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Why does Plot only sometimes use different colors for each curve

Plot draws list of curves in same color when not using EvaluateUsing Evaluate and Evaluated -> True in PlotPlot graphs in different colorsUse different ColorFunction for each function plottedPlot sublists as different colorsPlot graphs in different colorsWhy does the replace all (/.) evaluate differently in plotWant a different color for each curve displayed with ShowHow to Color each curve differently?How to plot data sets each of different length?ErrorListPlot with a different color for each element of the listPlotting with different color for a single curveHow to Plot several functions using differents colors , It is possible to use TeX or MathSymbols?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

2

$begingroup$

In this example Plot does not color each curve differently:

Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}]

but in this example it does:

Plot[{vx0 Sin[t] /. {vy0 -> 1, vx0 -> 1}, 

vy0 Cos[t] /. {vy0 -> 1, vx0 -> 1}}, {t, 0, 12}]

How do I automatically enable Mathematica to give a different color to each curve?

plotting

share|improve this question

asked 8 hours ago

kotoznakotozna

11655 bronze badges

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Evaluate?${}$
  $endgroup$
  – AccidentalFourierTransform
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Mathematica guesses how many curves there are before the plot expression is evaluated. In most cases, you can get the result you want by wrapping the expression with Evaluate
  $endgroup$
  – mikado
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Evaluate works, is that the simplest solution? Why does it guess incorrectly in one case? It feels like excess coding just to plot two curves.
  $endgroup$
  – kotozna
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate: mathematica.stackexchange.com/questions/1731/… and mathematica.stackexchange.com/questions/43323/…
  $endgroup$
  – Michael E2
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

In this example Plot does not color each curve differently:

Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}]

but in this example it does:

Plot[{vx0 Sin[t] /. {vy0 -> 1, vx0 -> 1}, 

vy0 Cos[t] /. {vy0 -> 1, vx0 -> 1}}, {t, 0, 12}]

How do I automatically enable Mathematica to give a different color to each curve?

plotting

share|improve this question

asked 8 hours ago

kotoznakotozna

11655 bronze badges

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Evaluate?${}$
  $endgroup$
  – AccidentalFourierTransform
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Mathematica guesses how many curves there are before the plot expression is evaluated. In most cases, you can get the result you want by wrapping the expression with Evaluate
  $endgroup$
  – mikado
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Evaluate works, is that the simplest solution? Why does it guess incorrectly in one case? It feels like excess coding just to plot two curves.
  $endgroup$
  – kotozna
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate: mathematica.stackexchange.com/questions/1731/… and mathematica.stackexchange.com/questions/43323/…
  $endgroup$
  – Michael E2
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

In this example Plot does not color each curve differently:

Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}]

but in this example it does:

Plot[{vx0 Sin[t] /. {vy0 -> 1, vx0 -> 1}, 

vy0 Cos[t] /. {vy0 -> 1, vx0 -> 1}}, {t, 0, 12}]

How do I automatically enable Mathematica to give a different color to each curve?

plotting

share|improve this question

asked 8 hours ago

kotoznakotozna

11655 bronze badges

$endgroup$

Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}]

Plot[{vx0 Sin[t] /. {vy0 -> 1, vx0 -> 1}, 

vy0 Cos[t] /. {vy0 -> 1, vx0 -> 1}}, {t, 0, 12}]

share|improve this question

asked 8 hours ago

kotoznakotozna

11655 bronze badges

share|improve this question

asked 8 hours ago

kotoznakotozna

11655 bronze badges

asked 8 hours ago

kotoznakotozna

11655 bronze badges

asked 8 hours ago

kotoznakotozna

11655 bronze badges

2 Answers
2

active

oldest

votes

5

$begingroup$

Compare the FullForm of your two inputs:

FullForm[Hold[Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}]]]

(* 

   Hold[Plot[ReplaceAll[List[Times[vx0,Sin[t]],Times[vy0,Cos[t]]],

   List[Rule[vy0,1],Rule[vx0,1]]],List[t,0,12]]]

*)



FullForm[Hold[Plot[{vx0 Sin[t] /. {vy0 -> 1, vx0 -> 1}, vy0 Cos[t] /.

