How to efficiently check conditions inside a function?

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up vote
3
down vote

favorite

Consider a function as

f[n_]:=Module[x,y, 
x=Table[Sin[i],i,0,Pi,Pi/(n-1)];
y=Table[Cos[i],i,0,Pi,Pi/(n-1)];
Transpose[x,y]
]

Now, this function will work for all the values of n>1. However, for n=1, I want the function to return 1,1. If I use If-else inside the function, I get the result, but, it gives me the error message as well.

How can I resolve this issue?

function-construction

share|improve this question

asked 2 hours ago

Majis

1,288313

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  To be structurally consistent, f[1] should be 1,1, i.e., a list of points. To be efficient, f[1]=1,1; f[n_Integer]:=Sin[#], Cos[#]& /@ Range[0, Pi, Pi/(n-1)]
  – Bob Hanlon
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

Consider a function as

f[n_]:=Module[x,y, 
x=Table[Sin[i],i,0,Pi,Pi/(n-1)];
y=Table[Cos[i],i,0,Pi,Pi/(n-1)];
Transpose[x,y]
]

Now, this function will work for all the values of n>1. However, for n=1, I want the function to return 1,1. If I use If-else inside the function, I get the result, but, it gives me the error message as well.

How can I resolve this issue?

function-construction

share|improve this question

asked 2 hours ago

Majis

1,288313

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  To be structurally consistent, f[1] should be 1,1, i.e., a list of points. To be efficient, f[1]=1,1; f[n_Integer]:=Sin[#], Cos[#]& /@ Range[0, Pi, Pi/(n-1)]
  – Bob Hanlon
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Consider a function as

f[n_]:=Module[x,y, 
x=Table[Sin[i],i,0,Pi,Pi/(n-1)];
y=Table[Cos[i],i,0,Pi,Pi/(n-1)];
Transpose[x,y]
]

Now, this function will work for all the values of n>1. However, for n=1, I want the function to return 1,1. If I use If-else inside the function, I get the result, but, it gives me the error message as well.

How can I resolve this issue?

function-construction

share|improve this question

asked 2 hours ago

Majis

1,288313

Consider a function as

f[n_]:=Module[x,y, 
x=Table[Sin[i],i,0,Pi,Pi/(n-1)];
y=Table[Cos[i],i,0,Pi,Pi/(n-1)];
Transpose[x,y]
]

Now, this function will work for all the values of n>1. However, for n=1, I want the function to return 1,1. If I use If-else inside the function, I get the result, but, it gives me the error message as well.

How can I resolve this issue?

function-construction

function-construction

share|improve this question

asked 2 hours ago

Majis

1,288313

share|improve this question

asked 2 hours ago

Majis

1,288313

share|improve this question

share|improve this question

asked 2 hours ago

Majis

1,288313

asked 2 hours ago

Majis

1,288313

asked 2 hours ago

Majis

1,288313

1,288313

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  To be structurally consistent, f[1] should be 1,1, i.e., a list of points. To be efficient, f[1]=1,1; f[n_Integer]:=Sin[#], Cos[#]& /@ Range[0, Pi, Pi/(n-1)]
  – Bob Hanlon
  1 hour ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  To be structurally consistent, f[1] should be 1,1, i.e., a list of points. To be efficient, f[1]=1,1; f[n_Integer]:=Sin[#], Cos[#]& /@ Range[0, Pi, Pi/(n-1)]
  – Bob Hanlon
  1 hour ago
  

  
  
  
  

2

2

To be structurally consistent, f[1] should be 1,1, i.e., a list of points. To be efficient, f[1]=1,1; f[n_Integer]:=Sin[#], Cos[#]& /@ Range[0, Pi, Pi/(n-1)]
– Bob Hanlon
1 hour ago

