Creating a sum from a list within a list in Python?

Question

I'm not too sure how to ask this, and I'm not all that experienced with programming, so you'll have to forgive me. Anyways, I have an issue. Basically, I need to come up with the sum of some numbers and also the average.

The program is supposed to have the user inputs values. They input a month and a number associated with that month, and then I need to get the average. So I have one big list, and then lists within that list. It basically looks like this:

    months = [["January", 3.45], ["February", 7.1865], ["March", 4.56]]

What I'm wondering is, how do I single out the second element in each list? I was thinking that I could use a for loop and compiling the numbers into a separate list, but I tried that and I couldn't it to calculate correctly.

DSM · Accepted Answer · 2012-11-09 02:51:23Z

6

Might as well make this an answer and not a comment. The zip function works kind of like a zipper, combining matching elements:

>>> zip([1,2,3],[4,5,6])
<zip object at 0xb6e5beec>
>>> list(zip([1,2,3],[4,5,6]))
[(1, 4), (2, 5), (3, 6)]

and with the * operator, which basically turns f(*[a,b]) into f(a,b), we can go backwards:

>>> list(zip(*(zip([1,2,3],[4,5,6]))))
[(1, 2, 3), (4, 5, 6)]

So we can use zip to break apart your lists:

>>> list(zip(*months))
[('January', 'February', 'March'), (3.45, 7.1865, 4.56)]
>>> monthnames, numbers = zip(*months)
>>> monthnames
('January', 'February', 'March')
>>> numbers
(3.45, 7.1865, 4.56)

This is a little less efficient if you only care about one of them, but is a useful idiom to be familiar with.

answered Nov 9, 2012 at 2:51

DSM

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2 Comments

mgilson Over a year ago

Clever. I wouldn't have thought of zip for this one.

mgilson Over a year ago

Might want to make a comment that on py2k, zip(*months) returns a list whereas on py3k, it's an iterator "zip object".

mgilson · Accepted Answer · 2012-11-09 02:56:54Z

6

Nobody's said this one yet:

sum(x[1] for x in monthlist)

As mentioned in the comments below, if you know each element in the monthlist is iterable and has exactly 2 elements, you can make this a little more explicit by unpacking the tuples as you go:

sum(value for month,value in monthlist)

This doesn't create an intermediate list just to pass to sum. That's the beauty of generators. The real beauty here is if monthlist were some sort of lazy iterator (e.g. a file object). Then you could sum over it without storing more than one element in memory at a time:

#code to sum the first column from a file:
with open(file) as f:
    first_col_sum = sum(float(line.split()[0]) for line in f)

edited Nov 9, 2012 at 2:56

answered Nov 9, 2012 at 2:49

mgilson

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2 Comments

DSM Over a year ago

I like the named version too: sum(num for name, num in monthlist), although it hardcodes the structure more than x[1].

mgilson Over a year ago

@DSM -- Added that version as well (thanks). Usually in that case I'd use sum(num for _,num in monthlist), but now I'm starting to rethink that. After all, why not give myself a reminder about what monthlist actually stores?

Blender · Accepted Answer · 2012-11-09 02:40:36Z

5

List comprehensions can help you here:

names = [item[0] for item in months]
numbers = [item[1] for item in months]

If you're using just plain for loops, things get much messier:

names = []
numbers = []

for item in months:
    names.append(item[0])
    numbers.append(item[1])

answered Nov 9, 2012 at 2:40

Blender

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Comments

Tim · Accepted Answer · 2012-11-09 02:40:34Z

3

To extract the second element in each list:

numbers = [i[1] for i in months]

If you want to get the sum of all the numbers:

numbersum = sum(numbers)

answered Nov 9, 2012 at 2:40

Tim

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Comments

Artsiom Rudzenka · Accepted Answer · 2012-11-09 02:44:23Z

1

One more option with itemgetter:

from operator import itemgetter

months = [["January", 3.45], ["February", 7.1865], ["March", 4.56]]
sum(map(itemgetter(1), months)) # what means sum all items where each item is retrieved from month using index 1

answered Nov 9, 2012 at 2:44

Artsiom Rudzenka

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Comments

AlwaysBTryin · Accepted Answer · 2012-11-09 03:21:04Z

0

Adding because I don't see it already.

>>> map(None,*months)
[('January', 'February', 'March'), (3.4500000000000002, 7.1864999999999997, 4.5599999999999996)]

So map(None, *months)[1] is the list you seek.

answered Nov 9, 2012 at 3:21

AlwaysBTryin

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2 Comments

Marcin Over a year ago

This is because map(None, *x) is the same as zip(*x).

DSM Over a year ago

I don't think this will actually work in Python 3. list(map(None, *months)) gives TypeError: 'NoneType' object is not callable; I think they removed the None shortcut.

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