Fundamentals of Computer Science I: Media Computing (CS151.02 2007F)
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Reference: [Scheme Report (R5RS)] [Scheme Reference] [DrScheme Manual]
Related Courses: [CSC151.01 2007F (Davis)] [CSC151 2007S (Rebelsky)] [CSCS151 2005S (Stone)]
Summary: Vectors are data structures that are very similar to lists in that they arrange data in linear fashion. Vectors differ from lists in two significant ways: Unlike lists, vectors are indexed and vectors are mutable.
As you've seen in many of the procedures and programs we've written so far, there are many problems in which we have to deal with collections of information. We have now learned two techniques for representing collections of data:
Both of these ways to represent collections have some similar deficiencies. In particular, it is relatively expensive to get a particular element of a list (or file) and it is equally expensive to change a particular element. Why is it expensive to get an element (say, the tenth element)? In the case of a list, we need to cdr through the list until we reach the element. In the case of a file, we need to read through the preceding elements. Changing an element may be even worse, because once we've reached the position, we need to build the list (or file) back to a new form.
Does this mean that lists, files, and other similar structures are inappropriate ways to represent collections? Certainly not. Rather, they work very well for some purposes (e.g., it is easy to extend a list; files persist between invocations of Scheme) and less well for other purposes (e.g., extracting and changing).
To resolve these deficiencies, Scheme provides an alternate mechanism for representing collections, the vector.
You may have noted that when we use lists to group data (e.g., the
information for a spot or a shape), we need to use list-ref
or
repeated calls to cdr
to get later elements of the list.
Unfortunately, list-ref
works by cdr'ing down the list.
Hence, it takes about five steps to get to the fifth element of the
list and about one hundred steps to get to the one hundredth element of
a list. Similarly, to get to the fifth element of a file, we'll need
to read the preceding elements and to get to the hundredth element,
we'll also need to read through the preceding elements. It would be
nicer if we could access any element of the group of data in the same
amount of time (preferably a small amount of time).
For example, if we need to translate a palette index to the
corresponding color, it would be nice if we could look up color number
212 just as fast as color number 0.
Vectors address this problem. Vectors contain a fixed number of elements and provide indexed access (also called random access) to those elements, in the sense that each element, regardless of its position in the vector, can be recovered in the same amount of time. In this respect, a vector differs from a list or a file: The initial element of a list is immediately accessible, but subsequent elements are increasingly difficult and time-consuming to access.
You may have also noted that we occasionally want to change an element of a group of data (e.g., to change a student's grade in the structure we use to represent that student; to change a color in a color palette). When we use lists, we essentially need to build a new list to change one element. When we use files, we often have to build a new file, copying both preceding and subsequent values.
Vectors are mutable data structures: It is possible to replace an element of a vector with a different value, just as one can take out the contents of a container and put in something else instead. It's still the same vector after the replacement, just as the container retains its identity no matter how often its contents are changed.
The particular values that a vector contains at some particular moment constitute its state. One could summarize the preceding paragraph by saying that the state of a vector can change and that state changes do not affect the underlying identity of the vector.
When showing a vector, Scheme shows each of its elements, enclosed
in parentheses, with an extra character, #
, in front of
the left parenthesis. For instance, here's how Scheme shows a vector
containing the strings "alpha"
, "beta"
,
and "gamma"
, in that order:
#("alpha" "beta" "gamma")
The mesh (also called pound, sharp, or hash) character distinguishes the vector from the list containing the same elements.
Some implementations of Scheme permit us to use vector literals, in which a programmer can use a similar syntax to specify a vector when writing a Scheme program or typing commands and definitions into the Scheme interactive interface. In some such implementations, the literal begins with the hash mark. In others, the programmer must place a single quotation mark before the mesh so that Scheme will not try to evaluate the vector as if it were some exotic kind of procedure call.
We traditionally recommend that you avoid using this notation just as we recommend that you avoid the corresponding list literal notation for lists.
Standard Scheme provides the following fundamental procedures for creating vectors and selecting and replacing their elements:
vector
The constructor vector
takes any number of arguments and
assembles them into a vector, which it returns.
