diff1(x) in Python: True if all adjacent items differ by 1, False otherwise.

[HenHanna]

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)))

[Paul Rubin]

Post by HenHanna
if len(x) <= 1: return True
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:]))

[HenHanna]

Post by Paul Rubin

Post by HenHanna
if len(x) <= 1: return True
if abs(x[i] - x[i+1]) != 1: return False
return True

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

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

[Kaz Kylheku]

Post by HenHanna

Post by Paul Rubin

Post by HenHanna
if len(x) <= 1: return True
if abs(x[i] - x[i+1]) != 1: return False
return True

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. :)

[HenHanna]

     if len(x) <= 1: return True
         if abs(x[i] - x[i+1]) != 1:    return False
     return True

   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.

[Lawrence D'Oliveiro]

Post by HenHanna
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]()

[HenHanna]

     if len(x) <= 1: return True
         if abs(x[i] - x[i+1]) != 1:    return False
     return True

   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.

[Michael F. Stemper]

     if len(x) <= 1: return True
         if abs(x[i] - x[i+1]) != 1:    return False
     return True

   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.