[ << ] | [ < ] | [ Up ] | [ > ] | [ >> ] | [Top] | [Contents] | [Index] | [ ? ] |
A list represents a sequence of zero or more elements (which may be any Lisp objects). The important difference between lists and vectors is that two or more lists can share part of their structure; in addition, you can insert or delete elements in a list without copying the whole list.
1.1 Lists and Cons Cells | How lists are made out of cons cells. | |
1.2 Lists as Linked Pairs of Boxes | Graphical notation to explain lists. | |
1.3 Predicates on Lists | Is this object a list? Comparing two lists. | |
1.4 Accessing Elements of Lists | Extracting the pieces of a list. | |
1.5 Building Cons Cells and Lists | Creating list structure. | |
1.6 Modifying Existing List Structure | Storing new pieces into an existing list. | |
1.7 Using Lists as Sets | A list can represent a finite mathematical set. | |
1.8 Association Lists | A list can represent a finite relation or mapping. |
[ << ] | [ < ] | [ Up ] | [ > ] | [ >> ] | [Top] | [Contents] | [Index] | [ ? ] |
Lists in Lisp are not a primitive data type; they are built up from cons cells. A cons cell is a data object which represents an ordered pair. It records two Lisp objects, one labeled as the CAR, and the other labeled as the CDR. (These names are traditional.)
A list is made by chaining cons cells together, one cons cell per
element. By convention, the CARs of the cons cells are the
elements of the list, and the CDRs are used to chain the list: the
CDR of each cons cell is the following cons cell. The CDR of
the last cons cell is nil
. This asymmetry between the CAR
and the CDR is entirely a matter of convention; at the level of
cons cells, the CAR and CDR slots have the same
characteristics.
The symbol nil
is considered a list as well as a symbol; it is
the list with no elements. For convenience, the symbol nil
is
considered to have nil
as its CDR (and also as its
CAR).
The CDR of any nonempty list l is a list containing all the elements of l except the first.
[ << ] | [ < ] | [ Up ] | [ > ] | [ >> ] | [Top] | [Contents] | [Index] | [ ? ] |
A cons cell can be illustrated as a pair of boxes. The first box
represents the CAR and the second box represents the CDR.
Here is an illustration of the two-element list, (tulip lily)
,
made from two cons cells:
--------------- --------------- | car | cdr | | car | cdr | | tulip | o---------->| lily | nil | | | | | | | --------------- ---------------
Each pair of boxes represents a cons cell. Each box “refers to”,
“points to” or “contains” a Lisp object. (These terms are
synonymous.) The first box, which is the CAR of the first cons
cell, contains the symbol tulip
. The arrow from the CDR of
the first cons cell to the second cons cell indicates that the CDR
of the first cons cell points to the second cons cell.
The same list can be illustrated in a different sort of box notation like this:
___ ___ ___ ___ |___|___|--> |___|___|--> nil | | | | --> tulip --> lily
Here is a more complex illustration, this time of the three-element
list, ((pine needles) oak maple)
, the first element of which is
a two-element list:
___ ___ ___ ___ ___ ___ |___|___|--> |___|___|--> |___|___|--> nil | | | | | | | --> oak --> maple | | ___ ___ ___ ___ --> |___|___|--> |___|___|--> nil | | | | --> pine --> needles
The same list is represented in the first box notation like this:
-------------- -------------- -------------- | car | cdr | | car | cdr | | car | cdr | | o | o------->| oak | o------->| maple | nil | | | | | | | | | | | -- | --------- -------------- -------------- | | | -------------- ---------------- | | car | cdr | | car | cdr | ------>| pine | o------->| needles | nil | | | | | | | -------------- ----------------
@xref{List Type}, for the read and print syntax of lists, and for more “box and arrow” illustrations of lists.
[ << ] | [ < ] | [ Up ] | [ > ] | [ >> ] | [Top] | [Contents] | [Index] | [ ? ] |
The following predicates test whether a Lisp object is an atom, is a cons
cell or is a list, or whether it is the distinguished object nil
.
(Many of these tests can be defined in terms of the others, but they are
used so often that it is worth having all of them.)
This function returns t
if object is a cons cell, nil
otherwise. nil
is not a cons cell, although it is a list.