>
(vector "alpha" "beta" "gamma")
#("alpha" "beta" "gamma")
>
(vector)
; the empty vector -- no elements!#()
>
(define beta 2)
>
(vector "alpha" beta (list "gamma" 3) (vector "delta" 4) (vector "epsilon"))
#("alpha" 2 ("gamma" 3) #("delta" 4) #("epsilon"))
As the last example shows, Scheme vectors, like Scheme lists, can be heterogeneous, containing elements of various types.
make-vector
The make-vector
procedure takes two arguments,
a natural number k
and a Scheme value obj
,
and returns a k
-element vector in which each position is
occupied by obj
.
>
(make-vector 12 "foo")
#("foo" "foo" "foo" "foo" "foo" "foo" "foo" "foo" "foo" "foo" "foo" "foo")
>
(make-vector 4 0)
#(0 0 0 0)
>
(make-vector 0 4)
; the empty vector, again#()
The second argument is optional; if you omit it, the value that
initially occupies each of the positions in the array is left
unspecified. Various implementations of Scheme have different
ways of filling them up, so you should omit the second argument of
make-vector
only when you intend to replace the
contents of the vector right away.
vector?
The type predicate vector?
takes any Scheme value as argument
and determines whether it is a vector.
>
(vector? (vector "alpha" "beta" "gamma"))
#t
>
(vector? (list "alpha" "beta" "gamma"))
; a list, not a vector#f
>
(vector? "alpha beta gamma")
; a string, not a vector#f
vector-length
The vector-length
procedure takes one argument,
which must be a vector, and returns the number of elements in the
vector.
>
(vector-length (vector 3 1 4 1 5 9))
6
>
(vector-length (vector "alpha" "beta" "gamma"))
3
>
(vector-length (vector))
0
vector-ref
The selector vector-ref
takes two arguments --
a vector vec
and a natural number k
(which
must be less than the length of vec
). It returns the
element of vec
that is preceded by exactly k
other elements. In other words, if k
is 0, you get the
element that begins the vector; if k
is 1, you get
the element after that; and so on.
>
(vector-ref (vector 3 1 4 1 5 9) 4)
5
>
(vector-ref (vector "alpha" "beta" "gamma") 0)
alpha
>
(vector-ref (vector "alpha" "beta" "gamma") 3)
vector-ref: out of bounds: 3
vector-set!
All of the previous procedures look a lot like list procedures, except
that many are more efficient (e.g., vector?
and
vector-length
take a constant number of steps;
list?
takes a number of steps proportional to the
the length of the list and list-ref
takes a number
of steps proportional to the index). Now let's see a procedure that's
much different. We can use procedures to change vectors.
The mutator vector-set!
takes three arguments
-- a vector vec
, a natural number k
(which
must be less than the length of vec
), and a Scheme value
obj
-- and replaces the element of vec
that is currently in the position indicated by k
with
obj
. This changes the state of the vector irreversibly;
there is no way to find out what used to be in that position after it
has been replaced.
As you may recall, it is a Scheme convention to place an exclamation point meaning “Proceed with caution!” at the end of the name of any procedure that makes such an irreversible change in the state of an object.
The value returned by vector-set!
is unspecified;
one calls vector-set!
only for its side effect
on the state of its first argument.
>
(define sample-vector (vector "alpha" "beta" "gamma" "delta" "epsilon"))
>
sample-vector
#("alpha" "beta" "gamma" "delta" "epsilon")
>
(vector-set! sample-vector 2 "zeta")
>
sample-vector
; same vector, now with changed contents#("alpha" "beta" "zeta" "delta" "epsilon")
>
(vector-set! sample-vector 0 "foo")
>
sample-vector
; changed contents again#("foo" "beta" zeta "delta" "epsilon")
>
(vector-set! sample-vector 2 -38.72)
>
sample-vector
; and again#("foo" "beta" -38.72 "delta" "epsilon")
Vectors introduced into a Scheme program by means
of the mesh-and-parentheses notation are supposed to be
“immutable”: applying vector-set!
to such a vector is an error, and the contents of such vectors are
therefore constant. (Warning! Some implementations
of Scheme, including the ones we use, don't enforce this rule.)
vector->list
and list->vector
The vector->list
procedure takes any vector as
argument and returns a list containing the same elements in the same
order; the list->vector
procedure performs
the converse operation.
