Turn the following values into percentage [closed]

I have the following data:
{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}
{5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 18800, 19216}
My data is already in Matrix form in Mathematica, is there a possibility to change the second raw into percentages? I want the first data: 5914 to be 100% and calculate how much did the data grow over the years based on the first year. So the last number should be: 324%.
How could this be done automatically and for a huge set of data?
Thank you!

edited Jan 5 at 14:52

J. M. is slightly pensive♦

97.9k10304464

asked Jan 4 at 14:06

user62202

241

closed as off-topic by Szabolcs, Thies Heidecke, Bob Hanlon, m_goldberg, bbgodfrey Jan 6 at 1:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Szabolcs, Thies Heidecke, Bob Hanlon, m_goldberg, bbgodfrey

If this question can be reworded to fit the rules in the help center, please edit the question.

$begingroup$
try this data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}, {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 18800, 19216}}; secondData = data[[2]]; secondData = IntegerPart[N[#*100/First[secondData]]]& /@ secondData; data[[2]] = secondData;
$endgroup$
– Xminer
Jan 4 at 14:13

1

$begingroup$
Just divide your array with the first element? arr/First[arr]. You might want to go through some basic tutorials: wolfram.com/language/elementary-introduction/2nd-ed
$endgroup$
– Szabolcs
Jan 4 at 14:14

add a comment |

edited Jan 5 at 14:52

J. M. is slightly pensive♦

97.9k10304464

asked Jan 4 at 14:06

user62202

241

closed as off-topic by Szabolcs, Thies Heidecke, Bob Hanlon, m_goldberg, bbgodfrey Jan 6 at 1:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Szabolcs, Thies Heidecke, Bob Hanlon, m_goldberg, bbgodfrey

If this question can be reworded to fit the rules in the help center, please edit the question.

$begingroup$
try this data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}, {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 18800, 19216}}; secondData = data[[2]]; secondData = IntegerPart[N[#*100/First[secondData]]]& /@ secondData; data[[2]] = secondData;
$endgroup$
– Xminer
Jan 4 at 14:13

1

$begingroup$
Just divide your array with the first element? arr/First[arr]. You might want to go through some basic tutorials: wolfram.com/language/elementary-introduction/2nd-ed
$endgroup$
– Szabolcs
Jan 4 at 14:14

add a comment |

edited Jan 5 at 14:52

J. M. is slightly pensive♦

97.9k10304464

asked Jan 4 at 14:06

user62202

241

list-manipulation numerics

edited Jan 5 at 14:52

J. M. is slightly pensive♦

97.9k10304464

asked Jan 4 at 14:06

user62202

241

edited Jan 5 at 14:52

J. M. is slightly pensive♦

97.9k10304464

asked Jan 4 at 14:06

user62202

241

edited Jan 5 at 14:52

J. M. is slightly pensive♦

97.9k10304464

edited Jan 5 at 14:52

J. M. is slightly pensive♦

97.9k10304464

edited Jan 5 at 14:52

J. M. is slightly pensive♦

97.9k10304464

asked Jan 4 at 14:06

user62202

241

asked Jan 4 at 14:06

user62202

241

asked Jan 4 at 14:06

user62202

241

closed as off-topic by Szabolcs, Thies Heidecke, Bob Hanlon, m_goldberg, bbgodfrey Jan 6 at 1:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Szabolcs, Thies Heidecke, Bob Hanlon, m_goldberg, bbgodfrey

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Szabolcs, Thies Heidecke, Bob Hanlon, m_goldberg, bbgodfrey Jan 6 at 1:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Szabolcs, Thies Heidecke, Bob Hanlon, m_goldberg, bbgodfrey

If this question can be reworded to fit the rules in the help center, please edit the question.

$begingroup$
try this data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}, {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 18800, 19216}}; secondData = data[[2]]; secondData = IntegerPart[N[#*100/First[secondData]]]& /@ secondData; data[[2]] = secondData;
$endgroup$
– Xminer
Jan 4 at 14:13

1

$begingroup$
Just divide your array with the first element? arr/First[arr]. You might want to go through some basic tutorials: wolfram.com/language/elementary-introduction/2nd-ed
$endgroup$
– Szabolcs
Jan 4 at 14:14

add a comment |

$begingroup$
try this data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}, {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 18800, 19216}}; secondData = data[[2]]; secondData = IntegerPart[N[#*100/First[secondData]]]& /@ secondData; data[[2]] = secondData;
$endgroup$
– Xminer
Jan 4 at 14:13