This function returns t
if object is an atom, nil
otherwise. All objects except cons cells are atoms. The symbol
nil
is an atom and is also a list; it is the only Lisp object
which is both.
(atom object) ≡ (not (consp object))
This function returns t
if object is a cons cell or
nil
. Otherwise, it returns nil
.
(listp '(1)) ⇒ t
(listp '()) ⇒ t
This function is the opposite of listp
: it returns t
if
object is not a list. Otherwise, it returns nil
.
(listp object) ≡ (not (nlistp object))
This function returns t
if object is nil
, and
returns nil
otherwise. This function is identical to not
,
but as a matter of clarity we use null
when object is
considered a list and not
when it is considered a truth value
(see not
in @ref{Combining Conditions}).
(null '(1)) ⇒ nil
(null '()) ⇒ t
[ << ] | [ < ] | [ Up ] | [ > ] | [ >> ] | [Top] | [Contents] | [Index] | [ ? ] |
This function returns the value pointed to by the first pointer of the cons cell cons-cell. Expressed another way, this function returns the CAR of cons-cell.
As a special case, if cons-cell is nil
, then car
is defined to return nil
; therefore, any list is a valid argument
for car
. An error is signaled if the argument is not a cons cell
or nil
.
(car '(a b c)) ⇒ a
(car '()) ⇒ nil
This function returns the value pointed to by the second pointer of the cons cell cons-cell. Expressed another way, this function returns the CDR of cons-cell.
As a special case, if cons-cell is nil
, then cdr
is defined to return nil
; therefore, any list is a valid argument
for cdr
. An error is signaled if the argument is not a cons cell
or nil
.
(cdr '(a b c)) ⇒ (b c)
(cdr '()) ⇒ nil
This function lets you take the CAR of a cons cell while avoiding
errors for other data types. It returns the CAR of object if
object is a cons cell, nil
otherwise. This is in contrast
to car
, which signals an error if object is not a list.
(car-safe object) ≡ (let ((x object)) (if (consp x) (car x) nil))
This function lets you take the CDR of a cons cell while
avoiding errors for other data types. It returns the CDR of
object if object is a cons cell, nil
otherwise.
This is in contrast to cdr
, which signals an error if
object is not a list.
(cdr-safe object) ≡ (let ((x object)) (if (consp x) (cdr x) nil))
This function returns the nth element of list. Elements
are numbered starting with zero, so the CAR of list is
element number zero. If the length of list is n or less,
the value is nil
.
If n is less than zero, then the first element is returned.
(nth 2 '(1 2 3 4)) ⇒ 3
(nth 10 '(1 2 3 4)) ⇒ nil
(nth -3 '(1 2 3 4)) ⇒ 1 (nth n x) ≡ (car (nthcdr n x))
This function returns the nth cdr of list. In other words, it removes the first n links of list and returns what follows.
If n is less than or equal to zero, then all of list is
returned. If the length of list is n or less, the value is
nil
.
(nthcdr 1 '(1 2 3 4)) ⇒ (2 3 4)
(nthcdr 10 '(1 2 3 4)) ⇒ nil
(nthcdr -3 '(1 2 3 4)) ⇒ (1 2 3 4)
[ << ] | [ < ] | [ Up ] | [ > ] | [ >> ] | [Top] | [Contents] | [Index] | [ ? ] |
Many functions build lists, as lists reside at the very heart of Lisp.
cons
is the fundamental list-building function; however, it is
interesting to note that list
is used more times in the source
code for Emacs than cons
.
This function is the fundamental function used to build new list structure. It creates a new cons cell, making object1 the CAR, and object2 the CDR. It then returns the new cons cell. The arguments object1 and object2 may be any Lisp objects, but most often object2 is a list.
(cons 1 '(2)) ⇒ (1 2)
(cons 1 '()) ⇒ (1)
(cons 1 2) ⇒ (1 . 2)
cons
is often used to add a single element to the front of a
list. This is called consing the element onto the list. For
example:
(setq list (cons newelt list))
Note that there is no conflict between the variable named list
used in this example and the function named list
described below;
any symbol can serve both functions.
This function creates a list with objects as its elements. The
resulting list is always nil
-terminated. If no objects
are given, the empty list is returned.