>
(vector->list (vector 31 27 16))
(31 27 16)
>
(vector->list (vector))
()
>
(list->vector (list #\a #\b #\c))
#(#\a #\b #\c)
>
(list->vector (list 31 27 16))
#(31 27 16)
vector-fill!
The vector-fill!
procedure takes two arguments,
the first of which must be a vector. It changes the state of that
vector, replacing each of the elements it formerly contained with the
second argument.
>
(define sample-vector (vector "rho" "sigma" "tau" "upsilon"))
>
sample-vector
; original vector#("rho" "sigma" "tau" "upsilon")
>
(vector-fill! sample-vector "kappa")
>
sample-vector
; same vector, now with changed contents#("kappa" "kappa" "kappa" "kappa")
The vector-fill!
procedure is invoked only for
its side effect and returns an unspecified value.
Some older implementations of Scheme may lack
the list->vector
,
vector->list
, and
vector-fill!
procedures, but it is straightforward
to define them in terms of the others.
list->vector
Since we are writing this procedure ourselves, we should begin by documenting it.
;;; Procedure: ;;; list->vector ;;; Parameters: ;;; lst, a list. ;;; Purpose: ;;; Convert the list to a vector. ;;; Produces: ;;; vec, a vector ;;; Preconditions: ;;; lst is a list ;;; Postconditions: ;;; vec is a vector. ;;; The length of vec equals the length of lst. ;;; The ith element of vec equals the ith element of lst for ;;; all "reasonable" i.
In most implementations, we will need to recurse over the list, adding elements to a corresponding vector. We will also need to build that vector first. Since we only want to build one vector, we should use the husk-and-kernel technique. The husk creates the vector and tells the kernel and then calls the kernel.
However, we may need other parameters for the kernel, so it is time to
think a little bit about the kernel. Since we want to copy all the
elements of the list to the vector, we probably need to repeat some
basic step, and the only way we know how to do repetition is recursion.
To keep track of what we are copying, we will probably need a counter that
we pass to the helper. Let us call that counter pos
, since
it keeps track of a position in the list or vector.
What should we copy from at each step? We could copy the element at
position pos
of the list into position pos
of the vector. However, that is somewhat inefficient because it requires
us to step through to the ith element of the list.
Since we no longer need a value from the list once we have copied it, we can cdr through the list as we step through the vector. Now, our goal is to copy the initial element of the list into the appropriate position of the vector.
When do we stop? When we run out of elements to copy.
Given all of the above, here is how to set things up.
(define list->vector (lambda (lst) (list->vector-kernel! lst ; Copy the whole list 0 ; Starting at position 0 (make-vector (length lst) null) ; Into a new vector of the appropriate length )))
The kernel is a little bit more complicated. We need to keep track of where we are in the vector. We may also want to carefully specify what this kernel is supposed to do. (Such specification is not always necessary for helpers, but I think it clarifies things in this case.)
;;; Procedure: ;;; list->vector-kernel! ;;; Parameters: ;;; lst, a list to copy ;;; pos, a position in the vector ;;; vec, a vector ;;; Purpose ;;; Copy values from lst into positions pos, pos+1, ... ;;; Produces: ;;; vec, the same vector but with different contents. ;;; Preconditions: ;;; The length of vec = pos + the length of list. [Unverified] ;;; pos >= 0. [Unverified] ;;; Postconditions: ;;; Element pos of vec now contains element 0 of list. ;;; Element pos+1 of vec now contains element 1 of lst. ;;; Element pos+2 of vec now contains element 2 of lst. ;;; ... ;;; The last element of vec now contains the last element of lst. (define list->vector-kernel! (lambda (lst pos vec) ; We don't bother to verify the preconditions because this ; procedure should only be called by something that has already ; verified the preconditions (explicitly or implicitly). (cond ; If there's nothing left to copy, stop. ((null? lst) vec) ; Otherwise, copy the initial element of the list and ; then copy the remaining elements (else (vector-set! vec pos (car lst)) (list->vector-kernel! (cdr lst) (+ pos 1) vec)))))
Is this the only way to write the list->vector
procedure? No. More advanced students might know about the
apply
procedure which makes life significantly
easier.
