Extract list element in Python

Question

This is my first python program. I use the below code to generate combinations for a given range.

for k in range(0, items+1):        
    for r in range(0, items+1):        
        if (r-k) > 0:
            res = [x for x in itertools.combinations(range(k, r), r-k)]            
            print res

Let say items=4, the code produce 10 combinations

            #     
            # [(0,)]
            # [(0, 1)]
            # [(0, 1, 2)]
            # [(0, 1, 2, 3)]
            # [(1,)]
            # [(1, 2)]
            # [(1, 2, 3)]
            # [(2,)]
            # [(2, 3)]
            # [(3,)]
            #

My questions are

(a) How can I retrieve each element in each combinations, let say, in [(1, 2, 3)], how can I retrieve value at offset 0 (i.e. 1) ?
(b) How can I store return value from itertools.combinations into a list array in "res" (eg, res[0] = [(0,)] , res[1] = [(0, 1)] ?
(c) Let say I want to use map(), How can I make the value eg [(0, 1)] as key, and assign a random value to this key?

That list comprehension isn't really doing anything for you. Why not simply print itertools.combinations(range(k, r), r - k)? — Ismail Badawi
– Ismail Badawi, Commented Jun 25, 2013 at 6:29
@isbadawi Because itertools.combinations returns a generator — TerryA
– TerryA, Commented Jun 25, 2013 at 6:30

TerryA · Accepted Answer · 2013-06-25 09:14:36Z

4

a) Use indexes:
```
>>> [(1, 2, 3)][0][0]
1
```
b) I don't 100% understand this question, but instead of using a list comprehension as you have done, you can use list(itertools.combinations(...))
c) I think you are misunderstanding what map() does. map(str, [1, 2, 3]) is equivalent to:
```
[str(i) for i in [1, 2, 3]]
```

If you want to give [(0, 1)] a value, you can use a dictionary, but you have to use (0, 1) instead of [(0, 1)] because you would otherwise get TypeError: unhashable type: 'list'. If you want a "random value", I guess you can use the random module:

import random
{(0, 1) : random.randint(1,10)} # This is just an example, of course

To store all the outputs in one list, you can use a massive list comprehension:

>>> [list(itertools.combinations(range(x, i), i-x)) for x in range(0, items+1) for i in range(0, items+1) if (i-x) > 0]
[[(0,)], [(0, 1)], [(0, 1, 2)], [(0, 1, 2, 3)], [(1,)], [(1, 2)], [(1, 2, 3)], [(2,)], [(2, 3)], [(3,)]]

edited Jun 25, 2013 at 9:14

answered Jun 25, 2013 at 6:29

TerryA

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2 Comments

twfx Over a year ago

For (b), I want to store all generated combinations into an array, now, "res" is not an array. I tried res[i] = list(itertools.combinations(range(k, r), r-k)) , where i=0, res = [], but it gives "IndexError: list assignment index out of range".

twfx Over a year ago

arh...Sorry, I am coding python with a C mind. I used res.append(list(itertools.combinations(range(k, r), r-k))) . It works thanks.

Blckknght · Accepted Answer · 2013-06-25 07:59:46Z

I initially posted this as a comment, but I think it's really an answer.

You're misapplying itertools.combinations, and it's getting in the way of your ability to get your data in a convenient way. itertools.combination(range(k, r), r-k) is always going to yield exactly one value, range(k,r). That's unnecessary; you should just use the range directly.

However, the for and if statements you use to produce your k and r values are a combination. Instead of the three levels of statements in your original code, you can use for k, r in itertools.combinations(range(items+1), 2). I suspect you were trying to do this when you put the itertools.combinations call in your code, but it ended up in the wrong place.

So, to finally get around to your actual question: To access your data, you need to put all your items into a single data structure, rather than printing them. A list comprehension is an easy way to do that, especially once you've simplified the logic producing them, as my suggested changes above do:

import itertools

items = 4

results = [range(k, r) for k,r in itertools.combinations(range(items+1),2)]
print results
# prints [[0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [1], [1, 2], [1, 2, 3], [2], [2, 3], [3]]

If you need to access the individual items, you use indexing into the results list, then into its member lists, as necessary.

print results[6]    # prints [1, 2, 3]
print results[6][0] # prints 1

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