(list 1 2 3 4 5) ⇒ (1 2 3 4 5)
(list 1 2 '(3 4 5) 'foo) ⇒ (1 2 (3 4 5) foo)
(list) ⇒ nil
This function creates a list of length length, in which all the
elements have the identical value object. Compare
make-list
with make-string
(@pxref{Creating Strings}).
(make-list 3 'pigs) ⇒ (pigs pigs pigs)
(make-list 0 'pigs) ⇒ nil
This function returns a list containing all the elements of sequences. The sequences may be lists, vectors, strings, or integers. All arguments except the last one are copied, so none of them are altered.
The final argument to append
may be any object but it is
typically a list. The final argument is not copied or converted; it
becomes part of the structure of the new list.
Here is an example:
(setq trees '(pine oak)) ⇒ (pine oak) (setq more-trees (append '(maple birch) trees)) ⇒ (maple birch pine oak)
trees ⇒ (pine oak) more-trees ⇒ (maple birch pine oak)
(eq trees (cdr (cdr more-trees))) ⇒ t
You can see what happens by looking at a box diagram. The variable
trees
is set to the list (pine oak)
and then the variable
more-trees
is set to the list (maple birch pine oak)
.
However, the variable trees
continues to refer to the original
list:
more-trees trees | | | ___ ___ ___ ___ -> ___ ___ ___ ___ --> |___|___|--> |___|___|--> |___|___|--> |___|___|--> nil | | | | | | | | --> maple -->birch --> pine --> oak
An empty sequence contributes nothing to the value returned by
append
. As a consequence of this, a final nil
argument
forces a copy of the previous argument.
trees ⇒ (pine oak)
(setq wood (append trees ())) ⇒ (pine oak)
wood ⇒ (pine oak)
(eq wood trees) ⇒ nil
This once was the standard way to copy a list, before the function
copy-sequence
was invented. @xref{Sequences Arrays Vectors}.
With the help of apply
, we can append all the lists in a list of
lists:
(apply 'append '((a b c) nil (x y z) nil)) ⇒ (a b c x y z)
If no sequences are given, nil
is returned:
(append) ⇒ nil
In the special case where one of the sequences is an integer
(not a sequence of integers), it is first converted to a string of
digits making up the decimal print representation of the integer. This
special case exists for compatibility with Mocklisp, and we don’t
recommend you take advantage of it. If you want to convert an integer
in this way, use format
(@pxref{Formatting Strings}) or
number-to-string
(@pxref{String Conversion}).
(setq trees '(pine oak)) ⇒ (pine oak)
(char-to-string 54) ⇒ "6"
(setq longer-list (append trees 6 '(spruce))) ⇒ (pine oak 54 spruce)
(setq x-list (append trees 6 6)) ⇒ (pine oak 54 . 6)
See nconc
in Functions that Rearrange Lists, for another way to join lists
without copying.
This function creates a new list whose elements are the elements of list, but in reverse order. The original argument list is not altered.
(setq x '(1 2 3 4)) ⇒ (1 2 3 4)
(reverse x) ⇒ (4 3 2 1) x ⇒ (1 2 3 4)
[ << ] | [ < ] | [ Up ] | [ > ] | [ >> ] | [Top] | [Contents] | [Index] | [ ? ] |
You can modify the CAR and CDR contents of a cons cell with the
primitives setcar
and setcdr
.
Common Lisp note: Common Lisp uses functions
rplaca
andrplacd
to alter list structure; they change structure the same way assetcar
andsetcdr
, but the Common Lisp functions return the cons cell whilesetcar
andsetcdr
return the new CAR or CDR.
1.6.1 Altering List Elements with setcar | Replacing an element in a list. | |
1.6.2 Altering the CDR of a List | Replacing part of the list backbone. This can be used to remove or add elements. | |
1.6.3 Functions that Rearrange Lists | Reordering the elements in a list; combining lists. |
[ << ] | [ < ] | [ Up ] | [ > ] | [ >> ] | [Top] | [Contents] | [Index] | [ ? ] |
setcar
Changing the CAR of a cons cell is done with setcar
and
replaces one element of a list with a different element.
This function stores object as the new CAR of cons, replacing its previous CAR. It returns the value object. For example:
(setq x '(1 2)) ⇒ (1 2)
(setcar x '4) ⇒ 4
x ⇒ (4 2)
When a cons cell is part of the shared structure of several lists, storing a new CAR into the cons changes one element of each of these lists. Here is an example:
;; Create two lists that are partly shared.
