How to get the nth value of a function with arbitrary number of argumentsHow to distribute a generic function of two arguments (without evaluating the arguments)How to take the derivative w.r.t. an arbitrary function?How can I get the number of slots in Function?General function taking general number of argumentsOptional arguments with lists as function parameterFunction with lazy and non-lazy arguments
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How to get the nth value of a function with arbitrary number of arguments
How to distribute a generic function of two arguments (without evaluating the arguments)How to take the derivative w.r.t. an arbitrary function?How can I get the number of slots in Function?General function taking general number of argumentsOptional arguments with lists as function parameterFunction with lazy and non-lazy arguments
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$begingroup$
I know that I can define a function of a List and get the nth argument like:
f[x_List]:=x[[1]]+x[[2]]
What if the argument of f is not a list? I.e. how to define an equivalent function for
f[x__Integer]:=?
This is probably extremely basic, but I couldn't figure it out from the Slot documentation.
functions
asked Oct 14 at 11:40
user366202user366202
755 bronze badges
$endgroup$
add a comment
|
$begingroup$
I know that I can define a function of a List and get the nth argument like:
f[x_List]:=x[[1]]+x[[2]]
What if the argument of f is not a list? I.e. how to define an equivalent function for
f[x__Integer]:=?
This is probably extremely basic, but I couldn't figure it out from the Slot documentation.
functions
asked Oct 14 at 11:40
user366202user366202
755 bronze badges
$endgroup$
add a comment
|
$begingroup$
I know that I can define a function of a List and get the nth argument like:
f[x_List]:=x[[1]]+x[[2]]
What if the argument of f is not a list? I.e. how to define an equivalent function for
f[x__Integer]:=?
This is probably extremely basic, but I couldn't figure it out from the Slot documentation.
functions
asked Oct 14 at 11:40
user366202user366202
755 bronze badges
$endgroup$
I know that I can define a function of a List and get the nth argument like:
f[x_List]:=x[[1]]+x[[2]]
What if the argument of f is not a list? I.e. how to define an equivalent function for
f[x__Integer]:=?
This is probably extremely basic, but I couldn't figure it out from the Slot documentation.
functions
functions
asked Oct 14 at 11:40
user366202user366202
755 bronze badges
asked Oct 14 at 11:40
user366202user366202
755 bronze badges
asked Oct 14 at 11:40
user366202user366202
755 bronze badges
asked Oct 14 at 11:40
user366202user366202
755 bronze badges
asked Oct 14 at 11:40
user366202user366202
755 bronze badges
755 bronze badges
add a comment
|
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
With f[x__Integer] := ... you can use
x[[i]]
to get the ith argument.
answered Oct 14 at 11:43
SzabolcsSzabolcs
173k18 gold badges473 silver badges1008 bronze badges
$endgroup$
add a comment
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$begingroup$
Szabolcs's answer (wrapping the sequence x with List and using Part) is the way.
An alternative is to use Indexed
ClearAll[f2]
f2[x__Integer] := Indexed[x, 1] + Indexed[x, 2]
f2[1, 2, 3, 4, 5]
3
You can also use Slot (#):
ClearAll[f3, f4]
f3[x__Integer] := Slot[1] + Slot[2] &[x]
f3[1, 2, 3, 4, 5]
3
f4[x__Integer] := #1 + #2 &[x]
f4[1, 2, 3, 4, 5]
3
but you can use a pure function directly to define your function:
f5 = #1 + #2 &
answered Oct 14 at 12:24
kglrkglr
223k10 gold badges253 silver badges511 bronze badges
$endgroup$
$begingroup$
Thanks. Is the pure function method likely to be faster than the Indexed/List methods?
$endgroup$
– user366202
Oct 14 at 14:24
$begingroup$
@user366202, I am not sure.
$endgroup$
– kglr
Oct 14 at 14:26
add a comment
|
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
With f[x__Integer] := ... you can use
x[[i]]
to get the ith argument.
answered Oct 14 at 11:43
SzabolcsSzabolcs
173k18 gold badges473 silver badges1008 bronze badges
$endgroup$
add a comment
|
$begingroup$
With f[x__Integer] := ... you can use
x[[i]]
to get the ith argument.
answered Oct 14 at 11:43
SzabolcsSzabolcs
173k18 gold badges473 silver badges1008 bronze badges
$endgroup$
add a comment
|
$begingroup$
With f[x__Integer] := ... you can use
x[[i]]
to get the ith argument.
answered Oct 14 at 11:43
SzabolcsSzabolcs
173k18 gold badges473 silver badges1008 bronze badges
$endgroup$
With f[x__Integer] := ... you can use
x[[i]]
to get the ith argument.
answered Oct 14 at 11:43
SzabolcsSzabolcs
173k18 gold badges473 silver badges1008 bronze badges
answered Oct 14 at 11:43
SzabolcsSzabolcs
173k18 gold badges473 silver badges1008 bronze badges
answered Oct 14 at 11:43
SzabolcsSzabolcs
173k18 gold badges473 silver badges1008 bronze badges
answered Oct 14 at 11:43
SzabolcsSzabolcs
173k18 gold badges473 silver badges1008 bronze badges
173k18 gold badges473 silver badges1008 bronze badges
add a comment
|
add a comment
|
$begingroup$
Szabolcs's answer (wrapping the sequence x with List and using Part) is the way.