(define list->vector (lambda (lst) (apply vector lst)))
In other words: Call the vector
procedure, giving it the
elements of lst
as its arguments. Don't worry if you don't
understand this now. We will return to the idea later in the semester.
vector->list
Now let us consider how to go in the opposite direction. Once again, we will probably need to recurse to step through positions and need to keep track of the position with an extra variable. Should we create a list first and then populate it? No. We do not typically mutate lists. So, we will update the list “on the fly”, adding elements and extending the list at each step. Let's start with the helper. The helper will build a list from a subrange of the vector.
;;; Procedure: ;;; vector->list-kernel ;;; Parameters: ;;; vec, a vector ;;; start, an integer ;;; finish, an integer ;;; Purpose: ;;; Builds a list that contains elements start ... finish-1 of vec ;;; Produces: ;;; lst, a new list. ;;; Preconditions: ;;; start >= 0 ;;; finish >= start ;;; finish <= length of vec ;;; Postconditions: ;;; The element at position i of lst is the element at position ;;; i+start of vec for all reasonable i. ;;; Does not change vec. (define vector->list-kernel (lambda (vec start finish) ; We stop at the finish. (if (= start finish) ; There are no more elements to put into a list, so use ; the empty list. null ; Otherwise, add the element at position start to a list of ; the remaining elements. (cons (vector-ref vec start) (vector->list-kernel vec (+ 1 start) finish)))))
Now, we just have to set things up for this kernel. We start at position 0. We end just before the length of the vector. So, here goes ...
;;; Procedure: ;;; vector->list ;;; Parameters: ;;; vec, a vector ;;; Purpose: ;;; Builds a list that contains the elements of vec in the same order. ;;; Produces: ;;; lst, a new list. ;;; Preconditions: ;;; [None] ;;; Postconditions: ;;; The element at position i of lst is the element at position ;;; i of vec for all reasonable i. ;;; Does not change vec. (define vector->list (lambda (vec) (vector->list-kernel vec 0 (vector-length vec))))
Each time we learn a new structure, we learn techniques for recursing over that structure. As the previous examples suggested, recursion over vectors is relatively straightforward, but usually requires that we have a helper procedure that includes additional parameters - the current position in the vector and the length of the vector. (Why do we include the length? So that we don't have to recompute it each time.)
The test for the base case is then to check whether the current position equals the length and the “simplify” step is to add 1 to the position.
(define vector-proc (lambda (vec other) (vector-proc-kernel vec other 0 (vector-length vec)))) (define vector-proc-kernel (lambda (vec other pos len) (if (= pos len) (base-case vec other) (combine (vector-ref vec pos) (vector-proc-kernel vec other (+ post 1) len)))))
At times, it's better to start at the end of the vector and work backwards. In this strategy, we get the base case when the position reaches 0 and we simplify by subtracting 1.
(define vector-proc (lambda (vec other) (vector-proc-kernel vec other (- (vector-length vec) 1)))) (define vector-proc-kernel (lambda (vec other pos) (if (< pos 0) (base-case vec other) (combine (vector-ref vec pos) (vector-proc-kernel vec other (- pos 1))))))
(list->vector lst)
(vector val_1 ... val_n)
(vector? val)
(vector-fill vec val)
(vector-ref vec pos)
(vector-set! vec pos val)
(vector->list vec)
Primary: [Front Door] [Glance] - [Academic Honesty] [Instructions]
Current: [Outline] [EBoard] [Reading] [Lab] [Assignment]
Groupings: [Assignments] [EBoards] [Examples] [Exams] [Handouts] [Labs] [Outlines] [Projects] [Readings] [Reference]
Reference: [Scheme Report (R5RS)] [Scheme Reference] [DrScheme Manual]
Related Courses: [CSC151.01 2007F (Davis)] [CSC151 2007S (Rebelsky)] [CSCS151 2005S (Stone)]
Copyright © 2007 Janet Davis, Matthew Kluber, and Samuel A. Rebelsky. (Selected materials copyright by John David Stone and Henry Walker and used by permission.)
This material is based upon work partially supported by the National Science Foundation under Grant No. CCLI-0633090. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
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