How can i iteratively apply elements from the 2 lists into a new function?

Question

Suppose I have 2 lists, for which the i-th element of the first lists corresponds to the i-th element of the second list. How can i iteratively apply elements from the 2 lists into a different function?

def GetLists(n):
    List1 = []
    List2 = []
    n = int(input("how many terms: "))
    for i in range(1,n):
        val1 = float(input("what is val1: "))
        val2 = float(input("what is the corresponding val2: "))
        List1.append(val1)
        List2.append(val2)
    return List1, List2

def newfunc(ListA, ListB, var):
    # take i-th elements from lists A and B where (a,b) are the i-th elements of A and B
    # want output = sum(a * var ** b) for index in len(A) if len(A) == len(B)

What is the most pythonic way to do this? If possible, I would like to do so without importing external modules.

Edit: I checked other solutions. The "duplicate" answer requires importing modules; I was trying to do without. Also, I am trying to do an operation that returns an output rather than print values, which complicates the use of zip beyond the level shown in the duplicate answer.

Are you looking for `for a, b in zip(ListA, ListB):`? Also, note that what you've posted isn't Python - case matters. — jonrsharpe, Jan 09 '17 at 08:46
@jonrsharpe Which part isn't python (which part is case-dependent)? Please excuse my ignorance, I am still a beginner. — , Jan 09 '17 at 09:20
`Def` -> `def`, `For` -> `for`. If you're new, **read a tutorial**. And this is a duplicate, `zip` doesn't require an import and you can do whatever you want with `a` and `b` once you have them. — jonrsharpe, Jan 09 '17 at 09:21
Ahhh, my careless mistake. I was typing on a mobile phone. I will edit it now. — , Jan 09 '17 at 09:22
As for being a duplicate, the "original" requires importing itertools. It also prints and does not do an operation on the zipped elements. — , Jan 09 '17 at 09:25
1. No it doesn't, it shows an *option* that uses `itertools`. 2. So what? SO isn't a code-writing service, you are expected to be able to adapt generic answers to your specific needs. — jonrsharpe, Jan 09 '17 at 09:26
Well, I specifically asked for an answer that does not import modules to do this. I could not adapt the methods in the "original" to my problem because I am not trying to just print my result. I don't treat SO as a code-writing service. If you feel so inclined, go to my profile and check out my other 2 most recent posts here to see how many ways I tried to this. — , Jan 09 '17 at 09:31
**The duplicate does not require you to import modules.** `zip` is built-in. Again, it simply shows *options*, alternative approaches, that do use imports (and only from the standard library, which you should *already have available*) and explains under what circumstances you'd use them. I see no point discussing this further. — jonrsharpe, Jan 09 '17 at 09:34
Then I guess we both feel that way. Perhaps we should explore that; it's not clear to me what you do not understand. You keep saying it's not a duplicate because you don't want to import any modules. 1. Why don't you want to import any modules? Are you aware there's a difference between third-party modules and the standard library? What constraints are you operating under? That would be useful context to [edit] in. 2. Even so, as I keep saying, the duplicate doesn't *require* any imports. Why do you think it does? The answer you've actually accepted **uses `zip`, as the duplicate does**. — jonrsharpe, Jan 09 '17 at 09:40
I feel that the best way to learn a program is to learn the basics. I already took a class and want to take it further on my own to apply my skills for physics research. The class I took requires that no modules be imported, and I learned a bit more this way (like how to linspace without numpy). I was not aware of differences between libraries, though now that I know I will look into it (known unknowns vs unknown unknowns). I was under the impression that itertools was an import. The use of zip isn't alien to me, I tried unsuccessfully to use it in an operation instead of printing a result. — , Jan 09 '17 at 09:46
`itertools` *is* an import, but from the [standard library](https://docs.python.org/3/library/), which is shipped with most Python distributions (*"batteries included"*). `numpy` is a third-party module, which you need to *install* separately before importing. — jonrsharpe, Jan 09 '17 at 09:47
That is not say imports are not useful; they are. But I want to explore them after the basics. As proof that I tried zip and did not find it helpful, here is one of my recent posts (mentioned in comment above) --> [proof](http://stackoverflow.com/questions/41522534/how-do-i-combine-the-ith-element-of-2-lists-as-an-input-for-a-function) — , Jan 09 '17 at 09:49
I found your last few comments helpful and will investigate further. Thanks — , Jan 09 '17 at 09:50

score 0 · Accepted Answer · edited Jun 20 '20 at 09:12

0

take i-th elements from lists A and B where (a,b) are the i-th elements of A and B

want output = sum(a * var ** b) for index in len(A) if len(A) == len(B)

Is this what you're looking for? It will zip two lists of the same length together, calculating f(a, b, i) = a[i] * var ** b[i] for some constant var and for each i where 0 <= i < len(a). It then returns the sum.

def process(list_a, list_b, var):
    if len(list_a) != len(list_b):
        raise ValueError('Lists are not equal in length')

    def combine():
        for a, b in zip(list_a, list_b):
            yield a * var ** b

    return sum(combine())

print(process([5, 2, 3], [2, 2, 3], 10))

Output

This output is the result of (1 * 10 ** 2) + (2 * 10 ** 2) + (3 * 10 ** 3).

Edit

The approach above decouples the combination logic - which is the focus of your question - from the summation logic. An alternative approach which is arguably more Pythonic (as per your requirements) and a lot shorter would be to use a generator expression, as suggested in the comments for this answer:

def process(list_a, list_b, var):
    if len(list_a) != len(list_b):
        raise ValueError('Lists are not equal in length')
    
    return sum(a * var ** b for a, b in zip(list_a, list_b))

print(process([1, 2, 3], [2, 2, 3], 10))

Essentially, the expression within sum acts as an anonymous replacement for the generator function combine that I defined in the previous approach.

edited Jun 20 '20 at 09:12

Community

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answered Jan 09 '17 at 08:55

Tagc

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It's a different formula than the one I was looking for but I should be able to apply the same principle to mine. Thank you! Edit: The formula is corrected, works like a charm. – Jan 09 '17 at 08:59
@mikey A little sleepy and misread the formula. Is the formula in my edited post correct? – Tagc Jan 09 '17 at 09:00
Yes, it is correct. – Jan 09 '17 at 09:01
Can I ask what the downvote is for? – Tagc Jan 09 '17 at 09:14
I upvoted your answer because the code works. Must have been someone else. – Jan 09 '17 at 09:18
@mikey No worries, I was addressing that "someone else". :) – Tagc Jan 09 '17 at 09:19
This is a perfect use-case for a [generator expression](https://www.python.org/dev/peps/pep-0289/). simply `return sum(a*var**b for a,b in zip(list_a, list_b))` – juanpa.arrivillaga Jan 09 '17 at 09:58
@juanpa.arrivillaga I considered doing that (and I knew someone would eventually make this remark). In this case I wanted to completely decouple the "zip/combination" logic (focus of question) from the summation logic to make it clearer for OP (hopefully). If you post your approach, I'll upvote it. – Tagc Jan 09 '17 at 10:01
@Tagc it was a suggesting for an addition to your question. I don't think there is any need for an additional answer. – juanpa.arrivillaga Jan 09 '17 at 10:04
@juanpa.arrivillaga Sure thing, done. – Tagc Jan 09 '17 at 10:11
I appreciate seeing both methods. Thanks! – Jan 09 '17 at 11:52

How can i iteratively apply elements from the 2 lists into a new function?

1 Answers1