1

$begingroup$
Just divide your array with the first element? arr/First[arr]. You might want to go through some basic tutorials: wolfram.com/language/elementary-introduction/2nd-ed
$endgroup$
– Szabolcs
Jan 4 at 14:14

try this

data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016,     2017}, {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 18800,     19216}};  secondData = data[[2]]; secondData = IntegerPart[N[#*100/First[secondData]]]& /@ secondData; data[[2]] = secondData;

– Xminer
Jan 4 at 14:13

try this

data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016,     2017}, {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 18800,     19216}};  secondData = data[[2]]; secondData = IntegerPart[N[#*100/First[secondData]]]& /@ secondData; data[[2]] = secondData;

– Xminer
Jan 4 at 14:13

Just divide your array with the first element? arr/First[arr]. You might want to go through some basic tutorials: wolfram.com/language/elementary-introduction/2nd-ed

– Szabolcs
Jan 4 at 14:14

add a comment |

3 Answers
3

active

oldest

votes

(I get to show this first...)

data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 

    2017}, {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 

    18800, 19216}};

{data[[1]], PercentForm[N[data[[2]]]/data[[2, 1]]]}

enter image description here

(...coming soon, in version 12)

answered Jan 4 at 16:16

Daniel Lichtblau

47.2k276164

3

$begingroup$
@RolfMertig The decimal point disparity was a design choice (at the insistence of the boss), to only put in a decimal when the value does not coerce to an integer. As for using N, we opted not to coerce exact values, other than integers, into percentages. So that was also a design choice.
$endgroup$
– Daniel Lichtblau
Jan 4 at 17:17

1

$begingroup$
Greetings to the boss: wrong design choice. It was always a pain to construct nicely aligned and consisent tables of real numbers. And what is even worse is that the documentation does not fit the behaviour of Mathematica anymore (at least in such type of details).I do remeber a time when the documentation coincided with the software (which was one reason I switched from Macsyma to Mathematica ...).
$endgroup$
– Rolf Mertig
Jan 4 at 21:59

1

$begingroup$
@RolfMertig I agree the alignment of numbers is not always easy. But I don't see what exact fractions have to do with formatting reals. On the topic of formatting reals, DecimalForm has seen some fixes..
$endgroup$
– Daniel Lichtblau
Jan 4 at 23:04

1

$begingroup$
@ΑλέξανδροςΖεγγ Did you run the code I posted?
$endgroup$
– Daniel Lichtblau
Jan 5 at 16:04

1

$begingroup$
Is there already a timeline for v12.0? If you have any say in this, please just make sure there's a release candidate sent to testers before a 12.0 final suddenly appears ...
$endgroup$
– Szabolcs
Jan 7 at 11:38

|
show 2 more comments

data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}, {5914, 6143,6182, 18000, 18173, 18344, 18454, 18506, 18800, 19216}}

There's no need to define additional variables,try

data[[2]] = 100 data[[2]]/data[[2, 1]]//Round[#,1]& (*//IntegerPart*)  ;

data

(*{{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}, 

{100,103, 104, 304, 307, 310, 312, 312, 317, 324}}*)

edited Jan 4 at 14:46

answered Jan 4 at 14:26

Ulrich Neumann

9,548617

$begingroup$
In many applications, Round would be more appropriate than IntegerPart
$endgroup$
– Bob Hanlon
Jan 4 at 14:35

$begingroup$
@ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
$endgroup$
– Ulrich Neumann
Jan 4 at 14:45

$begingroup$
@Ulrich If it is about speed, I'd prefer data[[2]] = Round[data[[2]] (100. /data[[2, 1]]), 1]. The converts to machine precision numbers and reduces the number of scalar-vector multiplications from 2 to 1.
$endgroup$
– Henrik Schumacher
Jan 4 at 15:01

$begingroup$
@ Henrik Thanks, my actual answer seems to be the power-version!
$endgroup$
– Ulrich Neumann
Jan 4 at 16:19

add a comment |

Similar to the above:

k = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 

   2017} , {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 

   18800, 19216}}



k[[2]]=Flatten[Floor[100*Rest[k]/Flatten[Rest[k]][[1]]]]

Gives

{{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017},
{100, 103, 104, 304, 307, 310, 312, 312, 317, 324}}

answered Jan 4 at 15:20

GerardF123

1547

add a comment |

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

(I get to show this first...)

data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 

    2017}, {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 

    18800, 19216}};

