Computer Science Fundamentals (CS153 2003S)
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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.
Contents
You may have noted that when we use lists to group data (e.g., the
information on a course), we need to use listref
or
repeated calls to cdr
to get later elements of the list.
Unfortunately, listref
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.
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).
Vectors contain a fixed number of elements and provide random access (also called indexed 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: The initial element of a list is immediately accessible, but subsequent elements are increasingly difficult and timeconsuming to get at.
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). When we use lists, we essentially need to build a new list to change one element.
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 symbols 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.
We can use the same syntax to specify a vector when writing a Scheme program or typing commands and definitions into the Scheme interactive interface, except that we have to 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. (DrScheme will not make this mistake even if you forget the single quotation mark, but not all implementations of Scheme are so generous.)
I recommend that you avoid using this vector literal notation (that is, with the quotation mark), just as I recommend that you avoid the corresponding list literal notation for lists.
In the interaction window, when DrScheme shows a vector as the value of some toplevel expression that the user supplied, it does something more ambitious, but potentially rather confusing: It inserts a numeral between the mesh character and the left parenthesis to indicate the number of elements in the vector, thus:
#3(alpha beta gamma)
However, you can instruct DrScheme to use the straight meshandparentheses representation, with no numeral, by giving the following command at the beginning of your program or interactive session:
(printvectorlength #f)
That is, Don't print the lengths of vectors
.
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.
> (printvectorlength #f) > (vector 'alpha 'beta 'gamma)) #(alpha beta gamma) > (vector) ; the empty vector  no elements! #() > (vector 'alpha "beta" '(gamma 3) '#(delta 4) (vector 'epsilon)) #(alpha "beta" (gamma 3) #(delta 4) #(epsilon))
As the last example shows, Scheme vectors can be heterogeneous, containing elements of various types, just like Scheme lists.
makevector
The makevector
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
.
> (makevector 12 'foo) #(foo foo foo foo foo foo foo foo foo foo foo foo) > (makevector 4 0) #(0 0 0 0) > (makevector 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 makevector
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? '#(alpha beta gamma)) ; discouraged notation #t > (vector? (list 'alpha 'beta 'gamma)) ; a list, not a vector #f > (vector? "alpha beta gamma") ; a string, not a vector #f > (vector? '#(#f)) ; a oneelement vector (the element is #f) #t
vectorlength
The vectorlength
procedure takes one argument, which must be
a vector, and returns the number of elements in the vector.
> (vectorlength (vector 3 1 4 1 5 9)) 6 > (vectorlength (vector 'alpha 'beta 'gamma)) 3 > (vectorlength (vector)) 0
vectorref
The selector vectorref
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.)
> (vectorref (vector 3 1 4 1 5 9) 4) 5 > (vectorref (vector 'alpha 'beta 'gamma) 0) alpha > (vectorref (vector 'alpha 'beta 'gamma) 3) vectorref: index 3 out of range [0, 2] for vector: #(alpha beta gamma)
vectorset!
All of the previous procedures look a lot like list procedures, except
that many are more efficient (e.g., vector?
takes a constant
number of steps; list?
takes a number of steps proportional
to the the length of the list). Now let's see a procedure that's much
different. We can actually use procedures to change vectors.
The mutator vectorset!
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.
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 vectorset!
is unspecified; one calls
vectorset!
only for its side effect on the state of its first
argument.
> (define samplevector (vector 'alpha 'beta 'gamma 'delta 'epsilon)) > samplevector #(alpha beta gamma delta epsilon) > (vectorset! samplevector 2 'zeta) > samplevector ; same vector, now with changed contents #(alpha beta zeta delta epsilon) > (vectorset! samplevector 0 "foo") > samplevector ; changed contents again #("foo" beta zeta delta epsilon) > (vectorset! samplevector 2 38.72) > samplevector ; and again #("foo" beta 38.72 delta epsilon)
Vectors introduced into a Scheme program by means of the
meshandparentheses notation are immutable
 applying
vectorset!
to such a vector is an error, and the contents of
such vectors are therefore constant. (Warning! Some implementations
of Scheme, including DrScheme, don't enforce this rule.)
vector>list
and list>vector
The vector>list
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 '#(31 27 16)) (31 27 16) > (vector>list (vector)) () > (list>vector '(#\a #\b #\c)) #(#\a #\b #\c) > (list>vector (list 31 27 16)) #(31 27 16)
vectorfill!
