Using abs() in a list in Python

Question

I am trying to create a list of integers and then scan it in order to find the minimum absolute value of the substractions of the elements of the list. I have created the list, but there is problem in the code which finds the minimum absolute value, as the result it shows is not correct. I think it is probably in the possitions of the elements of the list during the loops. Can you help me find it?

For example, when I create a list Α = [2, 7, 5, 9, 3, 1, 2], the result of min should be 0, but it is 1.

Here is my code:

min=1000
for i in range (1, N-1):
    for j in range (i+1, N):
        if (abs (A [i-1] - A [j-1])<min):
            min = abs (A [i-1] - A [j-1])
print ("%d" %min)

Python indexing starts from `0`. `range(1, N - 1)` starts from `1`. — Cong Ma, Feb 09 '18 at 10:43
The minimum *difference* is 0 in this case, because `2-2 == 0`. — quamrana, Feb 09 '18 at 10:43
How does N relate to the length of your list (A)? I'm guessing that you may be excluding a check of the last element, where `2-2` would be 0, because `range(x,y)` excludes y. — Andy G, Feb 09 '18 at 10:46
@CongMa: that is not an error in Mike's code. He uses `A[i-1]` and so starting at 0 would throw an error. — Jongware, Feb 09 '18 at 10:50
I'm pretty sure you're never reaching the last element, debug your code to see what's going on. — Oleg, Feb 09 '18 at 10:50
The answers so far are offering alternative approaches but not explaining the problem with the OP's code. — Andy G, Feb 09 '18 at 10:52
@AndyG N is the number of elements of the list which is given by the user. There at the for loop I wrote range(1, N+1) and I placed each element at A[i-1]. — Mike Johnson, Feb 09 '18 at 10:56

Ma0 · Answer 1 · 2018-02-09T11:55:08.350

5

You can do it like this:

A = [2, 7, 5, 9, 3, 1, 2]

temp = sorted(A)
min_diff = min([abs(i - j) for i, j in zip(temp [:-1], temp [1:])])

print(min_diff)  # -> 0

Sorting makes sure that the element pair (i, j) which produce the overall smallest difference would be a pair of consecutive elements. That makes the number of checks you have to perform much less than the brute force approach of all possible combinations.

Something a bit more clever that short-circuits:

A = [2, 7, 5, 9, 3, 1, 2]


def find_min_diff(my_list):
    if len(set(my_list)) != len(my_list):  # See note 1
        return 0
    else:
        temp = sorted(my_list)
        my_min = float('inf')
        for i, j in zip(temp [:-1], temp [1:]):
            diff = abs(i - j)
            if diff < my_min:
                my_min = diff
        return my_min

print(find_min_diff(A))  # -> 0

Notes:

1: Converting to set removes the duplicates so if the corresponding set has less elements than the original list it means that there is at least one duplicate value. But that necessarily means that the min absolute difference is 0 and we do not have to look any further.

I would be willing to bet that this is the fastest approach for all lists that would return 0.

edited Feb 09 '18 at 11:55

answered Feb 09 '18 at 10:48

Ma0

15,057
4
35
65

Sorting itself incurs a number of checks ;) – Andy G Feb 09 '18 at 10:54
@AndyG True but I would argue it is faster. – Ma0 Feb 09 '18 at 10:54
3

"As a result, you can traverse the list just once instead of twice." I find this a bit misleading, though, as this does not reduces the complexity from O(2n) to O(n), but from O(n²/2) to O(nlogn+n) – tobias_k Feb 09 '18 at 10:55
@tobias_k I was in the process of updating that part; you are totally right. – Ma0 Feb 09 '18 at 10:56
Sorting is good, numpy is better: Ev. Kounis: 131 µs, Espoir: 75 µs, jp_data: 71 ms, quamrana: 225 ms (A an array of 1000 random integers) - Oh, just noticed, that the numpy solution now also uses pre-sorting. So sorting was the major advance here. – Mr. T Feb 09 '18 at 11:40
1

It is. The second version clocks in at 26 µs (the same array as before) – Mr. T Feb 09 '18 at 11:56
Oh, wait. This is not a fair comparison, since you shortcut 0 and an array with 1000 random integers between zero and 100000 rarely has only unique elements. – Mr. T Feb 09 '18 at 12:01
1

@Piinthesky And why is that not fair? Short-circuiting is not a privileged I enjoy alone. – Ma0 Feb 09 '18 at 12:02
1

Let's summarise: Your solution is better on average and for shorter lists (probably numpy overhead). Numpy is better for longer lists and no repetitions. – Mr. T Feb 09 '18 at 12:08

Andy G · Accepted Answer · 2018-02-09T11:47:22.357

4

You should not be subtracting 1 from j in the inner loop as you end up skipping the comparison of the last 2. It is better to make the adjustments in the loop ranges, rather than subtracting 1 (or not) in the loop code:

A = [2, 7, 5, 9, 3, 1, 2]

N = 7

mint = 1000

for i in range (0, N-1):
    for j in range (i+1, N):
        if (abs(A[i] - A[j]) < mint):
            mint = abs(A[i] - A[j])
            print(i, j)
            print(mint)
print(mint) # 0

I have also avoided the use of a built-in function name min.