To be structurally consistent, f[1] should be 1,1, i.e., a list of points. To be efficient, f[1]=1,1; f[n_Integer]:=Sin[#], Cos[#]& /@ Range[0, Pi, Pi/(n-1)]
– Bob Hanlon
1 hour ago

add a comment | 

3 Answers
3

active

oldest

votes

up vote
3
down vote

ClearAll[f]
f[n_] := Transpose[Table[Sin[i], i, n /. 1 -> π / 2, _ :> 0, π, π/(n - 1)], 
 Table[Cos[i], i, 0, π, π/(n - 1)]]

f[1]

1, 1

f[3]

0, 1, 1, 0, 0, -1

share|improve this answer

answered 33 mins ago

kglr

161k8184384

add a comment | 

up vote
2
down vote

f[n_] := Module[x, y, 
 If[n == 1, 1, 1, 
 x = Table[Sin[i], i, 0, Pi, Pi/(n - 1)]; 
 y = Table[Cos[i], i, 0, Pi, Pi/(n - 1)]; 
 Transpose[x, y]]]

f[1]
(* 1,1 *)

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Mariusz Iwaniuk

6,17311026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. It's really a great way. Exactly the kind of solution I was looking for.
  – Majis
  2 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

Mathematica is an expression rewriting language. Use that (and eliminate unnecessary code):

f[n_] := Transpose[Table[Sin[i], i, 0, Pi, Pi/(n - 1)],
 Table[Cos[i], i, 0, Pi, Pi/(n - 1)]]
f[1] = 1, 1;

When more than one rewrite is possible, Mathematica prefers the one with the more specific trigger, f[1] in this case

share|improve this answer

answered 1 hour ago

John Doty

6,1491924

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

ClearAll[f]
f[n_] := Transpose[Table[Sin[i], i, n /. 1 -> π / 2, _ :> 0, π, π/(n - 1)], 
 Table[Cos[i], i, 0, π, π/(n - 1)]]

f[1]

1, 1

f[3]

0, 1, 1, 0, 0, -1

share|improve this answer

answered 33 mins ago

kglr

161k8184384

add a comment | 

up vote
3
down vote

ClearAll[f]
f[n_] := Transpose[Table[Sin[i], i, n /. 1 -> π / 2, _ :> 0, π, π/(n - 1)], 
 Table[Cos[i], i, 0, π, π/(n - 1)]]

f[1]

1, 1

f[3]

0, 1, 1, 0, 0, -1

share|improve this answer

answered 33 mins ago

kglr

161k8184384

add a comment | 

up vote
3
down vote

up vote
3
down vote

ClearAll[f]
f[n_] := Transpose[Table[Sin[i], i, n /. 1 -> π / 2, _ :> 0, π, π/(n - 1)], 
 Table[Cos[i], i, 0, π, π/(n - 1)]]

f[1]

1, 1

f[3]

0, 1, 1, 0, 0, -1

share|improve this answer

answered 33 mins ago

kglr

161k8184384

ClearAll[f]
f[n_] := Transpose[Table[Sin[i], i, n /. 1 -> π / 2, _ :> 0, π, π/(n - 1)], 
 Table[Cos[i], i, 0, π, π/(n - 1)]]

f[1]

1, 1

f[3]

0, 1, 1, 0, 0, -1

share|improve this answer

answered 33 mins ago

kglr

161k8184384

share|improve this answer

share|improve this answer

answered 33 mins ago

kglr

161k8184384

answered 33 mins ago

kglr

161k8184384

answered 33 mins ago

kglr

161k8184384

161k8184384

add a comment | 

add a comment | 

up vote
2
down vote

f[n_] := Module[x, y, 
 If[n == 1, 1, 1, 
 x = Table[Sin[i], i, 0, Pi, Pi/(n - 1)]; 
 y = Table[Cos[i], i, 0, Pi, Pi/(n - 1)]; 
 Transpose[x, y]]]

f[1]
(* 1,1 *)

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Mariusz Iwaniuk

6,17311026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. It's really a great way. Exactly the kind of solution I was looking for.
  – Majis
  2 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

f[n_] := Module[x, y, 
 If[n == 1, 1, 1, 
 x = Table[Sin[i], i, 0, Pi, Pi/(n - 1)]; 
 y = Table[Cos[i], i, 0, Pi, Pi/(n - 1)]; 
 Transpose[x, y]]]

f[1]
(* 1,1 *)