The vectorfill!
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 samplevector (vector 'rho 'sigma 'tau 'upsilon)) > samplevector ; original vector #(rho sigma tau upsilon) > (vectorfill! samplevector 'kappa) > samplevector ; same vector, now with changed contents #(kappa kappa kappa kappa)
The vectorfill!
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
vectorfill!
procedures, but it is straightforward to define
them in terms of the others.
list>vector
;;; 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'll need to recurse over the list, adding elements to a corresponding vector. We'll also need to build that vector, first. Since we only want to build one vector, we should use the huskandkernel technique (which we've previously used for error checking). 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's 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're copying, we'll probably need a counter that
we pass to the helper. I'll 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's 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've 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.
So, here's how to set things up.
(define list>vector (lambda (lst) (list>vectorkernel! lst ; Copy the whole list 0 ; Starting at position 0 (makevector (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>vectorkernel! ;;; 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>vectorkernel! (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). ; If there's nothing left to copy, stop. (if (null? lst) vec ; Otherwise, copy the initial element of the list and ; then copy the remaining elements (begin (vectorset! vec pos (car lst)) (list>vectorkernel! (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 (ls) (apply vector ls)))
In other words: Call the vector
procedure, giving it the
elements of ls
as its arguments.
vector>list
Now let's consider how to go in the opposite direction. Once again,
we'll 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 don't typically mutate lists. So,
we'll 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>listkernel ;;; Parameters: ;;; vec, a vector ;;; start, an integer ;;; finish, an integer ;;; Purpose: ;;; Builds a list that contains elements start ... finish1 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>listkernel (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 (vectorref vec start) (vector>listkernel 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>listkernel vec 0 (vectorlength vec))))
November 7, 1997 [John Stone and/or Henry Walker]
April 5, 2000 [John Stone]
http://www.cs.grinnell.edu/~stone/courses/scheme/spring2000/vectors.xhtml
.
Wednesday, 20 September 2000 [Samuel A. Rebelsky]
Wednesday, 6 November 2000 [Samuel A. Rebelsky]
http://www.cs.grinnell.edu/~rebelsky/Courses/CS151/2000F/Readings/vectors.html
.
Thursday, 22 February 2001 [Samuel A. Rebelsky]
Wednesday, 7 March 2001 [Samuel A. Rebelsky]
list>vector
and
vector>list
that requires less background.
http://www.cs.grinnell.edu/~rebelsky/Courses/CS151/2001S/Readings/vectors.html
.
Wednesday, 2 October 2002 [Samuel A. Rebelsky]
Monday, 7 October 2002 [Samuel A. Rebelsky]
http://www.cs.grinnell.edu/~rebelsky/Courses/CS151/2002F/Readings/vectors.html
.
Wednesday, 12 February 2003 [Samuel A. Rebelsky]
http://www.cs.grinnell.edu/~rebelsky/Courses/CS153/2003S/Readings/vectors.html
.
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[SamR]
Disclaimer:
I usually create these pages on the fly
, which means that I rarely
proofread them and they may contain bad grammar and incorrect details.
It also means that I tend to update them regularly (see the history for
more details). Feel free to contact me with any suggestions for changes.
This document was generated by
Siteweaver on Tue May 6 09:21:49 2003.
The source to the document was last modified on Wed Feb 12 08:41:51 2003.
This document may be found at http://www.cs.grinnell.edu/~rebelsky/Courses/CS153/2003S/Readings/vectors.html
.
; ; Check with Bobby
Samuel A. Rebelsky, rebelsky@grinnell.edu