Is there a way to shorten this specific code snippet?

Question

def cross(a, b, w):
    f1 = b - w
    f2 = a - w

    if f1 < 0:
        f1 = f1 * (-1)
    elif f2 < 0:
        f2 = f2 * (-1)
    else:
        pass

    return f1, f2
x = cross(10, 7, 5)
print(x)

I tried to make this more pythonic, e.g. shorten it and combine it with shorted one line if-conditions, though I couldn't find any way.. Does anyone know how to shorten it?

I want to make "both" the f1 and f2 positive if any of them is negative.

Use `return math.abs(b-w), math.abs(a-w)` as a one-liner solution — Yossi Levi, Sep 19 '20 at 16:46
OP's code has an `elif`. If both `f1` and `f2` are negative, only `f1` is turned positive. ALL the answers so far has different behavior. Test with `cross(1,2,5)`. — Aziz, Sep 19 '20 at 16:47
@velotoy, what is it you're trying to shorten? You can remove the `else: pass`, but everything else seems ok. Also, is it your intention to only change the sign of `f1` if both `f1` and `f2` are negative? — Aziz, Sep 19 '20 at 16:49
@Aziz oh I didn't quite catch that! My intentions are to only get positive values, e.g. if either f1 or f2 are negative, output the positive. I also thought of turning it into a string and stripping the "-" to achieve it.. Any opinions? — velotoy, Sep 19 '20 at 16:54
The `elif` part in your code results in `f2` remaining negative if `f1` was positive too. But since you want both to be positive, @VisheshMangla's original answer is correct. — Aziz, Sep 19 '20 at 16:59
if you want you can use this type of comparison in python (false,true)[Decision Check] ((-1,(b-c,c-a))[a — Deepak Tripathi, Sep 19 '20 at 19:26
Does this answer your question? [Does Python have a ternary conditional operator?](https://stackoverflow.com/questions/394809/does-python-have-a-ternary-conditional-operator) — Vishesh Mangla, Sep 27 '20 at 16:09

Vishesh Mangla · Answer 1 · 2020-09-19T17:24:27.120

0

Nothing pythonic here

def cross(a, b, w):
    f1 = b - w
    f2 = a - w

    f1 = f1 if f1>0 else -f1
    f2 = f2 if f2>0 else -f2
    
    return f1, f2

x = cross(10, 7, 5)

print(x)

But better will be just to return the abs value

def cross(a, b, w):
    return list(map(abs, ( b - w,  a - w)))

But if you were talking about if else like logic, well it can be reduced to one liner but it would be just more compact code but not efficient.

def cross(a, b, w):
    f1 = b - w
    f2 = a - w

 
    
    return abs(f1), abs(f2) if f1!= abs(f1) else f2 

x = cross(10, 7, 5)

print(x)

edited Sep 19 '20 at 17:24

answered Sep 19 '20 at 16:38

Vishesh Mangla

664
9
20

The behavior is different. If both `f1` and `f2` are negative, OP's code will only change `f1`. (I didn't downvote your answer, btw) – Aziz Sep 19 '20 at 16:44
oh ok, if else trap – Vishesh Mangla Sep 19 '20 at 16:46
OP's intention was to get both positive anyway. So I rolled back your original answer, since it was correct :) – Aziz Sep 19 '20 at 17:01
Wow, but how did you revert my answer? If I delete something I don't get the answer back. – Vishesh Mangla Sep 19 '20 at 17:12
hey but, w8 this removed my latest answer. That was more appropriate wrt to the code provided, leaving intentions apart. – Vishesh Mangla Sep 19 '20 at 17:15

Nour Alhadi Mahmoud · Answer 2 · 2020-09-19T16:51:51.340

0

Here you go, this will handle the tricky if/else:

def cross(a, b, w):
    f1 = w-b if (b-w) < 0 else b-w
    f2 = w-a if ((a-w) < 0 and (b-w) >= 0) else a-w
    return f1, f2
x = cross(10, 7, 5)
print(x)

edited Sep 19 '20 at 16:51

answered Sep 19 '20 at 16:42

Nour Alhadi Mahmoud

453
2
9

score 0 · Answer 3 · edited Sep 19 '20 at 17:34

0

def cross(a, b, w):
    f1 = abs(b - w)
    f2 = abs(a - w)
    return f1, f2
print(f1,f2)

>>>cross(10,7,5)

>>>output = (2, 5)

edited Sep 19 '20 at 17:34

Vishesh Mangla

664
9
20

answered Sep 19 '20 at 16:45

Chifrijo

66
7

score 0 · Answer 4 · answered Sep 19 '20 at 17:05

0

one-liner solution:

def cross(a, b, w):
    return (abs(b-w),a-w) if b-w<0 else (b-w, abs(a-w)) if a-w<0 else (b-w,a-w)

x = cross(1, 2, 5)
print(x)

output:

(3, -4)

answered Sep 19 '20 at 17:05

Yossi Levi

1,258
1
4
7

Is there a way to shorten this specific code snippet?

4 Answers4