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Mariusz Iwaniuk

6,17311026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. It's really a great way. Exactly the kind of solution I was looking for.
  – Majis
  2 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

f[n_] := Module[x, y, 
 If[n == 1, 1, 1, 
 x = Table[Sin[i], i, 0, Pi, Pi/(n - 1)]; 
 y = Table[Cos[i], i, 0, Pi, Pi/(n - 1)]; 
 Transpose[x, y]]]

f[1]
(* 1,1 *)

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Mariusz Iwaniuk

6,17311026

f[n_] := Module[x, y, 
 If[n == 1, 1, 1, 
 x = Table[Sin[i], i, 0, Pi, Pi/(n - 1)]; 
 y = Table[Cos[i], i, 0, Pi, Pi/(n - 1)]; 
 Transpose[x, y]]]

f[1]
(* 1,1 *)

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Mariusz Iwaniuk

6,17311026

share|improve this answer

share|improve this answer

edited 2 hours ago

edited 2 hours ago

edited 2 hours ago

answered 2 hours ago

Mariusz Iwaniuk

6,17311026

answered 2 hours ago

Mariusz Iwaniuk

6,17311026

answered 2 hours ago

Mariusz Iwaniuk

6,17311026

6,17311026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. It's really a great way. Exactly the kind of solution I was looking for.
  – Majis
  2 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. It's really a great way. Exactly the kind of solution I was looking for.
  – Majis
  2 hours ago
  

  
  
  
  

Thanks. It's really a great way. Exactly the kind of solution I was looking for.
– Majis
2 hours ago

Thanks. It's really a great way. Exactly the kind of solution I was looking for.
– Majis
2 hours ago

add a comment | 

up vote
2
down vote

Mathematica is an expression rewriting language. Use that (and eliminate unnecessary code):

f[n_] := Transpose[Table[Sin[i], i, 0, Pi, Pi/(n - 1)],
 Table[Cos[i], i, 0, Pi, Pi/(n - 1)]]
f[1] = 1, 1;

When more than one rewrite is possible, Mathematica prefers the one with the more specific trigger, f[1] in this case

share|improve this answer

answered 1 hour ago

John Doty

6,1491924

add a comment | 

up vote
2
down vote

Mathematica is an expression rewriting language. Use that (and eliminate unnecessary code):

f[n_] := Transpose[Table[Sin[i], i, 0, Pi, Pi/(n - 1)],
 Table[Cos[i], i, 0, Pi, Pi/(n - 1)]]
f[1] = 1, 1;

When more than one rewrite is possible, Mathematica prefers the one with the more specific trigger, f[1] in this case

share|improve this answer

answered 1 hour ago

John Doty

6,1491924

add a comment | 

up vote
2
down vote

up vote
2
down vote

Mathematica is an expression rewriting language. Use that (and eliminate unnecessary code):

f[n_] := Transpose[Table[Sin[i], i, 0, Pi, Pi/(n - 1)],
 Table[Cos[i], i, 0, Pi, Pi/(n - 1)]]
f[1] = 1, 1;

When more than one rewrite is possible, Mathematica prefers the one with the more specific trigger, f[1] in this case

share|improve this answer

answered 1 hour ago

John Doty

6,1491924

Mathematica is an expression rewriting language. Use that (and eliminate unnecessary code):

f[n_] := Transpose[Table[Sin[i], i, 0, Pi, Pi/(n - 1)],
 Table[Cos[i], i, 0, Pi, Pi/(n - 1)]]
f[1] = 1, 1;

When more than one rewrite is possible, Mathematica prefers the one with the more specific trigger, f[1] in this case

share|improve this answer

answered 1 hour ago

John Doty

6,1491924

share|improve this answer

share|improve this answer

answered 1 hour ago

John Doty

6,1491924

answered 1 hour ago

John Doty

6,1491924

answered 1 hour ago

John Doty

6,1491924

6,1491924

add a comment | 

add a comment | 

 

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