(setq x1 '(a b c))
⇒ (a b c)
(setq x2 (cons 'z (cdr x1)))
⇒ (z b c)
;; Replace the CAR of a shared link. (setcar (cdr x1) 'foo) ⇒ foo x1 ; Both lists are changed. ⇒ (a foo c) x2 ⇒ (z foo c)
;; Replace the CAR of a link that is not shared. (setcar x1 'baz) ⇒ baz x1 ; Only one list is changed. ⇒ (baz foo c) x2 ⇒ (z foo c)
Here is a graphical depiction of the shared structure of the two lists
x1 and x2, showing why replacing b
changes them both:
___ ___ ___ ___ ___ ___ x1---> |___|___|----> |___|___|--> |___|___|--> nil | --> | | | | | | --> a | --> b --> c | ___ ___ | x2--> |___|___|-- | | --> z
Here is an alternative form of box diagram, showing the same relationship:
x1: -------------- -------------- -------------- | car | cdr | | car | cdr | | car | cdr | | a | o------->| b | o------->| c | nil | | | | -->| | | | | | -------------- | -------------- -------------- | x2: | -------------- | | car | cdr | | | z | o---- | | | --------------
[ << ] | [ < ] | [ Up ] | [ > ] | [ >> ] | [Top] | [Contents] | [Index] | [ ? ] |
The lowest-level primitive for modifying a CDR is setcdr
:
This function stores object into the cdr of cons. The value returned is object, not cons.
Here is an example of replacing the CDR of a list with a different list. All but the first element of the list are removed in favor of a different sequence of elements. The first element is unchanged, because it resides in the CAR of the list, and is not reached via the CDR.
(setq x '(1 2 3)) ⇒ (1 2 3)
(setcdr x '(4)) ⇒ (4)
x ⇒ (1 4)
You can delete elements from the middle of a list by altering the
CDRs of the cons cells in the list. For example, here we delete
the second element, b
, from the list (a b c)
, by changing
the CDR of the first cell:
(setq x1 '(a b c)) ⇒ (a b c) (setcdr x1 (cdr (cdr x1))) ⇒ (c) x1 ⇒ (a c)
Here is the result in box notation:
-------------------- | | -------------- | -------------- | -------------- | car | cdr | | | car | cdr | -->| car | cdr | | a | o----- | b | o-------->| c | nil | | | | | | | | | | -------------- -------------- --------------
The second cons cell, which previously held the element b
, still
exists and its CAR is still b
, but it no longer forms part
of this list.
It is equally easy to insert a new element by changing CDRs:
(setq x1 '(a b c)) ⇒ (a b c) (setcdr x1 (cons 'd (cdr x1))) ⇒ (d b c) x1 ⇒ (a d b c)
Here is this result in box notation:
-------------- ------------- ------------- | car | cdr | | car | cdr | | car | cdr | | a | o | -->| b | o------->| c | nil | | | | | | | | | | | | --------- | -- | ------------- ------------- | | ----- -------- | | | --------------- | | | car | cdr | | -->| d | o------ | | | ---------------
[ << ] | [ < ] | [ Up ] | [ > ] | [ >> ] | [Top] | [Contents] | [Index] | [ ? ] |
Here are some functions that rearrange lists “destructively” by modifying the CDRs of their component cons cells. We call these functions “destructive” because the original lists passed as arguments to them are chewed up to produce a new list that is subsequently returned.
This function returns a list containing all the elements of lists.
Unlike append
(see section Building Cons Cells and Lists), the lists are
not copied. Instead, the last CDR of each of the
lists is changed to refer to the following list. The last of the
lists is not altered. For example:
(setq x '(1 2 3)) ⇒ (1 2 3)
(nconc x '(4 5)) ⇒ (1 2 3 4 5)
x ⇒ (1 2 3 4 5)
Since the last argument of nconc
is not itself modified, it is
reasonable to use a constant list, such as '(4 5)
, as is done in
the above example. For the same reason, the last argument need not be a
list:
(setq x '(1 2 3)) ⇒ (1 2 3)
(nconc x 'z) ⇒ (1 2 3 . z)
x ⇒ (1 2 3 . z)
A common pitfall is to use a quoted constant list as a non-last
argument to nconc
. If you do this, your program will change
each time you run it! Here is what happens:
(defun add-foo (x) ; This function should add
(nconc '(foo) x)) ; foo
to the front of its arg.
