in Python, is this something you'd write using all() ?

https://stackoverflow.com/questions/19389490/how-do-pythons-any-and-all-functions-work

def diff1(x):
if len(x) <= 1: return True
for i in range(len(x) - 1):
if abs(x[i] - x[i+1]) != 1: return False
return True

def pd(x): print (x, diff1(x))
pd( [0,3]) # ===> #f
pd( [2,3,4,3]) # ===> #t
pd( [1,2,3,4,3,2,3]) # ===> #t
pd( [1,2,3,4,5,6,7,8,9,0]) # ===> #f

(define (diff1p? x)
(or (null? x)
(null? (cdr x))
(and (= 1 (abs (- (car x) (cadr x)) ))
(diff1p? (cdr x)) )))

> (diff1p? '(0 3)) ===> #f
> (diff1p? '(2 3 4 3)) ===> #t
> (diff1p? '(1 2 3 4 5 6 7 8 9 8)) ===> #t
> (diff1p? '(1 2 3 4 5 6 7 8 9 8 0)) ===> #f

__________________________
find
find-tail
any
every
count

every pred clist1 clist2 . . . [Function]

[R7RS list] Applies pred across each element of clists, and returns
#f as soon as pred returns #f. If all application of pred
return a non-false value, [every] returns the last result of the
applications.

it sounds like it expands to a big (OR ........ )

______________________________ Not

(define (every? predicate lst) (and (map predicate lst)))

HenHanna <HenHanna@devnull.tb> writes:
> def diff1(x):
> if len(x) <= 1: return True
> for i in range(len(x) - 1):
> if abs(x[i] - x[i+1]) != 1: return False
> return True

def diff2(x):
return all(abs(a-b)==1) for a,b in zip(x,x[1:]))

On 6/15/2024 2:30 PM, Paul Rubin wrote:
> HenHanna <HenHanna@devnull.tb> writes:
>> def diff1(x):
>> if len(x) <= 1: return True
>> for i in range(len(x) - 1):
>> if abs(x[i] - x[i+1]) != 1: return False
>> return True
>
> def diff2(x):
> return all(abs(a-b)==1) for a,b in zip(x,x[1:]))

Does this work when x is [] or [1] ?

On 2024-06-16, HenHanna <HenHanna@devnull.tb> wrote:
> On 6/15/2024 2:30 PM, Paul Rubin wrote:
>> HenHanna <HenHanna@devnull.tb> writes:
>>> def diff1(x):
>>> if len(x) <= 1: return True
>>> for i in range(len(x) - 1):
>>> if abs(x[i] - x[i+1]) != 1: return False
>>> return True
>>
>> def diff2(x):
>> return all(abs(a-b)==1) for a,b in zip(x,x[1:]))
>
>
> Does this work when x is [] or [1] ?

Can we keep this abject filth out of the Lisp newsgroup?

Thanks. :)

On 6/15/2024 5:13 PM, HenHanna wrote:
> On 6/15/2024 2:30 PM, Paul Rubin wrote:
>> HenHanna <HenHanna@devnull.tb> writes:
>>> def diff1(x):
>>>      if len(x) <= 1: return True
>>>      for i in range(len(x) - 1):
>>>          if abs(x[i] - x[i+1]) != 1:    return False
>>>      return True
>>
>> def diff2(x):
>>    return all(abs(a-b)==1) for a,b in zip(x,x[1:]))
>
>
> Does this work when x is  []  or  [1]  ?

i'm pretty sure that x[1:] would raise an error.

Lispers can learn from Pythonicity and vice versa.

On Sat, 15 Jun 2024 17:52:18 -0700, HenHanna wrote:

> Lispers can learn from Pythonicity and vice versa.

diff2 = \
lambda x : \
(
lambda : True,
lambda : all(abs(a - b) == 1) for a, b in zip(x, x[1:]),
)[len(x) > 1]()

On 6/15/2024 5:52 PM, HenHanna wrote:
> On 6/15/2024 5:13 PM, HenHanna wrote:
>> On 6/15/2024 2:30 PM, Paul Rubin wrote:
>>> HenHanna <HenHanna@devnull.tb> writes:
>>>> def diff1(x):
>>>>      if len(x) <= 1: return True
>>>>      for i in range(len(x) - 1):
>>>>          if abs(x[i] - x[i+1]) != 1:    return False
>>>>      return True
>>>
>>> def diff2(x):
>>>    return all(abs(a-b)==1) for a,b in zip(x,x[1:]))
>>
>>
>> Does this work when x is  []  or  [1]  ?
>
>
> i'm pretty sure that  x[1:]  would  raise an error.

x[1] would raise an error,
but x[1:] is ok. (works like Lisp's CDR)... Nice!

>
>             Lispers can learn from Pythonicity and vice versa.
>

On 15/06/2024 19.13, HenHanna wrote:
> On 6/15/2024 2:30 PM, Paul Rubin wrote:
>> HenHanna <HenHanna@devnull.tb> writes:
>>> def diff1(x):
>>>      if len(x) <= 1: return True
>>>      for i in range(len(x) - 1):
>>>          if abs(x[i] - x[i+1]) != 1:    return False
>>>      return True
>>
>> def diff2(x):
>>    return all(abs(a-b)==1) for a,b in zip(x,x[1:]))
>
>
> Does this work when x is  []  or  [1]  ?

The easiest way to find out would be to type it in to REPL and
give it a whirl.

--
Michael F. Stemper
87.3% of all statistics are made up by the person giving them.