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An alternative to (%%…%) (k times)

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3

$begingroup$

Sometimes it may be useful to use (%%…%) (k times) to get the kth previous result in a notebook instead of setting a variable.

However, if I want to get the 8th previous result, for example, I would have to type (%) eight times (%%%%%%%%) and I could get lost. Needless to remember that every time (%) is needed (Shift)(5) must be pressed, which demands both hands and is somewhat inconvenient.

So, in the code below, I have defined a new command.

Now, when I want the previous 4th result I just type (o4) instead of (%%%%). If I wanted a previous result before the 9th, say the 14th, I would have to change the Range[9] in the command to Range[14] and generate (o1…o14). I could also obtain only the previous 3rd to the 7th values using Range[3, 7], getting (o3…o7).

This command should be run every time a new session of Mathematica is initiated. It could be written in an initialization cell in the notebook.
Remember that you should not redefine or clear (o1…o9) at the cost of getting unexpected errors. You must consider them as "system variables".

(* Define o1,o2,...,o9 as "%","%%",...,"%%%%%%%%%" *)

Map[ToExpression[StringJoin["o", ToString[#], " := ", "Out[$Line - ", ToString[#], "]"]] &, Range[9]];

Example:

Pi/6.

{Sin[%], Sin[o1]}

{Cos[%%], Cos[o2]}

{Tan[%%%], Tan[o3]}

{Sin[2 %%%%], Sin[2 o4]}

{Cos[2 %%%%%], Cos[2 o5]}

{Tan[2 %%%%%%], Tan[2 o6]}



0.523599

{0.5, 0.5}

{0.866025, 0.866025}

{0.57735, 0.57735}

{0.866025, 0.866025}

{0.5, 0.5}

{1.73205, 1.73205}

evaluation notebooks editing

share|improve this question

asked 11 hours ago

CaioCaio

4144 bronze badges

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Just use Out. See the "Details" section.
  $endgroup$
  – Alan
  11 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  As in Out[-5] (i.e., five lines ago). Though frankly, in such a situation, I would start with foo = Pi/6. and then use foo instead of % or Out.
  $endgroup$
  – Michael E2
  11 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe try With[ { foo = Pi/6 }, { {Sin[ foo ], Sin[ o1 ]}, ... } ] and use Column to arrange the output.
  $endgroup$
  – LouisB
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Out[-k] is not as convenient as the command presented. What I wanted to show is a simpler form of getting previous values writing only o1, o2 &c, with only 2 keystrokes instead of having to use 8 keystrokes every time I use Out[-k] or (k +1) keystrokes when I use (%).
  $endgroup$
  – Caio
  9 hours ago
  
  
  

  
  
  
  

add a comment
 | 

3

$begingroup$

Sometimes it may be useful to use (%%…%) (k times) to get the kth previous result in a notebook instead of setting a variable.

However, if I want to get the 8th previous result, for example, I would have to type (%) eight times (%%%%%%%%) and I could get lost. Needless to remember that every time (%) is needed (Shift)(5) must be pressed, which demands both hands and is somewhat inconvenient.

So, in the code below, I have defined a new command.

Now, when I want the previous 4th result I just type (o4) instead of (%%%%). If I wanted a previous result before the 9th, say the 14th, I would have to change the Range[9] in the command to Range[14] and generate (o1…o14). I could also obtain only the previous 3rd to the 7th values using Range[3, 7], getting (o3…o7).

This command should be run every time a new session of Mathematica is initiated. It could be written in an initialization cell in the notebook.
Remember that you should not redefine or clear (o1…o9) at the cost of getting unexpected errors. You must consider them as "system variables".

(* Define o1,o2,...,o9 as "%","%%",...,"%%%%%%%%%" *)

Map[ToExpression[StringJoin["o", ToString[#], " := ", "Out[$Line - ", ToString[#], "]"]] &, Range[9]];

Example:

Pi/6.

{Sin[%], Sin[o1]}

{Cos[%%], Cos[o2]}

{Tan[%%%], Tan[o3]}

{Sin[2 %%%%], Sin[2 o4]}

{Cos[2 %%%%%], Cos[2 o5]}

{Tan[2 %%%%%%], Tan[2 o6]}



0.523599

{0.5, 0.5}

{0.866025, 0.866025}

{0.57735, 0.57735}

{0.866025, 0.866025}

{0.5, 0.5}

{1.73205, 1.73205}

evaluation notebooks editing

share|improve this question

asked 11 hours ago

CaioCaio

4144 bronze badges

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Just use Out. See the "Details" section.
  $endgroup$
  – Alan
  11 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  As in Out[-5] (i.e., five lines ago). Though frankly, in such a situation, I would start with foo = Pi/6. and then use foo instead of % or Out.
  $endgroup$
  – Michael E2
  11 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe try With[ { foo = Pi/6 }, { {Sin[ foo ], Sin[ o1 ]}, ... } ] and use Column to arrange the output.
  $endgroup$
  – LouisB
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Out[-k] is not as convenient as the command presented. What I wanted to show is a simpler form of getting previous values writing only o1, o2 &c, with only 2 keystrokes instead of having to use 8 keystrokes every time I use Out[-k] or (k +1) keystrokes when I use (%).
  $endgroup$
  – Caio
  9 hours ago
  
  
  

  
  
  
  

add a comment
 | 

$begingroup$

Sometimes it may be useful to use (%%…%) (k times) to get the kth previous result in a notebook instead of setting a variable.

However, if I want to get the 8th previous result, for example, I would have to type (%) eight times (%%%%%%%%) and I could get lost. Needless to remember that every time (%) is needed (Shift)(5) must be pressed, which demands both hands and is somewhat inconvenient.

So, in the code below, I have defined a new command.

Now, when I want the previous 4th result I just type (o4) instead of (%%%%). If I wanted a previous result before the 9th, say the 14th, I would have to change the Range[9] in the command to Range[14] and generate (o1…o14). I could also obtain only the previous 3rd to the 7th values using Range[3, 7], getting (o3…o7).

This command should be run every time a new session of Mathematica is initiated. It could be written in an initialization cell in the notebook.
Remember that you should not redefine or clear (o1…o9) at the cost of getting unexpected errors. You must consider them as "system variables".

(* Define o1,o2,...,o9 as "%","%%",...,"%%%%%%%%%" *)

Map[ToExpression[StringJoin["o", ToString[#], " := ", "Out[$Line - ", ToString[#], "]"]] &, Range[9]];

Example:

Pi/6.

{Sin[%], Sin[o1]}

{Cos[%%], Cos[o2]}

{Tan[%%%], Tan[o3]}

{Sin[2 %%%%], Sin[2 o4]}

{Cos[2 %%%%%], Cos[2 o5]}

{Tan[2 %%%%%%], Tan[2 o6]}



0.523599

{0.5, 0.5}

{0.866025, 0.866025}

{0.57735, 0.57735}

{0.866025, 0.866025}

{0.5, 0.5}

{1.73205, 1.73205}

evaluation notebooks editing

share|improve this question

asked 11 hours ago

CaioCaio

4144 bronze badges

$endgroup$

(* Define o1,o2,...,o9 as "%","%%",...,"%%%%%%%%%" *)

Map[ToExpression[StringJoin["o", ToString[#], " := ", "Out[$Line - ", ToString[#], "]"]] &, Range[9]];

Pi/6.

{Sin[%], Sin[o1]}

{Cos[%%], Cos[o2]}

{Tan[%%%], Tan[o3]}

{Sin[2 %%%%], Sin[2 o4]}

{Cos[2 %%%%%], Cos[2 o5]}

{Tan[2 %%%%%%], Tan[2 o6]}



0.523599

{0.5, 0.5}

{0.866025, 0.866025}

{0.57735, 0.57735}

{0.866025, 0.866025}

{0.5, 0.5}

{1.73205, 1.73205}

share|improve this question

asked 11 hours ago

CaioCaio

4144 bronze badges

share|improve this question

asked 11 hours ago

CaioCaio

4144 bronze badges

asked 11 hours ago

CaioCaio

4144 bronze badges

asked 11 hours ago

CaioCaio

4144 bronze badges

1 Answer
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6

$begingroup$

% is just a shorthand for Out[-1] or even Out[]. And so on, Out[-k] gives the k-th previous result. Or from the details section of Out

Out[-k] is equivalent to %%...% (k times).

One way I've used Out, along with the way you can get multiple outputs by giving it a list of negative integers, before is testing timing and equality between different methods.

ClearSystemCache[]

Range[-10^6, 10^6]~Complement~{0} // AbsoluteTiming;

Join @@ {-Reverse@#, #} &@Range[10^6] // AbsoluteTiming;

Out[{-1, -2}][[;; , 1]]

SameQ @@ Out[{-1, -2} - 1][[;; , 2]]

{0.0201598, 0.0370587}

True

Or maybe a more clear example

a;

b;

Out[{-1, -2}]

{b, a}

share|improve this answer

answered 11 hours ago

That Gravity GuyThat Gravity Guy

2,40111 gold badge66 silver badges1515 bronze badges

$endgroup$

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6

$begingroup$

% is just a shorthand for Out[-1] or even Out[]. And so on, Out[-k] gives the k-th previous result. Or from the details section of Out

Out[-k] is equivalent to %%...% (k times).

One way I've used Out, along with the way you can get multiple outputs by giving it a list of negative integers, before is testing timing and equality between different methods.

ClearSystemCache[]

Range[-10^6, 10^6]~Complement~{0} // AbsoluteTiming;

Join @@ {-Reverse@#, #} &@Range[10^6] // AbsoluteTiming;

Out[{-1, -2}][[;; , 1]]

SameQ @@ Out[{-1, -2} - 1][[;; , 2]]

{0.0201598, 0.0370587}

True

Or maybe a more clear example

a;

b;

Out[{-1, -2}]

{b, a}

share|improve this answer

answered 11 hours ago

That Gravity GuyThat Gravity Guy

2,40111 gold badge66 silver badges1515 bronze badges

$endgroup$

add a comment
 | 

6

$begingroup$

% is just a shorthand for Out[-1] or even Out[]. And so on, Out[-k] gives the k-th previous result. Or from the details section of Out

Out[-k] is equivalent to %%...% (k times).

One way I've used Out, along with the way you can get multiple outputs by giving it a list of negative integers, before is testing timing and equality between different methods.

ClearSystemCache[]

Range[-10^6, 10^6]~Complement~{0} // AbsoluteTiming;

Join @@ {-Reverse@#, #} &@Range[10^6] // AbsoluteTiming;

Out[{-1, -2}][[;; , 1]]

SameQ @@ Out[{-1, -2} - 1][[;; , 2]]

{0.0201598, 0.0370587}

True

Or maybe a more clear example

a;

b;

Out[{-1, -2}]

{b, a}

share|improve this answer

answered 11 hours ago

That Gravity GuyThat Gravity Guy

2,40111 gold badge66 silver badges1515 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

% is just a shorthand for Out[-1] or even Out[]. And so on, Out[-k] gives the k-th previous result. Or from the details section of Out

Out[-k] is equivalent to %%...% (k times).

One way I've used Out, along with the way you can get multiple outputs by giving it a list of negative integers, before is testing timing and equality between different methods.

ClearSystemCache[]

Range[-10^6, 10^6]~Complement~{0} // AbsoluteTiming;

Join @@ {-Reverse@#, #} &@Range[10^6] // AbsoluteTiming;

Out[{-1, -2}][[;; , 1]]

SameQ @@ Out[{-1, -2} - 1][[;; , 2]]

{0.0201598, 0.0370587}

True

Or maybe a more clear example

a;

b;

Out[{-1, -2}]

{b, a}

share|improve this answer

answered 11 hours ago

That Gravity GuyThat Gravity Guy

2,40111 gold badge66 silver badges1515 bronze badges

$endgroup$

ClearSystemCache[]

Range[-10^6, 10^6]~Complement~{0} // AbsoluteTiming;

Join @@ {-Reverse@#, #} &@Range[10^6] // AbsoluteTiming;

Out[{-1, -2}][[;; , 1]]

SameQ @@ Out[{-1, -2} - 1][[;; , 2]]

a;

b;

Out[{-1, -2}]

share|improve this answer

answered 11 hours ago

That Gravity GuyThat Gravity Guy

2,40111 gold badge66 silver badges1515 bronze badges

answered 11 hours ago

That Gravity GuyThat Gravity Guy

2,40111 gold badge66 silver badges1515 bronze badges

answered 11 hours ago

That Gravity GuyThat Gravity Guy

2,40111 gold badge66 silver badges1515 bronze badges