(symbol-function 'add-foo) ⇒ (lambda (x) (nconc (quote (foo)) x))
(setq xx (add-foo '(1 2))) ; It seems to work.
⇒ (foo 1 2)
(setq xy (add-foo '(3 4))) ; What happened?
⇒ (foo 1 2 3 4)
(eq xx xy) ⇒ t
(symbol-function 'add-foo) ⇒ (lambda (x) (nconc (quote (foo 1 2 3 4) x)))
This function reverses the order of the elements of list.
Unlike reverse
, nreverse
alters its argument destructively
by reversing the CDRs in the cons cells forming the list. The cons
cell which used to be the last one in list becomes the first cell
of the value.
For example:
(setq x '(1 2 3 4)) ⇒ (1 2 3 4)
x ⇒ (1 2 3 4) (nreverse x) ⇒ (4 3 2 1)
;; The cell that was first is now last.
x
⇒ (1)
To avoid confusion, we usually store the result of nreverse
back in the same variable which held the original list:
(setq x (nreverse x))
Here is the nreverse
of our favorite example, (a b c)
,
presented graphically:
Original list head: Reversed list: ------------- ------------- ------------ | car | cdr | | car | cdr | | car | cdr | | a | nil |<-- | b | o |<-- | c | o | | | | | | | | | | | | | | ------------- | --------- | - | -------- | - | | | | ------------- ------------
This function sorts list stably, though destructively, and returns the sorted list. It compares elements using predicate. A stable sort is one in which elements with equal sort keys maintain their relative order before and after the sort. Stability is important when successive sorts are used to order elements according to different criteria.
The argument predicate must be a function that accepts two
arguments. It is called with two elements of list. To get an
increasing order sort, the predicate should return t
if the
first element is “less than” the second, or nil
if not.
The destructive aspect of sort
is that it rearranges the cons
cells forming list by changing CDRs. A nondestructive sort
function would create new cons cells to store the elements in their
sorted order. If you wish to sort a list without destroying the
original, copy it first with copy-sequence
.
The CARs of the cons cells are not changed; the cons cell that
originally contained the element a
in list still has
a
in its CAR after sorting, but it now appears in a
different position in the list due to the change of CDRs. For
example:
(setq nums '(1 3 2 6 5 4 0)) ⇒ (1 3 2 6 5 4 0)
(sort nums '<) ⇒ (0 1 2 3 4 5 6)
nums ⇒ (1 2 3 4 5 6)
Note that the list in nums
no longer contains 0; this is the same
cons cell that it was before, but it is no longer the first one in the
list. Don’t assume a variable that formerly held the argument now holds
the entire sorted list! Instead, save the result of sort
and use
that. Most often we store the result back into the variable that held
the original list:
(setq nums (sort nums '<))
@xref{Sorting}, for more functions that perform sorting.
See documentation
in @ref{Accessing Documentation}, for a
useful example of sort
.
[ << ] | [ < ] | [ Up ] | [ > ] | [ >> ] | [Top] | [Contents] | [Index] | [ ? ] |
A list can represent an unordered mathematical set—simply consider a
value an element of a set if it appears in the list, and ignore the
order of the list. To form the union of two sets, use append
(as
long as you don’t mind having duplicate elements). Other useful
functions for sets include memq
and delq
, and their
equal
versions, member
and delete
.
Common Lisp note: Common Lisp has functions
union
(which avoids duplicate elements) andintersection
for set operations, but GNU Emacs Lisp does not have them. You can write them in Lisp if you wish.
This function tests to see whether object is a member of
list. If it is, memq
returns a list starting with the
first occurrence of object. Otherwise, it returns nil
.
The letter ‘q’ in memq
says that it uses eq
to
compare object against the elements of the list. For example:
(memq 2 '(1 2 3 2 1)) ⇒ (2 3 2 1)
(memq '(2) '((1) (2))) ; (2)
and (2)
are not eq
.
⇒ nil
This function removes all elements eq
to object from
list. The letter ‘q’ in delq
says that it uses
eq
to compare object against the elements of the list, like
memq
.
