How to efficiently check conditions inside a function?

up vote
4
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?

asked Sep 19 at 12:46

Majis

1,313313

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)]
â€“Â Bob Hanlon
Sep 19 at 13:52

What error? Share the code, which has the problem.
â€“Â Johu
Sep 19 at 15:25

add a commentÂ |Â

up vote
4
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]
]

How can I resolve this issue?

asked Sep 19 at 12:46

Majis

1,313313

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)]
â€“Â Bob Hanlon
Sep 19 at 13:52

What error? Share the code, which has the problem.
â€“Â Johu
Sep 19 at 15:25

add a commentÂ |Â

up vote
4
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]
]

How can I resolve this issue?

asked Sep 19 at 12:46

Majis

1,313313

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]
]

How can I resolve this issue?

function-construction

asked Sep 19 at 12:46

Majis

1,313313

asked Sep 19 at 12:46

Majis

1,313313

asked Sep 19 at 12:46

Majis

1,313313

asked Sep 19 at 12:46

Majis

1,313313

asked Sep 19 at 12:46

Majis

1,313313

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)]
â€“Â Bob Hanlon
Sep 19 at 13:52

What error? Share the code, which has the problem.
â€“Â Johu
Sep 19 at 15:25

add a commentÂ |Â

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)]
â€“Â Bob Hanlon
Sep 19 at 13:52

What error? Share the code, which has the problem.
â€“Â Johu
Sep 19 at 15:25

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)]
â€“Â Bob Hanlon
Sep 19 at 13:52

What error? Share the code, which has the problem.
â€“Â Johu
Sep 19 at 15:25

add a commentÂ |Â

4 Answers
4

active

oldest

votes

up vote
3
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 *)

edited Sep 19 at 13:05

answered Sep 19 at 12:59

Mariusz Iwaniuk

6,32811027

Thanks. It's really a great way. Exactly the kind of solution I was looking for.
â€“Â Majis
Sep 19 at 13:04

add a commentÂ |Â

up vote
3
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

answered Sep 19 at 13:15

John Doty

6,2841924

add a commentÂ |Â

up vote
3
down vote

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

f[1]

1, 1

f[3]

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

answered Sep 19 at 14:31

kglr

163k8188387

add a commentÂ |Â

up vote
1
down vote

This is more efficient.

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

Table[f[i], i, 5]

1, 1, 
 0, 1, 0, -1, 
 0, 1, 1, 0, 0, -1, 
 0, 1, Sqrt[3]/2, 1/2, Sqrt[3]/2, -(1/2), 0, -1, 
 0, 1, 1/Sqrt[2], 1/Sqrt[2], 1, 0, 1/Sqrt[2], -(1/Sqrt[2]), 0, -1

answered Sep 19 at 17:32

m_goldberg

82.5k869190

add a commentÂ |Â

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
3
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 *)

edited Sep 19 at 13:05

answered Sep 19 at 12:59

Mariusz Iwaniuk

6,32811027

Thanks. It's really a great way. Exactly the kind of solution I was looking for.
â€“Â Majis
Sep 19 at 13:04

add a commentÂ |Â

up vote
3
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 *)

edited Sep 19 at 13:05

answered Sep 19 at 12:59

Mariusz Iwaniuk

6,32811027

Thanks. It's really a great way. Exactly the kind of solution I was looking for.
â€“Â Majis
Sep 19 at 13:04

add a commentÂ |Â

up vote
3
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 *)

edited Sep 19 at 13:05

answered Sep 19 at 12:59

Mariusz Iwaniuk

6,32811027

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 *)

edited Sep 19 at 13:05

answered Sep 19 at 12:59

Mariusz Iwaniuk

6,32811027

edited Sep 19 at 13:05

answered Sep 19 at 12:59

Mariusz Iwaniuk

6,32811027

answered Sep 19 at 12:59

Mariusz Iwaniuk

6,32811027

answered Sep 19 at 12:59

Mariusz Iwaniuk

6,32811027

Thanks. It's really a great way. Exactly the kind of solution I was looking for.
â€“Â Majis
Sep 19 at 13:04

add a commentÂ |Â

Thanks. It's really a great way. Exactly the kind of solution I was looking for.
â€“Â Majis
Sep 19 at 13:04

