.

For example, a source set ([1,2,3]) has the following 2-element subsets:

set([1,2]), set([1,3]), set([2,3])

Is there a way to do this in python?

edited Mar 19 '16 at 14:19

omerbp

2,49732038

asked Sep 11 '11 at 12:30

John Manak

7,4152562105

I have a set of values and would like to create list of all subsets containing 2 elements.

For example, a source set ([1,2,3]) has the following 2-element subsets:

set([1,2]), set([1,3]), set([2,3])

Is there a way to do this in python?

python set tuples subset

edited Mar 19 '16 at 14:19

omerbp

2,49732038

asked Sep 11 '11 at 12:30

John Manak

7,4152562105

edited Mar 19 '16 at 14:19

omerbp

2,49732038

asked Sep 11 '11 at 12:30

John Manak

7,4152562105

edited Mar 19 '16 at 14:19

omerbp

2,49732038

edited Mar 19 '16 at 14:19

omerbp

2,49732038

edited Mar 19 '16 at 14:19

omerbp

2,49732038

asked Sep 11 '11 at 12:30

John Manak

7,4152562105

asked Sep 11 '11 at 12:30

John Manak

7,4152562105

asked Sep 11 '11 at 12:30

John Manak

7,4152562105

add a comment |

3 Answers
3

active

oldest

votes

Seems like you want itertools.combinations:

>>> list(itertools.combinations((1, 2, 3), 2))

[(1, 2), (1, 3), (2, 3)]

If you want sets you'll have to convert them explicitly. If you don't mind an iterable instead of a list, and you're using Python 3, you can use map:

>>> s = set((1, 2, 3))

>>> map(set, itertools.combinations(s, 2))

<map object at 0x10cdc26d8>

To view all the results at once, you can pass the output of map to list. (In Python 2, the output of map is automatically a list.)

>>> list(map(set, itertools.combinations(s, 2)))

[{1, 2}, {1, 3}, {2, 3}]

However, if you know you'll need a list, a list comprehension is marginally better (h/t Jacob Bowyer):

>>> [set(i) for i in itertools.combinations(s, 2)]

[{1, 2}, {1, 3}, {2, 3}]

edited Nov 14 '18 at 16:36

answered Sep 11 '11 at 12:54

senderle

91.6k20165188

Damn it!, by the way your map can be done with a list comp [set(i) for i in itertools.combinations(s, 2))]
– Jakob Bowyer
Sep 11 '11 at 13:13

add a comment |

This is a subset of the power set of {1, 2, 3} (or whatever set) containing all two-element sets.

See the Python itertools documentation and search on the term "powerset" for a general answer to this problem.

answered Sep 11 '11 at 12:59

Alex Reynolds

67k47205295

add a comment |

Just to give another perspective, I looked for a way to iterate all subset of size 2 of {1.....N}, so I put itertools.combinations into test:

import itertools

from time import time





N = 7000

lst = [i for i in xrange(N)]



st = time()

c1 = 0

for x in itertools.combinations(lst, 2):

    c1 += 1

print "combinations: %f" % (time()-st)



st = time()

c2=0

for x in xrange(N):

    for y in xrange(x):

        c2 += 1

print "double loop: %f" % (time()-st)

print "c1=%d,c2=%d" % (c1,c2)



# prints:

#combinations: 4.247000

#double loop: 3.479000

# c1=24496500,c2=24496500

So I guess you should not always turn into the general solution.... If you know in advance the size of the subset you want, it should be more efficient to iterate using for loops.

Also note that you should not iterate over list(itertools.combinations(lst, 2)) since this move creates the list (and much slower than using the generator itself).

edited Mar 20 '16 at 8:05

answered Mar 19 '16 at 13:26

omerbp

2,49732038

2

These two tests don't do the same thing. itertools.combinations actually creates a tuple; your nested loop doesn't create a tuple.
– senderle
Dec 4 '17 at 15:30

2

I did a quick test. If you actually need to create tuples inside the nested loop, it's slower by 50%. Furthermore, if you don't need to use a for loop to process the output of itertools.combinations, you can get a substantial speedup with this generator expression: c3 = sum(1 for pair in itertools.combinations(lst, 2)). That runs about 40% faster than the fastest nested loop. There are always many subtleties to consider when optimizing this kind of code!
– senderle
Dec 4 '17 at 15:43

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

Seems like you want itertools.combinations:

>>> list(itertools.combinations((1, 2, 3), 2))

[(1, 2), (1, 3), (2, 3)]

If you want sets you'll have to convert them explicitly. If you don't mind an iterable instead of a list, and you're using Python 3, you can use map:

>>> s = set((1, 2, 3))

>>> map(set, itertools.combinations(s, 2))

<map object at 0x10cdc26d8>

To view all the results at once, you can pass the output of map to list. (In Python 2, the output of map is automatically a list.)