When delq
deletes elements from the front of the list, it does so
simply by advancing down the list and returning a sublist that starts
after those elements:
(delq 'a '(a b c)) ≡ (cdr '(a b c))
When an element to be deleted appears in the middle of the list, removing it involves changing the CDRs (see section Altering the CDR of a List).
(setq sample-list '(1 2 3 (4))) ⇒ (1 2 3 (4))
(delq 1 sample-list) ⇒ (2 3 (4))
sample-list ⇒ (1 2 3 (4))
(delq 2 sample-list) ⇒ (1 3 (4))
sample-list ⇒ (1 3 (4))
Note that (delq 2 sample-list)
modifies sample-list
to
splice out the second element, but (delq 1 sample-list)
does not
splice anything—it just returns a shorter list. Don’t assume that a
variable which formerly held the argument list now has fewer
elements, or that it still holds the original list! Instead, save the
result of delq
and use that. Most often we store the result back
into the variable that held the original list:
(setq flowers (delq 'rose flowers))
In the following example, the (4)
that delq
attempts to match
and the (4)
in the sample-list
are not eq
:
(delq '(4) sample-list) ⇒ (1 3 (4))
The following two functions are like memq
and delq
but use
equal
rather than eq
to compare elements. They are new in
Emacs 19.
The function member
tests to see whether object is a member
of list, comparing members with object using equal
.
If object is a member, memq
returns a list starting with
its first occurrence in list. Otherwise, it returns nil
.
Compare this with memq
:
(member '(2) '((1) (2))) ; (2)
and (2)
are equal
.
⇒ ((2))
(memq '(2) '((1) (2))) ; (2)
and (2)
are not eq
.
⇒ nil
;; Two strings with the same contents are equal
.
(member "foo" '("foo" "bar"))
⇒ ("foo" "bar")
This function removes all elements equal
to object from
list. It is to delq
as member
is to memq
: it
uses equal
to compare elements with object, like
member
; when it finds an element that matches, it removes the
element just as delq
would. For example:
(delete '(2) '((2) (1) (2))) ⇒ '((1))
Common Lisp note: The functions
member
anddelete
in GNU Emacs Lisp are derived from Maclisp, not Common Lisp. The Common Lisp versions do not useequal
to compare elements.
[ << ] | [ < ] | [ Up ] | [ > ] | [ >> ] | [Top] | [Contents] | [Index] | [ ? ] |
An association list, or alist for short, records a mapping from keys to values. It is a list of cons cells called associations: the CAR of each cell is the key, and the CDR is the associated value. (This usage of “key” is not related to the term “key sequence”; it means any object which can be looked up in a table.)
Here is an example of an alist. The key pine
is associated with
the value cones
; the key oak
is associated with
acorns
; and the key maple
is associated with seeds
.
'((pine . cones) (oak . acorns) (maple . seeds))
The associated values in an alist may be any Lisp objects; so may the
keys. For example, in the following alist, the symbol a
is
associated with the number 1
, and the string "b"
is
associated with the list (2 3)
, which is the CDR of
the alist element:
((a . 1) ("b" 2 3))
Sometimes it is better to design an alist to store the associated value in the CAR of the CDR of the element. Here is an example:
'((rose red) (lily white) (buttercup yellow)))
Here we regard red
as the value associated with rose
. One
advantage of this method is that you can store other related
information—even a list of other items—in the CDR of the
CDR. One disadvantage is that you cannot use rassq
(see
below) to find the element containing a given value. When neither of
these considerations is important, the choice is a matter of taste, as
long as you are consistent about it for any given alist.
Note that the same alist shown above could be regarded as having the
associated value in the CDR of the element; the value associated
with rose
would be the list (red)
.
Association lists are often used to record information that you might otherwise keep on a stack, since new associations may be added easily to the front of the list. When searching an association list for an association with a given key, the first one found is returned, if there is more than one.
In Emacs Lisp, it is not an error if an element of an association list is not a cons cell. The alist search functions simply ignore such elements. Many other versions of Lisp signal errors in such cases.
Note that property lists are similar to association lists in several respects. A property list behaves like an association list in which each key can occur only once. @xref{Property Lists}, for a comparison of property lists and association lists.