  {vy0 -> 1, vx0 -> 1}}, {t, 0, 12}]]]

(*

    Hold[Plot[List[ReplaceAll[Times[vx0,Sin[t]],List[Rule[vy0,1],Rule[vx0,1]]],

   ReplaceAll[Times[vy0,Cos[t]],List[Rule[vy0,1],Rule[vx0,1]]]],List[t,0,12]]]

*)

For the second, the head of the first argument to Plot is List. When Plot sees this, it uses the length of the list as the number of colors to choose. For the first, the head is ReplaceAll. In this case, it doesn't anticipate that there will be more than one plot, so it only chooses one color.

Evaluate lets ReplaceAll do its job before Plot sees the argument. This yields a list, which allows plot to determine the number of colors to choose.

share|improve this answer

edited 6 hours ago

answered 6 hours ago

John DotyJohn Doty

8,34811 gold badge1212 silver badges2424 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

See

• for an explanation of the general problem: Plot draws list of curves in same color when not using Evaluate
  


• for this specific problem: Plot graphs in different colors
  


• for the differences below: Using Evaluate and Evaluated -> True in Plot
  

Note the difference if t has a value:

t = 2.;



Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}, 

 Evaluated -> True]

Plot[Evaluate[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}], {t, 0, 12}]

share|improve this answer

answered 3 hours ago

Michael E2Michael E2

156k1313 gold badges214214 silver badges506506 bronze badges

$endgroup$

add a comment | 

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5

$begingroup$

Compare the FullForm of your two inputs:

FullForm[Hold[Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}]]]

(* 

   Hold[Plot[ReplaceAll[List[Times[vx0,Sin[t]],Times[vy0,Cos[t]]],

   List[Rule[vy0,1],Rule[vx0,1]]],List[t,0,12]]]

*)



FullForm[Hold[Plot[{vx0 Sin[t] /. {vy0 -> 1, vx0 -> 1}, vy0 Cos[t] /.

  {vy0 -> 1, vx0 -> 1}}, {t, 0, 12}]]]

(*

    Hold[Plot[List[ReplaceAll[Times[vx0,Sin[t]],List[Rule[vy0,1],Rule[vx0,1]]],

   ReplaceAll[Times[vy0,Cos[t]],List[Rule[vy0,1],Rule[vx0,1]]]],List[t,0,12]]]

*)

For the second, the head of the first argument to Plot is List. When Plot sees this, it uses the length of the list as the number of colors to choose. For the first, the head is ReplaceAll. In this case, it doesn't anticipate that there will be more than one plot, so it only chooses one color.

Evaluate lets ReplaceAll do its job before Plot sees the argument. This yields a list, which allows plot to determine the number of colors to choose.

share|improve this answer

edited 6 hours ago

answered 6 hours ago

John DotyJohn Doty

8,34811 gold badge1212 silver badges2424 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

Compare the FullForm of your two inputs:

FullForm[Hold[Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}]]]

(* 

   Hold[Plot[ReplaceAll[List[Times[vx0,Sin[t]],Times[vy0,Cos[t]]],

   List[Rule[vy0,1],Rule[vx0,1]]],List[t,0,12]]]

*)



FullForm[Hold[Plot[{vx0 Sin[t] /. {vy0 -> 1, vx0 -> 1}, vy0 Cos[t] /.

  {vy0 -> 1, vx0 -> 1}}, {t, 0, 12}]]]

(*

    Hold[Plot[List[ReplaceAll[Times[vx0,Sin[t]],List[Rule[vy0,1],Rule[vx0,1]]],

   ReplaceAll[Times[vy0,Cos[t]],List[Rule[vy0,1],Rule[vx0,1]]]],List[t,0,12]]]

*)

For the second, the head of the first argument to Plot is List. When Plot sees this, it uses the length of the list as the number of colors to choose. For the first, the head is ReplaceAll. In this case, it doesn't anticipate that there will be more than one plot, so it only chooses one color.

Evaluate lets ReplaceAll do its job before Plot sees the argument. This yields a list, which allows plot to determine the number of colors to choose.

share|improve this answer

edited 6 hours ago

answered 6 hours ago

John DotyJohn Doty

8,34811 gold badge1212 silver badges2424 bronze badges

$endgroup$

add a comment | 

$begingroup$

Compare the FullForm of your two inputs:

FullForm[Hold[Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}]]]

(* 

   Hold[Plot[ReplaceAll[List[Times[vx0,Sin[t]],Times[vy0,Cos[t]]],

   List[Rule[vy0,1],Rule[vx0,1]]],List[t,0,12]]]

*)



FullForm[Hold[Plot[{vx0 Sin[t] /. {vy0 -> 1, vx0 -> 1}, vy0 Cos[t] /.

  {vy0 -> 1, vx0 -> 1}}, {t, 0, 12}]]]

(*

    Hold[Plot[List[ReplaceAll[Times[vx0,Sin[t]],List[Rule[vy0,1],Rule[vx0,1]]],

   ReplaceAll[Times[vy0,Cos[t]],List[Rule[vy0,1],Rule[vx0,1]]]],List[t,0,12]]]

*)

For the second, the head of the first argument to Plot is List. When Plot sees this, it uses the length of the list as the number of colors to choose. For the first, the head is ReplaceAll. In this case, it doesn't anticipate that there will be more than one plot, so it only chooses one color.

Evaluate lets ReplaceAll do its job before Plot sees the argument. This yields a list, which allows plot to determine the number of colors to choose.

share|improve this answer

edited 6 hours ago

answered 6 hours ago

John DotyJohn Doty

8,34811 gold badge1212 silver badges2424 bronze badges

$endgroup$

FullForm[Hold[Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}]]]

(* 

   Hold[Plot[ReplaceAll[List[Times[vx0,Sin[t]],Times[vy0,Cos[t]]],

   List[Rule[vy0,1],Rule[vx0,1]]],List[t,0,12]]]

*)



FullForm[Hold[Plot[{vx0 Sin[t] /. {vy0 -> 1, vx0 -> 1}, vy0 Cos[t] /.

  {vy0 -> 1, vx0 -> 1}}, {t, 0, 12}]]]

(*

    Hold[Plot[List[ReplaceAll[Times[vx0,Sin[t]],List[Rule[vy0,1],Rule[vx0,1]]],

   ReplaceAll[Times[vy0,Cos[t]],List[Rule[vy0,1],Rule[vx0,1]]]],List[t,0,12]]]

*)

share|improve this answer

edited 6 hours ago

answered 6 hours ago

John DotyJohn Doty

8,34811 gold badge1212 silver badges2424 bronze badges

answered 6 hours ago

John DotyJohn Doty

8,34811 gold badge1212 silver badges2424 bronze badges

answered 6 hours ago

John DotyJohn Doty

8,34811 gold badge1212 silver badges2424 bronze badges

1

$begingroup$

See

• for an explanation of the general problem: Plot draws list of curves in same color when not using Evaluate
  


• for this specific problem: Plot graphs in different colors
  


• for the differences below: Using Evaluate and Evaluated -> True in Plot
  

Note the difference if t has a value:

t = 2.;



Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}, 

 Evaluated -> True]

Plot[Evaluate[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}], {t, 0, 12}]

share|improve this answer

answered 3 hours ago

Michael E2Michael E2

156k1313 gold badges214214 silver badges506506 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

See

• for an explanation of the general problem: Plot draws list of curves in same color when not using Evaluate
  


• for this specific problem: Plot graphs in different colors
  


• for the differences below: Using Evaluate and Evaluated -> True in Plot
  

Note the difference if t has a value:

t = 2.;



Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}, 

 Evaluated -> True]

Plot[Evaluate[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}], {t, 0, 12}]

share|improve this answer

answered 3 hours ago

Michael E2Michael E2

156k1313 gold badges214214 silver badges506506 bronze badges

$endgroup$

add a comment | 

$begingroup$

See

• for an explanation of the general problem: Plot draws list of curves in same color when not using Evaluate
  


• for this specific problem: Plot graphs in different colors
  


• for the differences below: Using Evaluate and Evaluated -> True in Plot
  

Note the difference if t has a value:

t = 2.;



Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}, 

 Evaluated -> True]

Plot[Evaluate[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}], {t, 0, 12}]

share|improve this answer

answered 3 hours ago

Michael E2Michael E2

156k1313 gold badges214214 silver badges506506 bronze badges

$endgroup$

t = 2.;



Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}, 

 Evaluated -> True]

Plot[Evaluate[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}], {t, 0, 12}]

share|improve this answer

answered 3 hours ago

Michael E2Michael E2

156k1313 gold badges214214 silver badges506506 bronze badges

answered 3 hours ago

Michael E2Michael E2

156k1313 gold badges214214 silver badges506506 bronze badges

answered 3 hours ago

Michael E2Michael E2

156k1313 gold badges214214 silver badges506506 bronze badges