>>> list(map(set, itertools.combinations(s, 2)))

[{1, 2}, {1, 3}, {2, 3}]

However, if you know you'll need a list, a list comprehension is marginally better (h/t Jacob Bowyer):

>>> [set(i) for i in itertools.combinations(s, 2)]

[{1, 2}, {1, 3}, {2, 3}]

edited Nov 14 '18 at 16:36

answered Sep 11 '11 at 12:54

senderle

91.6k20165188

Damn it!, by the way your map can be done with a list comp [set(i) for i in itertools.combinations(s, 2))]
– Jakob Bowyer
Sep 11 '11 at 13:13

add a comment |

Seems like you want itertools.combinations:

>>> list(itertools.combinations((1, 2, 3), 2))

[(1, 2), (1, 3), (2, 3)]

If you want sets you'll have to convert them explicitly. If you don't mind an iterable instead of a list, and you're using Python 3, you can use map:

>>> s = set((1, 2, 3))

>>> map(set, itertools.combinations(s, 2))

<map object at 0x10cdc26d8>

To view all the results at once, you can pass the output of map to list. (In Python 2, the output of map is automatically a list.)

>>> list(map(set, itertools.combinations(s, 2)))

[{1, 2}, {1, 3}, {2, 3}]

However, if you know you'll need a list, a list comprehension is marginally better (h/t Jacob Bowyer):

>>> [set(i) for i in itertools.combinations(s, 2)]

[{1, 2}, {1, 3}, {2, 3}]

edited Nov 14 '18 at 16:36

answered Sep 11 '11 at 12:54

senderle

91.6k20165188

Damn it!, by the way your map can be done with a list comp [set(i) for i in itertools.combinations(s, 2))]
– Jakob Bowyer
Sep 11 '11 at 13:13

add a comment |

Seems like you want itertools.combinations:

>>> list(itertools.combinations((1, 2, 3), 2))

[(1, 2), (1, 3), (2, 3)]

If you want sets you'll have to convert them explicitly. If you don't mind an iterable instead of a list, and you're using Python 3, you can use map:

>>> s = set((1, 2, 3))

>>> map(set, itertools.combinations(s, 2))

<map object at 0x10cdc26d8>

To view all the results at once, you can pass the output of map to list. (In Python 2, the output of map is automatically a list.)

>>> list(map(set, itertools.combinations(s, 2)))

[{1, 2}, {1, 3}, {2, 3}]

However, if you know you'll need a list, a list comprehension is marginally better (h/t Jacob Bowyer):

>>> [set(i) for i in itertools.combinations(s, 2)]

[{1, 2}, {1, 3}, {2, 3}]

edited Nov 14 '18 at 16:36

answered Sep 11 '11 at 12:54

senderle

91.6k20165188

Seems like you want itertools.combinations:

>>> list(itertools.combinations((1, 2, 3), 2))

[(1, 2), (1, 3), (2, 3)]

If you want sets you'll have to convert them explicitly. If you don't mind an iterable instead of a list, and you're using Python 3, you can use map:

>>> s = set((1, 2, 3))

>>> map(set, itertools.combinations(s, 2))

<map object at 0x10cdc26d8>

To view all the results at once, you can pass the output of map to list. (In Python 2, the output of map is automatically a list.)

>>> list(map(set, itertools.combinations(s, 2)))

[{1, 2}, {1, 3}, {2, 3}]

However, if you know you'll need a list, a list comprehension is marginally better (h/t Jacob Bowyer):

>>> [set(i) for i in itertools.combinations(s, 2)]

[{1, 2}, {1, 3}, {2, 3}]

edited Nov 14 '18 at 16:36

answered Sep 11 '11 at 12:54

senderle

91.6k20165188

edited Nov 14 '18 at 16:36

answered Sep 11 '11 at 12:54

senderle

91.6k20165188

answered Sep 11 '11 at 12:54

senderle

91.6k20165188

answered Sep 11 '11 at 12:54

senderle

91.6k20165188

Damn it!, by the way your map can be done with a list comp [set(i) for i in itertools.combinations(s, 2))]
– Jakob Bowyer
Sep 11 '11 at 13:13

add a comment |

Damn it!, by the way your map can be done with a list comp [set(i) for i in itertools.combinations(s, 2))]
– Jakob Bowyer
Sep 11 '11 at 13:13

Damn it!, by the way your map can be done with a list comp [set(i) for i in itertools.combinations(s, 2))]
– Jakob Bowyer
Sep 11 '11 at 13:13

add a comment |

This is a subset of the power set of {1, 2, 3} (or whatever set) containing all two-element sets.

