Functional Problem Solving (CSC 151 2014F) : Readings
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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.
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 several techniques for representing collections of data:
Representing a collection as a list has some problems. In particular, it is relatively expensive to get a particular element 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 tree, we need to figure out how many values are in the left subtree to decide where to look. Changing an element may be even worse, because once we've reached the position, we need to build the structure back to a new form.
Does this mean that lists 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) 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.
While some older implementations of Scheme may lack the
list->vector
,
vector->list
, and
vector-fill!
procedures, it is
straightforward to define them in terms of the others. You may
(optionally) read
about how to do so.
We frequently store only one type in a collection. For example, just as we had number trees, we could restrict the values in a vector to be only numbers, perhaps for keeping tallies of colors bearing certain properties (e.g., bright versus dark).
Say we had such a number vector--how could we increment the tally in one of the positions? After some reflection, it seems We need three steps, one to get the current value in the vector, another to increment that value, and the last step to make that value the new entry in the given position. We put these steps together as follows.
;;; Procedure: ;;; number-vector-increment-index! ;;; Parameters: ;;; vec, a vector ;;; index, an integer ;;; Purpose: ;;; Increment the value at a vector position ;;; Produces: ;;; [Nothing; called for side effect.] ;;; Preconditions: ;;; (vector-ref vec index) is a number ;;; Postconditions: ;;; Let val be (vector-ref vec index) before the procedure call. After the ;; call (vector-ref vec index) produces val+1. (define number-vector-increment-index! (lambda (vec index) (vector-set! vec index (increment (vector-ref vec index)))))
The fact that vectors allow mutation makes the increment
straightforward. What if we wanted to increment every value in a
vector? With lists we might think about using basic list recursion
to pass over the list and add one to each item. (And then we might
think better of it and use map
). There is no
analog for car
/cdr
that we
might use to process lists, neither is there an analog for
map
.
Fortunately, indexing is fast in vectors, so we can use numeric
recursion to iterate over all the positions in a vector, applying
our increment function along the way. Thus, we might track
pos
, the current position to modify in the
vector, starting at zero and ending when we reach the length of the
vector. We can encapsulate this repeated mutation with a named let
;;; Procedure: ;;; number-vector-increment! ;;; Parameters: ;;; vec, a vector ;;; Purpose: ;;; Increment the value at all vector positions ;;; Produces: ;;; [Nothing; called for side effect.] ;;; Preconditions: ;;; (vector-ref vec index) for 0 <= index < (vector-length vec) is a number. ;;; number-vect-increment-index! is defined ;;; Postconditions: ;;; Let val be (vector-ref vec index) before the procedure call. After the ;; call (vector-ref vec index) produces val+1. (define number-vector-increment! (lambda (vec) (let ([len (vector-length vec)]) ; unchanging value, tells recursion to stop (let kernel! ([pos 0]) ; Start the recursion at the first position (when (< pos len) ; When the position is valid, (number-vector-increment-index! vec pos) ; increment the number at pos (kernel! (+ 1 pos))))))) ; and process the rest of the vector
There are various ways to mutate all vector elements; the lab extras
will suggest some alternatives. Unfortunately, we do not always have
a special helper function like
number-vector-increment-index!
that allows us
to write such streamlined code. Instead we must combine the indexing
and mutation steps directly in the recursion.
As an example, suppose we wished to convert the tallies to
percentages by dividing each number by the sum of all the numbers in
the vector. Assuming you have a means of totalling these numbers (a
procedure you will write in the lab), we still need to iterate over
all vector positions, just as we did in
number-vector-increment!
only this time we use
the position variable pos
directly to index
the vector with vector-ref
, rather than with a
helper. Putting this together, we might write the following procedure.
