Fifth clear solution for All the Same by tom-tom - python coding challenges

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Fifth solution in Clear category for All the Same by tom-tom

def all_the_same(elements):
   return elements[1:] == elements[:-1]

Jan. 31, 2018

Comments:

tom-tom on Jan. 31, 2018, 10:13 p.m.
Inspired by <a href="https://py.checkio.org/mission/all-the-same/publications/Tinus_Trotyl/python-3/all-the-same/">this solution</a>

Tinus_Trotyl on Feb. 1, 2018, 1:05 a.m.
Thanks for the credits tho, but you raised it to the superpythonical level :)

J.W on March 23, 2018, 10:29 p.m.
Very smart. You must be very good at mathematics.

lynnlan on April 13, 2018, 2:53 a.m.
print(allthesame({1,1})) there is the error: TypeError: 'set' object is not subscriptable

Saingue_Lainglin on June 21, 2018, 11:30 p.m.
clever

ani231990 on July 11, 2018, 10:21 a.m.
WOW! It's cool)

berserk9779 on July 13, 2018, 5:46 p.m.
Great solution

mikrolab075 on July 20, 2018, 9:09 a.m.
Even if you are inspired by someone else's task, this is very cool

jfp053 on July 25, 2018, 10:33 p.m.
According to my tests, (see Comparison of Solutions), I come to the conclusion that this method tests all couples, even in the case of a list like [0, 1, 2, 3], In an optimized test, the test 0==1 should break and return False.

jqy1977017245 on Aug. 5, 2018, 1:02 a.m.
It's perfect.

AlexSIBER on Sept. 2, 2018, 2:14 p.m.
It's a perfect solution

kangmin on Sept. 28, 2018, 4:11 a.m.
excellent!

Duomim on Oct. 4, 2018, 2:29 p.m.
厉害,还可以这样解决,我只能想到Len(set(ls)) 拓展了我的思维

lukasz181818 on Oct. 6, 2018, 5:05 p.m.
It's great have opportunity to come back and see such good solutions

gnosis on Oct. 23, 2018, 3 a.m.
This even works with nested lists.

art.hur on Nov. 27, 2018, 10:26 a.m.
<pre class='brush: python'>elements = [X1,X2,X3,X4,X5] elements[:-1] # [X1,X2,X3,X4] elements[1:] # [X2,X3,X4,X5] == compares a pair and then moves further: compares X1 with X2 and if they are equal, it will then compare X2 and X3 and so on. This solution works because of transitivity. If X1 == X2 and X2 == X3, then X1 == X3. In reality, you're always comparing X1 (which is equal to X2, X3, X4) to the next element. </pre> Good to learn how == works on lists and it's always useful to review your slices. It's a clever one liner, but I don't think it's clear just because it has few lines.

tom-tom on Nov. 27, 2018, 7:50 p.m.
What do you mean 'in reality'? Equivalence relation is reflexive, symmetric and transitive. If you define `==` not as equivalence, you have to make all possible tests. Example: for `[X1, X2, X3]` if `==` means equivalence, you need 2 tests only: `X1 == X2` and `X2 == X3` (or `X1 == X3`) if not (you implemented `eq` in some weird way), 9 tests: `X1 == X1`, `X1 == X2`, `X1 == X3`, `X2 == X1`, `X2 == X2`, `X2 == X3`, `X3 == X1`, `X3 == X2`, and `X3 == X3`. Yes, you have to check whether X1, X2, and X3 are 'equal' to themselves!

mlb00 on March 23, 2019, 4:58 p.m.
thanks for this helpful explanation; didn't immediately understand why this worked!

Str3et on Dec. 22, 2018, 6:42 a.m.
awesome!

mdhirsch on March 14, 2019, 9:27 p.m.
This is a very clever and concise solution. However, I think it creates a two copies of the list (minus one element). So if the list is large this can be a very costly algorithm.

jukka.neppius on Sept. 7, 2019, 1:08 p.m.
If that's a problem, then: return elements == elements[1:] + elements[0:1] Removal of the first element might be slow.

Hi-Man10 on March 19, 2019, 11:59 a.m.
Good solution, i used for loop and if but this is pure perfect

ha123321ha on March 31, 2019, 3:29 p.m.
太强了

aakashsharma845 on April 3, 2019, 5:02 a.m.
please explain in detail that how it is working and what if i take elements as [1,2,2,1] then elements[1:] =[1,2,2,1] elements[:-1]=[1,2,2,1] then please explain the solution . as i m not an expert in python , i will be obliged if u reply with detailed explanation.

tom-tom on April 3, 2019, 6:54 p.m.
`elements[1:]` makes a copy of the list without the first item, as list index starts with 0. `elements[:-1]` makes a copy of the list without the last item, as item with index -1 (the last) is not included. When these two copies are compared, the first item is compared to the second, the second to the third, and so on. As a result, every item is compared to the next. If all comparisons give True, all items are equal.