See the Python itertools documentation and search on the term "powerset" for a general answer to this problem.

answered Sep 11 '11 at 12:59

Alex Reynolds

67k47205295

add a comment |

This is a subset of the power set of {1, 2, 3} (or whatever set) containing all two-element sets.

See the Python itertools documentation and search on the term "powerset" for a general answer to this problem.

answered Sep 11 '11 at 12:59

Alex Reynolds

67k47205295

add a comment |

This is a subset of the power set of {1, 2, 3} (or whatever set) containing all two-element sets.

See the Python itertools documentation and search on the term "powerset" for a general answer to this problem.

answered Sep 11 '11 at 12:59

Alex Reynolds

67k47205295

This is a subset of the power set of {1, 2, 3} (or whatever set) containing all two-element sets.

See the Python itertools documentation and search on the term "powerset" for a general answer to this problem.

answered Sep 11 '11 at 12:59

Alex Reynolds

67k47205295

answered Sep 11 '11 at 12:59

Alex Reynolds

67k47205295

answered Sep 11 '11 at 12:59

Alex Reynolds

67k47205295

answered Sep 11 '11 at 12:59

Alex Reynolds

67k47205295

add a comment |

Just to give another perspective, I looked for a way to iterate all subset of size 2 of {1.....N}, so I put itertools.combinations into test:

import itertools

from time import time





N = 7000

lst = [i for i in xrange(N)]



st = time()

c1 = 0

for x in itertools.combinations(lst, 2):

    c1 += 1

print "combinations: %f" % (time()-st)



st = time()

c2=0

for x in xrange(N):

    for y in xrange(x):

        c2 += 1

print "double loop: %f" % (time()-st)

print "c1=%d,c2=%d" % (c1,c2)



# prints:

#combinations: 4.247000

#double loop: 3.479000

# c1=24496500,c2=24496500

So I guess you should not always turn into the general solution.... If you know in advance the size of the subset you want, it should be more efficient to iterate using for loops.

Also note that you should not iterate over list(itertools.combinations(lst, 2)) since this move creates the list (and much slower than using the generator itself).

edited Mar 20 '16 at 8:05

answered Mar 19 '16 at 13:26

omerbp

2,49732038

2

These two tests don't do the same thing. itertools.combinations actually creates a tuple; your nested loop doesn't create a tuple.
– senderle
Dec 4 '17 at 15:30

2

I did a quick test. If you actually need to create tuples inside the nested loop, it's slower by 50%. Furthermore, if you don't need to use a for loop to process the output of itertools.combinations, you can get a substantial speedup with this generator expression: c3 = sum(1 for pair in itertools.combinations(lst, 2)). That runs about 40% faster than the fastest nested loop. There are always many subtleties to consider when optimizing this kind of code!
– senderle
Dec 4 '17 at 15:43

add a comment |

Just to give another perspective, I looked for a way to iterate all subset of size 2 of {1.....N}, so I put itertools.combinations into test:

import itertools

from time import time





N = 7000

lst = [i for i in xrange(N)]



st = time()

c1 = 0

for x in itertools.combinations(lst, 2):

    c1 += 1

print "combinations: %f" % (time()-st)



st = time()

c2=0

for x in xrange(N):

    for y in xrange(x):

        c2 += 1

print "double loop: %f" % (time()-st)

print "c1=%d,c2=%d" % (c1,c2)



# prints:

#combinations: 4.247000

#double loop: 3.479000

# c1=24496500,c2=24496500

So I guess you should not always turn into the general solution.... If you know in advance the size of the subset you want, it should be more efficient to iterate using for loops.