Thanks. It's really a great way. Exactly the kind of solution I was looking for.
â€“Â Majis
Sep 19 at 13:04

add a commentÂ |Â

up vote
3
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

answered Sep 19 at 13:15

John Doty

6,2841924

add a commentÂ |Â

up vote
3
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

answered Sep 19 at 13:15

John Doty

6,2841924

add a commentÂ |Â

up vote
3
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

answered Sep 19 at 13:15

John Doty

6,2841924

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

answered Sep 19 at 13:15

John Doty

6,2841924

answered Sep 19 at 13:15

John Doty

6,2841924

answered Sep 19 at 13:15

John Doty

6,2841924

answered Sep 19 at 13:15

John Doty

6,2841924

add a commentÂ |Â

up vote
3
down vote

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

f[1]

1, 1

f[3]

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

answered Sep 19 at 14:31

kglr

163k8188387

add a commentÂ |Â

up vote
3
down vote

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

f[1]

1, 1

f[3]

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

answered Sep 19 at 14:31

kglr

163k8188387

add a commentÂ |Â

up vote
3
down vote

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

f[1]

1, 1

f[3]

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

answered Sep 19 at 14:31

kglr

163k8188387

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

f[1]

1, 1

f[3]

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

answered Sep 19 at 14:31

kglr

163k8188387

answered Sep 19 at 14:31

kglr

163k8188387

answered Sep 19 at 14:31

kglr

163k8188387

answered Sep 19 at 14:31

kglr

163k8188387

add a commentÂ |Â

up vote
1
down vote

This is more efficient.

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

Table[f[i], i, 5]

1, 1, 
 0, 1, 0, -1, 
 0, 1, 1, 0, 0, -1, 
 0, 1, Sqrt[3]/2, 1/2, Sqrt[3]/2, -(1/2), 0, -1, 
 0, 1, 1/Sqrt[2], 1/Sqrt[2], 1, 0, 1/Sqrt[2], -(1/Sqrt[2]), 0, -1

answered Sep 19 at 17:32

m_goldberg

82.5k869190

add a commentÂ |Â

up vote
1
down vote

This is more efficient.

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

Table[f[i], i, 5]

1, 1, 
 0, 1, 0, -1, 
 0, 1, 1, 0, 0, -1, 
 0, 1, Sqrt[3]/2, 1/2, Sqrt[3]/2, -(1/2), 0, -1, 
 0, 1, 1/Sqrt[2], 1/Sqrt[2], 1, 0, 1/Sqrt[2], -(1/Sqrt[2]), 0, -1

answered Sep 19 at 17:32

m_goldberg

82.5k869190

add a commentÂ |Â

up vote
1
down vote

This is more efficient.

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

Table[f[i], i, 5]

1, 1, 
 0, 1, 0, -1, 
 0, 1, 1, 0, 0, -1, 
 0, 1, Sqrt[3]/2, 1/2, Sqrt[3]/2, -(1/2), 0, -1, 
 0, 1, 1/Sqrt[2], 1/Sqrt[2], 1, 0, 1/Sqrt[2], -(1/Sqrt[2]), 0, -1

answered Sep 19 at 17:32

m_goldberg

82.5k869190

This is more efficient.

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

Table[f[i], i, 5]

1, 1, 
 0, 1, 0, -1, 
 0, 1, 1, 0, 0, -1, 
 0, 1, Sqrt[3]/2, 1/2, Sqrt[3]/2, -(1/2), 0, -1, 
 0, 1, 1/Sqrt[2], 1/Sqrt[2], 1, 0, 1/Sqrt[2], -(1/Sqrt[2]), 0, -1

answered Sep 19 at 17:32

m_goldberg

82.5k869190

answered Sep 19 at 17:32

m_goldberg

82.5k869190

answered Sep 19 at 17:32

m_goldberg

82.5k869190

answered Sep 19 at 17:32

m_goldberg

82.5k869190

add a commentÂ |Â

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