An alternative is to use Indexed
ClearAll[f2]
f2[x__Integer] := Indexed[x, 1] + Indexed[x, 2]
f2[1, 2, 3, 4, 5]
3
You can also use Slot (#):
ClearAll[f3, f4]
f3[x__Integer] := Slot[1] + Slot[2] &[x]
f3[1, 2, 3, 4, 5]
3
f4[x__Integer] := #1 + #2 &[x]
f4[1, 2, 3, 4, 5]
3
but you can use a pure function directly to define your function:
f5 = #1 + #2 &
answered Oct 14 at 12:24
kglrkglr
223k10 gold badges253 silver badges511 bronze badges
$endgroup$
$begingroup$
Thanks. Is the pure function method likely to be faster than the Indexed/List methods?
$endgroup$
– user366202
Oct 14 at 14:24
$begingroup$
@user366202, I am not sure.
$endgroup$
– kglr
Oct 14 at 14:26
add a comment
|
$begingroup$
Szabolcs's answer (wrapping the sequence x with List and using Part) is the way.
An alternative is to use Indexed
ClearAll[f2]
f2[x__Integer] := Indexed[x, 1] + Indexed[x, 2]
f2[1, 2, 3, 4, 5]
3
You can also use Slot (#):
ClearAll[f3, f4]
f3[x__Integer] := Slot[1] + Slot[2] &[x]
f3[1, 2, 3, 4, 5]
3
f4[x__Integer] := #1 + #2 &[x]
f4[1, 2, 3, 4, 5]
3
but you can use a pure function directly to define your function:
f5 = #1 + #2 &
answered Oct 14 at 12:24
kglrkglr
223k10 gold badges253 silver badges511 bronze badges
$endgroup$
$begingroup$
Thanks. Is the pure function method likely to be faster than the Indexed/List methods?
$endgroup$
– user366202
Oct 14 at 14:24
$begingroup$
@user366202, I am not sure.
$endgroup$
– kglr
Oct 14 at 14:26
add a comment
|
$begingroup$
Szabolcs's answer (wrapping the sequence x with List and using Part) is the way.
An alternative is to use Indexed
ClearAll[f2]
f2[x__Integer] := Indexed[x, 1] + Indexed[x, 2]
f2[1, 2, 3, 4, 5]
3
You can also use Slot (#):
ClearAll[f3, f4]
f3[x__Integer] := Slot[1] + Slot[2] &[x]
f3[1, 2, 3, 4, 5]
3
f4[x__Integer] := #1 + #2 &[x]
f4[1, 2, 3, 4, 5]
3
but you can use a pure function directly to define your function:
f5 = #1 + #2 &
answered Oct 14 at 12:24
kglrkglr
223k10 gold badges253 silver badges511 bronze badges
$endgroup$
Szabolcs's answer (wrapping the sequence x with List and using Part) is the way.
An alternative is to use Indexed
ClearAll[f2]
f2[x__Integer] := Indexed[x, 1] + Indexed[x, 2]
f2[1, 2, 3, 4, 5]
3
You can also use Slot (#):
ClearAll[f3, f4]
f3[x__Integer] := Slot[1] + Slot[2] &[x]
f3[1, 2, 3, 4, 5]
3
f4[x__Integer] := #1 + #2 &[x]
f4[1, 2, 3, 4, 5]
3
but you can use a pure function directly to define your function:
f5 = #1 + #2 &
answered Oct 14 at 12:24
kglrkglr
223k10 gold badges253 silver badges511 bronze badges
answered Oct 14 at 12:24
kglrkglr
223k10 gold badges253 silver badges511 bronze badges
answered Oct 14 at 12:24
kglrkglr
223k10 gold badges253 silver badges511 bronze badges
answered Oct 14 at 12:24
kglrkglr
223k10 gold badges253 silver badges511 bronze badges
223k10 gold badges253 silver badges511 bronze badges
$begingroup$
Thanks. Is the pure function method likely to be faster than the Indexed/List methods?
$endgroup$
– user366202
Oct 14 at 14:24
$begingroup$
@user366202, I am not sure.
$endgroup$
– kglr
Oct 14 at 14:26
add a comment
|
$begingroup$
Thanks. Is the pure function method likely to be faster than the Indexed/List methods?
$endgroup$
– user366202
Oct 14 at 14:24
$begingroup$
@user366202, I am not sure.
$endgroup$
– kglr
Oct 14 at 14:26
$begingroup$
Thanks. Is the pure function method likely to be faster than the Indexed/List methods?
$endgroup$
– user366202
Oct 14 at 14:24
$begingroup$
Thanks. Is the pure function method likely to be faster than the Indexed/List methods?
$endgroup$
– user366202
Oct 14 at 14:24
$begingroup$
@user366202, I am not sure.
$endgroup$
– kglr
Oct 14 at 14:26
$begingroup$
@user366202, I am not sure.
$endgroup$
– kglr
Oct 14 at 14:26
add a comment
|
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