Also note that you should not iterate over list(itertools.combinations(lst, 2)) since this move creates the list (and much slower than using the generator itself).

edited Mar 20 '16 at 8:05

answered Mar 19 '16 at 13:26

omerbp

2,49732038

2

These two tests don't do the same thing. itertools.combinations actually creates a tuple; your nested loop doesn't create a tuple.
– senderle
Dec 4 '17 at 15:30

2

I did a quick test. If you actually need to create tuples inside the nested loop, it's slower by 50%. Furthermore, if you don't need to use a for loop to process the output of itertools.combinations, you can get a substantial speedup with this generator expression: c3 = sum(1 for pair in itertools.combinations(lst, 2)). That runs about 40% faster than the fastest nested loop. There are always many subtleties to consider when optimizing this kind of code!
– senderle
Dec 4 '17 at 15:43

add a comment |

Just to give another perspective, I looked for a way to iterate all subset of size 2 of {1.....N}, so I put itertools.combinations into test:

import itertools

from time import time





N = 7000

lst = [i for i in xrange(N)]



st = time()

c1 = 0

for x in itertools.combinations(lst, 2):

    c1 += 1

print "combinations: %f" % (time()-st)



st = time()

c2=0

for x in xrange(N):

    for y in xrange(x):

        c2 += 1

print "double loop: %f" % (time()-st)

print "c1=%d,c2=%d" % (c1,c2)



# prints:

#combinations: 4.247000

#double loop: 3.479000

# c1=24496500,c2=24496500

So I guess you should not always turn into the general solution.... If you know in advance the size of the subset you want, it should be more efficient to iterate using for loops.

Also note that you should not iterate over list(itertools.combinations(lst, 2)) since this move creates the list (and much slower than using the generator itself).

edited Mar 20 '16 at 8:05

answered Mar 19 '16 at 13:26

omerbp

2,49732038

Just to give another perspective, I looked for a way to iterate all subset of size 2 of {1.....N}, so I put itertools.combinations into test:

import itertools

from time import time





N = 7000

lst = [i for i in xrange(N)]



st = time()

c1 = 0

for x in itertools.combinations(lst, 2):

    c1 += 1

print "combinations: %f" % (time()-st)



st = time()

c2=0

for x in xrange(N):

    for y in xrange(x):

        c2 += 1

print "double loop: %f" % (time()-st)

print "c1=%d,c2=%d" % (c1,c2)



# prints:

#combinations: 4.247000

#double loop: 3.479000

# c1=24496500,c2=24496500

So I guess you should not always turn into the general solution.... If you know in advance the size of the subset you want, it should be more efficient to iterate using for loops.

Also note that you should not iterate over list(itertools.combinations(lst, 2)) since this move creates the list (and much slower than using the generator itself).

edited Mar 20 '16 at 8:05

answered Mar 19 '16 at 13:26

omerbp

2,49732038

edited Mar 20 '16 at 8:05

answered Mar 19 '16 at 13:26

omerbp

2,49732038

answered Mar 19 '16 at 13:26

omerbp

2,49732038

answered Mar 19 '16 at 13:26

omerbp

2,49732038

2

These two tests don't do the same thing. itertools.combinations actually creates a tuple; your nested loop doesn't create a tuple.
– senderle
Dec 4 '17 at 15:30

2

I did a quick test. If you actually need to create tuples inside the nested loop, it's slower by 50%. Furthermore, if you don't need to use a for loop to process the output of itertools.combinations, you can get a substantial speedup with this generator expression: c3 = sum(1 for pair in itertools.combinations(lst, 2)). That runs about 40% faster than the fastest nested loop. There are always many subtleties to consider when optimizing this kind of code!
– senderle
Dec 4 '17 at 15:43

add a comment |

2

These two tests don't do the same thing. itertools.combinations actually creates a tuple; your nested loop doesn't create a tuple.
– senderle
Dec 4 '17 at 15:30

2

I did a quick test. If you actually need to create tuples inside the nested loop, it's slower by 50%. Furthermore, if you don't need to use a for loop to process the output of itertools.combinations, you can get a substantial speedup with this generator expression: c3 = sum(1 for pair in itertools.combinations(lst, 2)). That runs about 40% faster than the fastest nested loop. There are always many subtleties to consider when optimizing this kind of code!
– senderle
Dec 4 '17 at 15:43

These two tests don't do the same thing. itertools.combinations actually creates a tuple; your nested loop doesn't create a tuple.
– senderle
Dec 4 '17 at 15:30

I did a quick test. If you actually need to create tuples inside the nested loop, it's slower by 50%. Furthermore, if you don't need to use a for loop to process the output of itertools.combinations, you can get a substantial speedup with this generator expression: c3 = sum(1 for pair in itertools.combinations(lst, 2)). That runs about 40% faster than the fastest nested loop. There are always many subtleties to consider when optimizing this kind of code!
– senderle
Dec 4 '17 at 15:43

add a comment |

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