aakashsharma845 on April 12, 2019, 12:19 p.m.
damn good .thanks a lot dear.

yersen on April 12, 2019, 1:47 p.m.
Great!

graulopezjavier on April 29, 2019, 2 a.m.
Pretty awesome solution

wq768458 on May 3, 2019, 2:17 p.m.
what about [1,2,1,1]?

tom-tom on May 5, 2019, 5:54 p.m.
What's wrong with it? `[1,2,1,1][1:]` makes `[2, 1, 1]` `[1,2,1,1][:-1]` makes `[1, 2, 1]` As the results are different, not all the elements are the same.

z.thami on May 22, 2019, 6:35 p.m.
Omg, this was so so so so much easier than mine! Nice work!

JYH89 on July 3, 2019, 4:50 a.m.
真的太棒了，your solution is very pythonic. if the list is [a,b,c,d],elements[1:] is [b,c,d] and elements[:-1] is [a,b,c] ,== compares elements[1] with elements[0],elements[2] with elements[1],elements[3] with elements[2].Amazing!

David_Dai845 on July 24, 2019, 3:19 p.m.
What a simple solution!

ame-lm on July 30, 2019, 9:10 a.m.
Sprised

gvswamy on Aug. 28, 2019, 1:12 p.m.
I can not see some code to review. Live example of zero coding solution. Excellent.

oldjaponec on Aug. 28, 2019, 8:47 p.m.
the best

edenzhan2333 on Aug. 29, 2019, 1:31 p.m.
Excellent!

nishith on Aug. 30, 2019, 7:47 a.m.
Not working on this one: [1, 1, 1, True, 1, 1, 1]

tom-tom on Aug. 30, 2019, 6:38 p.m.
Python disagrees with you: <pre class='brush: python'>>>> 1 == True True </pre> So, all elements in the given list are equal.

coder_du on Sept. 19, 2019, 7:54 a.m.
excellent

michael.orange34 on Sept. 29, 2019, 12:08 p.m.
clever

mazzi on Oct. 24, 2019, 9:38 a.m.
Wow, I'm really impressed with this solution!

Ocean5app on Oct. 30, 2019, 7:34 a.m.
Great! very clever

dusfkd on Nov. 2, 2019, 8:33 a.m.
Wow!! it is simple but very creative

Idletuner on Nov. 7, 2019, 5:44 p.m.
OMG ... I have still to learn a lot ... this one-liner is awesome.

art8008 on Nov. 16, 2019, 7:45 p.m.
Greate! Thank you!

Abdur-rahmaanJ on Dec. 2, 2019, 9:11 a.m.
haha palindrome checker!

marlotacla on Dec. 6, 2019, 2:08 p.m.
Sensational! It's to see solutions like this one that I try this exercises. We learn much more by trying then seeing the elegant solutions other devs reached than simply reading or attending courses.

work-hrf on Dec. 27, 2019, 5:26 a.m.
not None?

soufi on Jan. 25, 2020, 4:22 p.m.
i dont understand how this work on allthesame([]) isn't it suppose to be none ?

tom-tom on Jan. 25, 2020, 10:05 p.m.
As there are no different elements in the empty list, `allthesame([])` is `True`.

koborC on Dec. 14, 2020, 12:28 p.m.
every slice of an empty array is an empty array I think. So the code still works for an empty list. <pre class='brush: python'>l = [] l[3:] # -> [] l[:3] # -> [] l[1:5] # -> [] </pre> But I cant think of an empty list when I see the solution. Good Point! Thanks. =D

sudhunter on Feb. 5, 2020, 7:32 p.m.
ohh!

katharina.noreia on Feb. 27, 2020, 9:15 a.m.
brilliant!

Steapn0806 on March 24, 2020, 3:16 p.m.
return elements == [elements[0]] * len(elements)

tom-tom on March 24, 2020, 5:17 p.m.
No, if `elements` is empty you'll get an exception

Steapn0806 on March 24, 2020, 5:39 p.m.
try: return elements == [elements[0]] * len(elements) except: return True

Steapn0806 on March 24, 2020, 5:39 p.m.
It worked for me: return elements == [elements[0]] * len(elements)

tom-tom on March 24, 2020, 6:52 p.m.
For `[]`: <pre class='brush: python'>IndexError: list index out of range </pre>

Steapn0806 on March 24, 2020, 7:01 p.m.
try: return elements == [elements[0]] * len(elements) except: return True

yeyuhongcheng on Aug. 16, 2020, 8:21 a.m.
amazing

Spiritusvini on Nov. 4, 2020, 7:43 p.m.
Try it: <pre class='brush: python'>elements = [1, 2, 1] </pre>

ambulu1939 on Dec. 3, 2020, 5:49 p.m.
Brilliant

gitkitten on Jan. 25, 2021, 2:45 p.m.
oh this is smart. thanks

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