;;; Procedure: ;;; number-vector-divide! ;;; Parameters: ;;; vec, a vector ;;; divisor, a number ;;; Purpose: ;;; Divide the value at all vector positions by a given divisor ;;; Produces: ;;; [Nothing; called for side effect.] ;;; Preconditions: ;;; (vector-ref vec index) for 0 <= index < (vector-length vec) is a number. ;;; Postconditions: ;;; Let val be (vector-ref vec index) before the procedure call. After the ;; call (vector-ref vec index) produces val/divisor. (define number-vector-divide! (lambda (vec divisor) (let ([len (vector-length vec)]) ; unchanging and tells recursion to stop (let kernel! ([pos 0]) ; Start the recursion at the first position (when (< pos len) ; When the position is valid, (vector-set! vec ; Set the new value in the vector pos ; at the current position (/ (vector-ref vec pos) divisor)) ; to the quotient (kernel! (+ 1 pos))))))) ; and process the rest of the vector
Of course, we need not change the vector as we iterate over it. Perhaps we just want to find the largest value in a vector. We still need to iterate over all the positions, except we might now use the standard recursive pattern that requires us to use a combination step to get a complete answer from the partial (recursive) answer.
;;; Procedure: ;;; number-vector-largest ;;; Parameters: ;;; vec, a vector ;;; Purpose: ;;; Find the largest number in a vector ;;; Produces: ;;; largest, a number ;;; Preconditions: ;;; (vector-ref vec index) for 0 <= index < (vector-length vec) is a number. ;;; Postconditions: ;;; (vector-ref vec index) <= largest for 0 <= index < (vector-length vec) ;;; largest is a value in vec, i.e., there exists an integer index such that ;;; 0 <= index < (vector-length vec) and ;;; (vector-ref vec index) = largest (define number-vector-largest (lambda (vec) (let ([last (- (vector-length vec) 1)]) ; last position to test (let kernel ([pos 0]) ; Start the recursion at the first position (if (= pos last) ; We are at the last position, so return the number (vector-ref vec pos) ; Otherwise return the maximum of the current position and ; the largest number in the rest of the vector (max (vector-ref vec pos) (kernel (+ 1 pos))))))))
Each time we learn a new structure, we learn techniques for recursion 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. (We also typically precompute a stopping point so that we don't have to recompute it for each pass through the kernel.)
The test for the base case is then to check whether the current position has reached the stopping point and the “simplify” step is to add 1 to the position. As usual, the “combine” step is problem dependent.
(define vector-proc (lambda (vec other) (let ([len (vector-length vec)]) (let kernel ([pos 0]) (if (= pos len) (base-case vec other) (combine (vector-ref vec pos) (kernel (+ pos 1))))))))
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) (let kernel ([pos (- (vector-length vec) 1)]) (if (< pos 0) (base-case vec other) (combine (vector-ref vec pos) (kernel (- pos 1)))))))
Because vectors are mutable, we often use an imperative pattern with (perhaps multiple) operations that have side-effects.
(define vector-proc (lambda (vec other) (let ([len (vector-length vec)]) (let kernel! ([pos 0]) (cond [(= pos len) (base-case! vec other) ...] [else (operation! vec pos) ... (kernel! (+ 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)
a. In DrRacket's Interactions pane, type in a vector literal that denotes a vector containing just the two elements 3.14159 and 2.71828. How does MediaScript display the value of this vector?
b. Create a vector that contains the same two values by using the
vector
procedure.
c. Create a vector that contains the same two values by using the
make-vector
and vector-set!
procedures.
a. Make a copy of vectors-lab.rkt
,
which contains much of the code from the reading.
b. Try using number-vector-increment-index!
on
a vector from the previous check.
>
(define v1 (vector 3.14159 2.71828))
>
(number-vector-increment-index! v1 1)
>
v1
c. Use number-vector-increment!
on the vector
to verify it behaves as intended.
>
(define v2 (make-vector 2 3.14159))
>
(vector-set! v2 12.71828))
>
(number-vector-increment! v2)
>
v2
d. Checks 2.b and 2.c showed you how to do checks 1.b and 1.c, respectively. What do you suppose would happen if we tried these operations on the vector you created in check 1.a?
>
(number-vector-increment-index! '#(3.14159 2.71828) 1)
e. Verify your prediction. Why do you think this happens?