To avoid the arbitrary, magic, number 1000, you can perform an initial check against None:

A = [2, 7, 5, 9, 3, 1, 2]

N = 7

mint = None

for i in range (0, N-1):
    for j in range (i+1, N):
        if mint is None:
            mint = abs(A[i] - A[j])
        elif (abs(A[i] - A[j]) < mint):
            mint = abs(A[i] - A[j])
            print(i, j)
            print(mint)
print(mint) # 0

edited Feb 09 '18 at 11:47

answered Feb 09 '18 at 11:23

Andy G

19,232
5
47
69

2

This does not compare `A[-2]` with `A[-1]`. You should loop `for i in range(0, N-1)` and use `A[i]` instead of `A[i-1]`, too. – tobias_k Feb 09 '18 at 11:35
What if every difference is larger than 1000? – Mr. T Feb 09 '18 at 11:41
Thank you @tobias_k, I have corrected that. It also is a lot clearer, and easier to follow, if the range(s) specy what we will be comparing. – Andy G Feb 09 '18 at 11:42
@Piinthesky Start the min at 10,000 ;) or start it as `None` and perform an initial check. – Andy G Feb 09 '18 at 11:44
Start with the difference of the first two elements? – Mr. T Feb 09 '18 at 11:45
I've added an approach to my answer if we wish to avoid a magic number. – Andy G Feb 09 '18 at 11:47

score 3 · Answer 3 · answered Feb 09 '18 at 10:50

3

This is a brute-force solution:

from itertools import combinations

A = [2, 7, 5, 9, 3, 1, 2]

min(abs(i-j) for i, j in combinations(A, 2))  # 0

answered Feb 09 '18 at 10:50

jpp

159,742
34
281
339

Espoir Murhabazi · Answer 4 · 2018-02-09T11:34:30.770

3

using numpy

import numpy as np
A = [2, 7, 5, 9, 3, 1, 2]
v = np.abs(np.diff(np.sort(np.array(A))))
np.min(v)

out : 0

Or You can use numpy only for the diff part like this :

v = min(abs(np.diff(sorted(A))))

edited Feb 09 '18 at 11:34

answered Feb 09 '18 at 10:52

Espoir Murhabazi

5,973
5
42
73

What is `t` here? This only works if you sort the input beforehand, see Ev. Kounis' answer. – tobias_k Feb 09 '18 at 11:03
In order your code to work you should sort list first, otherwise it will produce incorrect result. So it should be min(abs(np.diff(sorted(t)))) – Vlad Bezden Feb 09 '18 at 11:06
Shorter, using `np` only for the `diff` part: `min(abs(np.diff(sorted(A))))` – tobias_k Feb 09 '18 at 11:31
it's also possible , want to propose a full numpy solution – Espoir Murhabazi Feb 09 '18 at 11:33

quamrana · Answer 5 · 2018-02-09T10:51:36.187

1

This is what you are looking for:

A = [2, 7, 5, 9, 3, 1, 2]

diffs = []
for index1, i in enumerate(A):
     for index2, j in enumerate(A):
         if index1 != index2:
             diffs.append(abs(i-j))             

print(min(diffs))

Output:

Updated to exclude subtraction of same items

edited Feb 09 '18 at 10:51

answered Feb 09 '18 at 10:46

quamrana

37,849
12
53
71

1

That would subtract each element from itself, so the answer would always be zero. – Andy G Feb 09 '18 at 10:47
Ok, but the OP didn't mention that. – quamrana Feb 09 '18 at 10:48
I suppose you could add an inner condition `i != j` – Andy G Feb 09 '18 at 10:50
@AndyG, but that would exclude the two different `2`s from the data the OP supplied. – quamrana Feb 09 '18 at 10:52
Yes, I meant to refer to the indexes, which you've adjusted your code to include. – Andy G Feb 09 '18 at 10:53
@quamrana OP didn't have to mention that since we are talking about absolute differences and if something like that was allowed you could simply return 0 without checking anything. – Ma0 Feb 09 '18 at 11:00

Using abs() in a list in Python

5 Answers5

Notes: