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How to substitute values from a list into a function?
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$begingroup$
I was hoping to take pairs of numbers from a list and substitute them into a function. So if my list was
list = {{1,2}, {3,4}, {5,6}}
and my function was
function = a x^b
The output I'm hoping to get is
result =1x^2 + 3x^4 + 5x^6
How would I best do this?
list-manipulation functions
asked 41 mins ago
PineapplePineapple
111
New contributor
Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I was hoping to take pairs of numbers from a list and substitute them into a function. So if my list was
list = {{1,2}, {3,4}, {5,6}}
and my function was
function = a x^b
The output I'm hoping to get is
result =1x^2 + 3x^4 + 5x^6
How would I best do this?
list-manipulation functions
asked 41 mins ago
PineapplePineapple
111
New contributor
Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I was hoping to take pairs of numbers from a list and substitute them into a function. So if my list was
list = {{1,2}, {3,4}, {5,6}}
and my function was
function = a x^b
The output I'm hoping to get is
result =1x^2 + 3x^4 + 5x^6
How would I best do this?
list-manipulation functions
asked 41 mins ago
PineapplePineapple
111
New contributor
Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I was hoping to take pairs of numbers from a list and substitute them into a function. So if my list was
list = {{1,2}, {3,4}, {5,6}}
and my function was
function = a x^b
The output I'm hoping to get is
result =1x^2 + 3x^4 + 5x^6
How would I best do this?
list-manipulation functions
list-manipulation functions
asked 41 mins ago
PineapplePineapple
111
New contributor
Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 41 mins ago
PineapplePineapple
111
New contributor
Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 41 mins ago
PineapplePineapple
111
New contributor
Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 41 mins ago
PineapplePineapple
111
asked 41 mins ago
PineapplePineapple
111
111
New contributor
Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
This is not a Function:
function = a x^b
But this is:
function = {a,b} [Function] a x^b
You can Apply it to each element of
list = {{1,2}, {3,4}, {5,6}}
with
function @@@ list
{x^3, 3 x^5, 5 x^7}
and sum it up with Total:
Total[ function @@@ list ]
x^3 + 3 x^5 + 5 x^7
answered 35 mins ago
Henrik SchumacherHenrik Schumacher
55.4k576154
$endgroup$
$begingroup$
Why would you do 'function = {a,b} [Function] a x^b' instead of function[a_, b_, x_] = a*x^b?
$endgroup$
– Pineapple
30 mins ago
$begingroup$
Because[Function]is easily entered with the escape sequence esc f n esc . AlsoFunctiondefines a pure function whilefunction[a_, b_, x_] = a*x^bdefines a replacement rule. However, they act the same way - most of the time. So it is a matter of taste.
$endgroup$
– Henrik Schumacher
26 mins ago
$begingroup$
@Pineapple Take a look at this tutorial on Defining Functions. Another good resource is the tutorial on Immediate and Delayed Definitions for the difference betweenSet(=) andSetDelayed(:=).
$endgroup$
– MarcoB
12 mins ago
add a comment |
$begingroup$
Total[#*x^#2&@@@list]
x^2 + 3 x^4 + 5 x^6
answered 31 mins ago
J42161217J42161217
3,935322
$endgroup$
add a comment |
$begingroup$
Total[(#[[1]] x^#[[2]]) & /@ list]
answered 36 mins ago
David G. StorkDavid G. Stork
24.5k22153
$endgroup$
$begingroup$
Amazing, thank you! Sorry - I'm very new to Mathematica.
$endgroup$
– Pineapple
31 mins ago
add a comment |
$begingroup$
FromCoefficientRules[Thread[list[[All, {2}]] -> list[[All, 1]]], x]
x^2 + 3 x^4 + 5 x^6
Internal`FromCoefficientList[Normal@SparseArray[1 + list[[All, {2}]] -> list[[All, 1]]], x]
x^2 + 3 x^4 + 5 x^6
answered 4 mins ago
kglrkglr
187k10203422
$endgroup$
add a comment |
$begingroup$
Inner[#1 x^#2 &, Sequence @@ Transpose@list, Plus]
x^2 + 3 x^4 + 5 x^6
answered 2 mins ago
MarcoBMarcoB
36.7k556113
$endgroup$
add a comment |
$begingroup$
Another way, perhaps easier on the eyes. Use a pattern to deconstruct the pairs in a function definition.
term[{a_, b_}] := a x^b
Then, Map it to the list.
Total[term /@ list]
(* x^2 + 3 x^4 + 5 x^6 *)
answered 28 secs ago
John DotyJohn Doty
7,17311024
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is not a Function:
function = a x^b
But this is:
function = {a,b} [Function] a x^b
You can Apply it to each element of
list = {{1,2}, {3,4}, {5,6}}
with
function @@@ list
{x^3, 3 x^5, 5 x^7}
and sum it up with Total:
Total[ function @@@ list ]
x^3 + 3 x^5 + 5 x^7
answered 35 mins ago
Henrik SchumacherHenrik Schumacher
55.4k576154
$endgroup$
$begingroup$
Why would you do 'function = {a,b} [Function] a x^b' instead of function[a_, b_, x_] = a*x^b?
$endgroup$
– Pineapple
30 mins ago
$begingroup$
Because[Function]is easily entered with the escape sequence esc f n esc . AlsoFunctiondefines a pure function whilefunction[a_, b_, x_] = a*x^bdefines a replacement rule. However, they act the same way - most of the time. So it is a matter of taste.
$endgroup$
– Henrik Schumacher
26 mins ago
$begingroup$
@Pineapple Take a look at this tutorial on Defining Functions. Another good resource is the tutorial on Immediate and Delayed Definitions for the difference betweenSet(=) andSetDelayed(:=).
$endgroup$
– MarcoB
12 mins ago
add a comment |
$begingroup$
This is not a Function:
function = a x^b
But this is:
function = {a,b} [Function] a x^b
You can Apply it to each element of
list = {{1,2}, {3,4}, {5,6}}
with
function @@@ list
{x^3, 3 x^5, 5 x^7}
and sum it up with Total:
Total[ function @@@ list ]
x^3 + 3 x^5 + 5 x^7
answered 35 mins ago
Henrik SchumacherHenrik Schumacher
55.4k576154
$endgroup$
$begingroup$
Why would you do 'function = {a,b} [Function] a x^b' instead of function[a_, b_, x_] = a*x^b?
$endgroup$
– Pineapple
30 mins ago
$begingroup$
Because[Function]is easily entered with the escape sequence esc f n esc . AlsoFunctiondefines a pure function whilefunction[a_, b_, x_] = a*x^bdefines a replacement rule. However, they act the same way - most of the time. So it is a matter of taste.
$endgroup$
– Henrik Schumacher
26 mins ago
$begingroup$
@Pineapple Take a look at this tutorial on Defining Functions. Another good resource is the tutorial on Immediate and Delayed Definitions for the difference betweenSet(=) andSetDelayed(:=).
$endgroup$
– MarcoB
12 mins ago
add a comment |
$begingroup$
This is not a Function:
function = a x^b
But this is:
function = {a,b} [Function] a x^b
You can Apply it to each element of
list = {{1,2}, {3,4}, {5,6}}
with
function @@@ list
{x^3, 3 x^5, 5 x^7}
and sum it up with Total:
Total[ function @@@ list ]
x^3 + 3 x^5 + 5 x^7
answered 35 mins ago
Henrik SchumacherHenrik Schumacher
55.4k576154
$endgroup$
This is not a Function:
function = a x^b
But this is:
function = {a,b} [Function] a x^b
You can Apply it to each element of
list = {{1,2}, {3,4}, {5,6}}
with
function @@@ list
{x^3, 3 x^5, 5 x^7}
and sum it up with Total:
Total[ function @@@ list ]
x^3 + 3 x^5 + 5 x^7
answered 35 mins ago
Henrik SchumacherHenrik Schumacher
55.4k576154
answered 35 mins ago
Henrik SchumacherHenrik Schumacher
55.4k576154
answered 35 mins ago
Henrik SchumacherHenrik Schumacher
55.4k576154
answered 35 mins ago
Henrik SchumacherHenrik Schumacher
55.4k576154
55.4k576154
$begingroup$
Why would you do 'function = {a,b} [Function] a x^b' instead of function[a_, b_, x_] = a*x^b?
$endgroup$
– Pineapple
30 mins ago
$begingroup$
Because[Function]is easily entered with the escape sequence esc f n esc . AlsoFunctiondefines a pure function whilefunction[a_, b_, x_] = a*x^bdefines a replacement rule. However, they act the same way - most of the time. So it is a matter of taste.
$endgroup$
– Henrik Schumacher
26 mins ago
$begingroup$
@Pineapple Take a look at this tutorial on Defining Functions. Another good resource is the tutorial on Immediate and Delayed Definitions for the difference betweenSet(=) andSetDelayed(:=).
$endgroup$
– MarcoB
12 mins ago
add a comment |
$begingroup$
Why would you do 'function = {a,b} [Function] a x^b' instead of function[a_, b_, x_] = a*x^b?
$endgroup$
– Pineapple
30 mins ago
$begingroup$
Because[Function]is easily entered with the escape sequence esc f n esc . AlsoFunctiondefines a pure function whilefunction[a_, b_, x_] = a*x^bdefines a replacement rule. However, they act the same way - most of the time. So it is a matter of taste.
$endgroup$
– Henrik Schumacher
26 mins ago
$begingroup$
@Pineapple Take a look at this tutorial on Defining Functions. Another good resource is the tutorial on Immediate and Delayed Definitions for the difference betweenSet(=) andSetDelayed(:=).
$endgroup$
– MarcoB
12 mins ago
$begingroup$
Why would you do 'function = {a,b} [Function] a x^b' instead of function[a_, b_, x_] = a*x^b?
$endgroup$
– Pineapple
30 mins ago
$begingroup$
Why would you do 'function = {a,b} [Function] a x^b' instead of function[a_, b_, x_] = a*x^b?
$endgroup$
– Pineapple
30 mins ago
$begingroup$
Because
[Function] is easily entered with the escape sequence esc f n esc . Also Function defines a pure function while function[a_, b_, x_] = a*x^b defines a replacement rule. However, they act the same way - most of the time. So it is a matter of taste.$endgroup$
– Henrik Schumacher
26 mins ago
$begingroup$
Because
[Function] is easily entered with the escape sequence esc f n esc . Also Function defines a pure function while function[a_, b_, x_] = a*x^b defines a replacement rule. However, they act the same way - most of the time. So it is a matter of taste.$endgroup$
– Henrik Schumacher
26 mins ago
$begingroup$
@Pineapple Take a look at this tutorial on Defining Functions. Another good resource is the tutorial on Immediate and Delayed Definitions for the difference between
Set (=) and SetDelayed (:=).$endgroup$
– MarcoB
12 mins ago
$begingroup$
@Pineapple Take a look at this tutorial on Defining Functions. Another good resource is the tutorial on Immediate and Delayed Definitions for the difference between
Set (=) and SetDelayed (:=).$endgroup$
– MarcoB
12 mins ago
add a comment |
$begingroup$
Total[#*x^#2&@@@list]
x^2 + 3 x^4 + 5 x^6
answered 31 mins ago
J42161217J42161217
3,935322
$endgroup$
add a comment |
$begingroup$
Total[#*x^#2&@@@list]
x^2 + 3 x^4 + 5 x^6
answered 31 mins ago
J42161217J42161217
3,935322
$endgroup$
add a comment |
$begingroup$
Total[#*x^#2&@@@list]
x^2 + 3 x^4 + 5 x^6
answered 31 mins ago
J42161217J42161217
3,935322
$endgroup$
Total[#*x^#2&@@@list]
x^2 + 3 x^4 + 5 x^6
answered 31 mins ago
J42161217J42161217
3,935322
answered 31 mins ago
J42161217J42161217
3,935322
answered 31 mins ago
J42161217J42161217
3,935322
answered 31 mins ago
J42161217J42161217
3,935322
3,935322
add a comment |
add a comment |
$begingroup$
Total[(#[[1]] x^#[[2]]) & /@ list]
answered 36 mins ago
David G. StorkDavid G. Stork
24.5k22153
$endgroup$
$begingroup$
Amazing, thank you! Sorry - I'm very new to Mathematica.
$endgroup$
– Pineapple
31 mins ago
add a comment |
$begingroup$
Total[(#[[1]] x^#[[2]]) & /@ list]
answered 36 mins ago
David G. StorkDavid G. Stork
24.5k22153
$endgroup$
$begingroup$
Amazing, thank you! Sorry - I'm very new to Mathematica.
$endgroup$
– Pineapple
31 mins ago
add a comment |
$begingroup$
Total[(#[[1]] x^#[[2]]) & /@ list]
answered 36 mins ago
David G. StorkDavid G. Stork
24.5k22153
$endgroup$
Total[(#[[1]] x^#[[2]]) & /@ list]
answered 36 mins ago
David G. StorkDavid G. Stork
24.5k22153
answered 36 mins ago
David G. StorkDavid G. Stork
24.5k22153
answered 36 mins ago
David G. StorkDavid G. Stork
24.5k22153
answered 36 mins ago
David G. StorkDavid G. Stork
24.5k22153
24.5k22153
$begingroup$
Amazing, thank you! Sorry - I'm very new to Mathematica.
$endgroup$
– Pineapple
31 mins ago
add a comment |
$begingroup$
Amazing, thank you! Sorry - I'm very new to Mathematica.
$endgroup$
– Pineapple
31 mins ago
$begingroup$
Amazing, thank you! Sorry - I'm very new to Mathematica.
$endgroup$
– Pineapple
31 mins ago
$begingroup$
Amazing, thank you! Sorry - I'm very new to Mathematica.
$endgroup$
– Pineapple
31 mins ago
add a comment |
$begingroup$
FromCoefficientRules[Thread[list[[All, {2}]] -> list[[All, 1]]], x]
x^2 + 3 x^4 + 5 x^6
Internal`FromCoefficientList[Normal@SparseArray[1 + list[[All, {2}]] -> list[[All, 1]]], x]
x^2 + 3 x^4 + 5 x^6
answered 4 mins ago
kglrkglr
187k10203422
$endgroup$
add a comment |
$begingroup$
FromCoefficientRules[Thread[list[[All, {2}]] -> list[[All, 1]]], x]
x^2 + 3 x^4 + 5 x^6
Internal`FromCoefficientList[Normal@SparseArray[1 + list[[All, {2}]] -> list[[All, 1]]], x]
x^2 + 3 x^4 + 5 x^6
answered 4 mins ago
kglrkglr
187k10203422
$endgroup$
add a comment |
$begingroup$
FromCoefficientRules[Thread[list[[All, {2}]] -> list[[All, 1]]], x]
x^2 + 3 x^4 + 5 x^6
Internal`FromCoefficientList[Normal@SparseArray[1 + list[[All, {2}]] -> list[[All, 1]]], x]
x^2 + 3 x^4 + 5 x^6
answered 4 mins ago
kglrkglr
187k10203422
$endgroup$
FromCoefficientRules[Thread[list[[All, {2}]] -> list[[All, 1]]], x]
x^2 + 3 x^4 + 5 x^6
Internal`FromCoefficientList[Normal@SparseArray[1 + list[[All, {2}]] -> list[[All, 1]]], x]
x^2 + 3 x^4 + 5 x^6
answered 4 mins ago
kglrkglr
187k10203422
answered 4 mins ago
kglrkglr
187k10203422
answered 4 mins ago
kglrkglr
187k10203422
answered 4 mins ago
kglrkglr
187k10203422
187k10203422
add a comment |
add a comment |
$begingroup$
Inner[#1 x^#2 &, Sequence @@ Transpose@list, Plus]
x^2 + 3 x^4 + 5 x^6
answered 2 mins ago
MarcoBMarcoB
36.7k556113
$endgroup$
add a comment |
$begingroup$
Inner[#1 x^#2 &, Sequence @@ Transpose@list, Plus]
x^2 + 3 x^4 + 5 x^6
answered 2 mins ago
MarcoBMarcoB
36.7k556113
$endgroup$
add a comment |
$begingroup$
Inner[#1 x^#2 &, Sequence @@ Transpose@list, Plus]
x^2 + 3 x^4 + 5 x^6
answered 2 mins ago
MarcoBMarcoB
36.7k556113
$endgroup$
Inner[#1 x^#2 &, Sequence @@ Transpose@list, Plus]
x^2 + 3 x^4 + 5 x^6
answered 2 mins ago
MarcoBMarcoB
36.7k556113
answered 2 mins ago
MarcoBMarcoB
36.7k556113
answered 2 mins ago
MarcoBMarcoB
36.7k556113
answered 2 mins ago
MarcoBMarcoB
36.7k556113
36.7k556113
add a comment |
add a comment |
$begingroup$
Another way, perhaps easier on the eyes. Use a pattern to deconstruct the pairs in a function definition.
term[{a_, b_}] := a x^b
Then, Map it to the list.
Total[term /@ list]
(* x^2 + 3 x^4 + 5 x^6 *)
answered 28 secs ago
John DotyJohn Doty
7,17311024
$endgroup$
add a comment |
$begingroup$
Another way, perhaps easier on the eyes. Use a pattern to deconstruct the pairs in a function definition.
term[{a_, b_}] := a x^b
Then, Map it to the list.
Total[term /@ list]
(* x^2 + 3 x^4 + 5 x^6 *)
answered 28 secs ago
John DotyJohn Doty
7,17311024
$endgroup$
add a comment |
$begingroup$
Another way, perhaps easier on the eyes. Use a pattern to deconstruct the pairs in a function definition.
term[{a_, b_}] := a x^b
Then, Map it to the list.
Total[term /@ list]
(* x^2 + 3 x^4 + 5 x^6 *)
answered 28 secs ago
John DotyJohn Doty
7,17311024
$endgroup$
Another way, perhaps easier on the eyes. Use a pattern to deconstruct the pairs in a function definition.
term[{a_, b_}] := a x^b
Then, Map it to the list.
Total[term /@ list]
(* x^2 + 3 x^4 + 5 x^6 *)
answered 28 secs ago
John DotyJohn Doty
7,17311024
answered 28 secs ago
John DotyJohn Doty
7,17311024
answered 28 secs ago
John DotyJohn Doty
7,17311024
answered 28 secs ago
John DotyJohn Doty
7,17311024
7,17311024
add a comment |
add a comment |
Pineapple is a new contributor. Be nice, and check out our Code of Conduct.
Pineapple is a new contributor. Be nice, and check out our Code of Conduct.
Pineapple is a new contributor. Be nice, and check out our Code of Conduct.
Pineapple is a new contributor. Be nice, and check out our Code of Conduct.
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