Remove first occurrence of element from list
$begingroup$
As explained here, I'm learning a limited subset of Scheme.
My task has been to sort a numerical list. To implement a simple insertion sort, I need to remove a single element from a list.
I've done the following (in the spirit of The Little Schemer):
(define remove-elem
(lambda (ll elem)
(cond
((null? ll) '())
((eq? (car ll) elem) (cdr ll))
(else
(cons (car ll) (remove-elem (cdr ll) elem))))
))
What are other ways of implementing this function?
scheme
edited 9 mins ago
Jamal♦
30.3k11118227
asked Oct 29 '13 at 22:17
ricardoricardo
185210
$endgroup$
add a comment |
$begingroup$
As explained here, I'm learning a limited subset of Scheme.
My task has been to sort a numerical list. To implement a simple insertion sort, I need to remove a single element from a list.
I've done the following (in the spirit of The Little Schemer):
(define remove-elem
(lambda (ll elem)
(cond
((null? ll) '())
((eq? (car ll) elem) (cdr ll))
(else
(cons (car ll) (remove-elem (cdr ll) elem))))
))
What are other ways of implementing this function?
scheme
edited 9 mins ago
Jamal♦
30.3k11118227
asked Oct 29 '13 at 22:17
ricardoricardo
185210
$endgroup$
add a comment |
$begingroup$
As explained here, I'm learning a limited subset of Scheme.
My task has been to sort a numerical list. To implement a simple insertion sort, I need to remove a single element from a list.
I've done the following (in the spirit of The Little Schemer):
(define remove-elem
(lambda (ll elem)
(cond
((null? ll) '())
((eq? (car ll) elem) (cdr ll))
(else
(cons (car ll) (remove-elem (cdr ll) elem))))
))
What are other ways of implementing this function?
scheme
edited 9 mins ago
Jamal♦
30.3k11118227
asked Oct 29 '13 at 22:17
ricardoricardo
185210
$endgroup$
As explained here, I'm learning a limited subset of Scheme.
My task has been to sort a numerical list. To implement a simple insertion sort, I need to remove a single element from a list.
I've done the following (in the spirit of The Little Schemer):
(define remove-elem
(lambda (ll elem)
(cond
((null? ll) '())
((eq? (car ll) elem) (cdr ll))
(else
(cons (car ll) (remove-elem (cdr ll) elem))))
))
What are other ways of implementing this function?
scheme
scheme
edited 9 mins ago
Jamal♦
30.3k11118227
asked Oct 29 '13 at 22:17
ricardoricardo
185210
edited 9 mins ago
Jamal♦
30.3k11118227
asked Oct 29 '13 at 22:17
ricardoricardo
185210
edited 9 mins ago
Jamal♦
30.3k11118227
edited 9 mins ago
Jamal♦
30.3k11118227
edited 9 mins ago
Jamal♦
30.3k11118227
30.3k11118227
asked Oct 29 '13 at 22:17
ricardoricardo
185210
asked Oct 29 '13 at 22:17
ricardoricardo
185210
asked Oct 29 '13 at 22:17
ricardoricardo
185210
185210
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
That looks mostly right to me. I'll suggest two things.
First, there is nice syntatic sugar for defining functions:
(define (remove-elem xs elem)
(cond ...)
)
Save yourself the lambda.
Second, you are using eq?. That is the wrong checker for this problem. That function returns true if the two objects are the exact same object in memory. It could be true for primitives like ints, but it's not necessarily true. What you want to use is eqv? which will guaranteed work for numbers.
answered Oct 29 '13 at 23:06
BarryBarry
17.2k13087
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
votes
$begingroup$
That looks mostly right to me. I'll suggest two things.
First, there is nice syntatic sugar for defining functions:
(define (remove-elem xs elem)
(cond ...)
)
Save yourself the lambda.
Second, you are using eq?. That is the wrong checker for this problem. That function returns true if the two objects are the exact same object in memory. It could be true for primitives like ints, but it's not necessarily true. What you want to use is eqv? which will guaranteed work for numbers.
answered Oct 29 '13 at 23:06
BarryBarry
17.2k13087
$endgroup$
add a comment |
$begingroup$
That looks mostly right to me. I'll suggest two things.
First, there is nice syntatic sugar for defining functions:
(define (remove-elem xs elem)
(cond ...)
)
Save yourself the lambda.
Second, you are using eq?. That is the wrong checker for this problem. That function returns true if the two objects are the exact same object in memory. It could be true for primitives like ints, but it's not necessarily true. What you want to use is eqv? which will guaranteed work for numbers.
answered Oct 29 '13 at 23:06
BarryBarry
17.2k13087
$endgroup$
add a comment |
$begingroup$
That looks mostly right to me. I'll suggest two things.
First, there is nice syntatic sugar for defining functions:
(define (remove-elem xs elem)
(cond ...)
)
Save yourself the lambda.
Second, you are using eq?. That is the wrong checker for this problem. That function returns true if the two objects are the exact same object in memory. It could be true for primitives like ints, but it's not necessarily true. What you want to use is eqv? which will guaranteed work for numbers.
answered Oct 29 '13 at 23:06
BarryBarry
17.2k13087
$endgroup$
That looks mostly right to me. I'll suggest two things.
First, there is nice syntatic sugar for defining functions:
(define (remove-elem xs elem)
(cond ...)
)
Save yourself the lambda.
Second, you are using eq?. That is the wrong checker for this problem. That function returns true if the two objects are the exact same object in memory. It could be true for primitives like ints, but it's not necessarily true. What you want to use is eqv? which will guaranteed work for numbers.
answered Oct 29 '13 at 23:06
BarryBarry
17.2k13087
answered Oct 29 '13 at 23:06
BarryBarry
17.2k13087
answered Oct 29 '13 at 23:06
BarryBarry
17.2k13087
answered Oct 29 '13 at 23:06
BarryBarry
17.2k13087
17.2k13087
add a comment |
add a comment |
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