This function returns the first association for key in
alist. It compares key against the alist elements using
equal
(@pxref{Equality Predicates}). It returns nil
if no
association in alist has a CAR equal
to key.
For example:
(setq trees '((pine . cones) (oak . acorns) (maple . seeds))) ⇒ ((pine . cones) (oak . acorns) (maple . seeds)) (assoc 'oak trees) ⇒ (oak . acorns) (cdr (assoc 'oak trees)) ⇒ acorns (assoc 'birch trees) ⇒ nil
Here is another example in which the keys and values are not symbols:
(setq needles-per-cluster '((2 . ("Austrian Pine" "Red Pine")) (3 . "Pitch Pine") (5 . "White Pine"))) (cdr (assoc 3 needles-per-cluster)) ⇒ "Pitch Pine" (cdr (assoc 2 needles-per-cluster)) ⇒ ("Austrian Pine" "Red Pine")
This function is like assoc
in that it returns the first
association for key in alist, but it makes the comparison
using eq
instead of equal
. assq
returns nil
if no association in alist has a CAR eq
to key.
This function is used more often than assoc
, since eq
is
faster than equal
and most alists use symbols as keys.
@xref{Equality Predicates}.
(setq trees '((pine . cones) (oak . acorns) (maple . seeds))) (assq 'pine trees) ⇒ (pine . cones)
On the other hand, assq
is not usually useful in alists where the
keys may not be symbols:
(setq leaves '(("simple leaves" . oak) ("compound leaves" . horsechestnut))) (assq "simple leaves" leaves) ⇒ nil (assoc "simple leaves" leaves) ⇒ ("simple leaves" . oak)
This function returns the first association with value value in
alist. It returns nil
if no association in alist has
a CDR eq
to value.
rassq
is like assq
except that the CDR of the
alist associations is tested instead of the CAR. You can
think of this as “reverse assq
”, finding the key for a given
value.
For example:
(setq trees '((pine . cones) (oak . acorns) (maple . seeds))) (rassq 'acorns trees) ⇒ (oak . acorns) (rassq 'spores trees) ⇒ nil
Note that rassq
cannot be used to search for a value stored in
the CAR of the CDR of an element:
(setq colors '((rose red) (lily white) (buttercup yellow))) (rassq 'white colors) ⇒ nil
In this case, the CDR of the association (lily white)
is not
the symbol white
, but rather the list (white)
. This can
be seen more clearly if the association is written in dotted pair
notation:
(lily white) ≡ (lily . (white))
This function returns a two-level deep copy of alist: it creates a new copy of each association, so that you can alter the associations of the new alist without changing the old one.
(setq needles-per-cluster '((2 . ("Austrian Pine" "Red Pine")) (3 . "Pitch Pine") (5 . "White Pine"))) ⇒ ((2 "Austrian Pine" "Red Pine") (3 . "Pitch Pine") (5 . "White Pine")) (setq copy (copy-alist needles-per-cluster)) ⇒ ((2 "Austrian Pine" "Red Pine") (3 . "Pitch Pine") (5 . "White Pine")) (eq needles-per-cluster copy) ⇒ nil (equal needles-per-cluster copy) ⇒ t (eq (car needles-per-cluster) (car copy)) ⇒ nil (cdr (car (cdr needles-per-cluster))) ⇒ "Pitch Pine" (eq (cdr (car (cdr needles-per-cluster))) (cdr (car (cdr copy)))) ⇒ t
[Top] | [Contents] | [Index] | [ ? ] |
This document was generated on January 16, 2023 using texi2html 5.0.
The buttons in the navigation panels have the following meaning:
Button | Name | Go to | From 1.2.3 go to |
---|---|---|---|
[ << ] | FastBack | Beginning of this chapter or previous chapter | 1 |
[ < ] | Back | Previous section in reading order | 1.2.2 |
[ Up ] | Up | Up section | 1.2 |
[ > ] | Forward | Next section in reading order | 1.2.4 |
[ >> ] | FastForward | Next chapter | 2 |
[Top] | Top | Cover (top) of document | |
[Contents] | Contents | Table of contents | |
[Index] | Index | Index | |
[ ? ] | About | About (help) |
where the Example assumes that the current position is at Subsubsection One-Two-Three of a document of the following structure:
This document was generated on January 16, 2023 using texi2html 5.0.