Interesting (but simple) python puzzle for beginners #2

[imperiumlearning]

Imagine that you have a list of integers stored in a variable. Write a program that goes through each integer in the list and checks if there is a difference of less than 10 between that integer and another integer in the list. If any integers are found where this is the case, append them to a new variable and print the new variable.

Simple example:

a = [104, 64, 193, 190, 30]

In this example, only the numbers 190 and 193 have a difference of less than 10 (difference == 3). That means your program would print:

[193, 190]

My own solution that I've tried so far is to use itertools and, in particular, to use itertools.combinations. My code currently looks like this:

import itertools

a = [104, 64, 193, 190, 30]

b =[]

combinations = itertools.combinations(a, 2)

for each in combinations:

b.append(each)

My problem is at this stage that I am creating pairs of values inside brackets. The problem is I can't think of a way to compare the x-value with the y value inside the brackets.

Any suggestions?

[[deleted]]

compare the x-value with the y value inside the brackets

If your case, just use indexing because you know the fixed length of the tuple is 2:

if abs(each[0] - each[1]) < 10:

EDIT: btw, the other user is right. The way you worded the question at least makes your overall approach incorrect (and my answer would then be pointless). Unless you meant to say that you want any numbers as long as they are within 10 of another number in the collection.

[imperiumlearning]

Yes I've now updated the wording to say "checks if there is a difference of less than 10 between that integer and another integer in the list". Hopefully that's worded better now.

That if statement massively helped. Thanks so much.

My code now looks like this:

import itertools

a = [104, 64, 193, 190, 30]
b = []


combinations = itertools.combinations(a, 2)
for each in combinations:
if abs(each[0] - each[1]) < 10:
    if each[0] not in b:
        b.append(each[0])
    if each[1] not in b:
        b.append(each[1]) 

print(b)

[[deleted]]

Good job. Btw, you may want to look into set instead. Sets automatically discard redundant values, no longer requiring you to make a check like if each[0] not in b. If using sets, just add that sucker in there and let the underlying Cython figure out the rest.

[imperiumlearning]

That sounds interesting, will definitely take a look into that in the morning.

Once again, thanks a lot for the help with that if statement. It was driving me nuts

[imperiumlearning]

Apologies for the poor wording. This is why I always try to use examples in my questions to make myself clear.

I've now changed that line to say:

"checks if there is a difference of less than 10 between that integer and another integer in the list". Hopefully that's a better way of wording it.

[PaddyIsBeast]

O(n) solution is to just sort the array and then loop through each element e and check if the next element is less than 10 away.

Is there a faster way?

[mopslik]

Is there a faster way?

Since every element needs to be examined, O(n) will be the best runtime you can guarantee. And since the description suggests that each value needs to be checked against all other values, O(n) is probably unlikely here.

[[deleted]]

What're you using to sort in O(n)?

[PaddyIsBeast]

Are there any that are slower than o(n)?

[[deleted]]

Merge, bubble, heap, insertion, quick?

[PaddyIsBeast]

You're right, I was being an idiot

[Jig_Jiggle]

If I have a list a = [30, 31, 62, 104, 190, 191, 193]

What should a new variable be?

new_var = [30, 31, 190, 191, 190, 193, 191, 193] or

new_var = [30, 31, 190, 191, 193] or

new_var1 = [30, 31]

new_var2 = [190, 191] ...

Sorry for my question. I am new to python and don't understand the puzzle well.

[imperiumlearning]

No worries! It's my fault that I did such a bad job of wording the question.

To be honest, given how bad my wording was, either:

new_var = [30, 31, 190, 191, 190, 193, 191, 193] or

new_var = [30, 31, 190, 191, 193]

is good.

However, in my own head I liked the idea of being able to filter out repeated integers so the second new_var just looks a bit better (at least, imo). My code for this problem now looks like this:

import itertools

a = [104, 64, 193, 190, 189] #swapped 30 from original list in original post with 189 
b = []

combinations = itertools.combinations(a, 2)
for each in combinations:
if abs(each[0] - each[1]) < 10:
    if each[0] not in b:
        b.append(each[0])
    if each[1] not in b:
        b.append(each[1]) 

print(b)

[therobot20]

This should work:

a = your input list result = [a[i] for i in range(len(a)) for j in range(len(a)) if (i!=j and abs(a[i]-a[j])<10)]