{data[[1]], PercentForm[N[data[[2]]]/data[[2, 1]]]}

enter image description here

(...coming soon, in version 12)

answered Jan 4 at 16:16

Daniel Lichtblau

47.2k276164

3

$begingroup$
@RolfMertig The decimal point disparity was a design choice (at the insistence of the boss), to only put in a decimal when the value does not coerce to an integer. As for using N, we opted not to coerce exact values, other than integers, into percentages. So that was also a design choice.
$endgroup$
– Daniel Lichtblau
Jan 4 at 17:17

1

$begingroup$
Greetings to the boss: wrong design choice. It was always a pain to construct nicely aligned and consisent tables of real numbers. And what is even worse is that the documentation does not fit the behaviour of Mathematica anymore (at least in such type of details).I do remeber a time when the documentation coincided with the software (which was one reason I switched from Macsyma to Mathematica ...).
$endgroup$
– Rolf Mertig
Jan 4 at 21:59

1

$begingroup$
@RolfMertig I agree the alignment of numbers is not always easy. But I don't see what exact fractions have to do with formatting reals. On the topic of formatting reals, DecimalForm has seen some fixes..
$endgroup$
– Daniel Lichtblau
Jan 4 at 23:04

1

$begingroup$
@ΑλέξανδροςΖεγγ Did you run the code I posted?
$endgroup$
– Daniel Lichtblau
Jan 5 at 16:04

1

$begingroup$
Is there already a timeline for v12.0? If you have any say in this, please just make sure there's a release candidate sent to testers before a 12.0 final suddenly appears ...
$endgroup$
– Szabolcs
Jan 7 at 11:38

|
show 2 more comments

(I get to show this first...)

data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 

    2017}, {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 

    18800, 19216}};

{data[[1]], PercentForm[N[data[[2]]]/data[[2, 1]]]}

enter image description here

(...coming soon, in version 12)

answered Jan 4 at 16:16

Daniel Lichtblau

47.2k276164

3

$begingroup$
@RolfMertig The decimal point disparity was a design choice (at the insistence of the boss), to only put in a decimal when the value does not coerce to an integer. As for using N, we opted not to coerce exact values, other than integers, into percentages. So that was also a design choice.
$endgroup$
– Daniel Lichtblau
Jan 4 at 17:17

1

$begingroup$
Greetings to the boss: wrong design choice. It was always a pain to construct nicely aligned and consisent tables of real numbers. And what is even worse is that the documentation does not fit the behaviour of Mathematica anymore (at least in such type of details).I do remeber a time when the documentation coincided with the software (which was one reason I switched from Macsyma to Mathematica ...).
$endgroup$
– Rolf Mertig
Jan 4 at 21:59

1

$begingroup$
@RolfMertig I agree the alignment of numbers is not always easy. But I don't see what exact fractions have to do with formatting reals. On the topic of formatting reals, DecimalForm has seen some fixes..
$endgroup$
– Daniel Lichtblau
Jan 4 at 23:04

1

$begingroup$
@ΑλέξανδροςΖεγγ Did you run the code I posted?
$endgroup$
– Daniel Lichtblau
Jan 5 at 16:04

1

$begingroup$
Is there already a timeline for v12.0? If you have any say in this, please just make sure there's a release candidate sent to testers before a 12.0 final suddenly appears ...
$endgroup$
– Szabolcs
Jan 7 at 11:38

|
show 2 more comments

(I get to show this first...)

data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 

    2017}, {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 

    18800, 19216}};

{data[[1]], PercentForm[N[data[[2]]]/data[[2, 1]]]}

enter image description here

(...coming soon, in version 12)

answered Jan 4 at 16:16

Daniel Lichtblau

47.2k276164

(I get to show this first...)

data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 

    2017}, {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 

    18800, 19216}};

{data[[1]], PercentForm[N[data[[2]]]/data[[2, 1]]]}

enter image description here

(...coming soon, in version 12)

answered Jan 4 at 16:16

Daniel Lichtblau

47.2k276164

answered Jan 4 at 16:16

Daniel Lichtblau

47.2k276164

answered Jan 4 at 16:16

Daniel Lichtblau

47.2k276164

answered Jan 4 at 16:16

Daniel Lichtblau

47.2k276164

3

$begingroup$
@RolfMertig The decimal point disparity was a design choice (at the insistence of the boss), to only put in a decimal when the value does not coerce to an integer. As for using N, we opted not to coerce exact values, other than integers, into percentages. So that was also a design choice.
$endgroup$
– Daniel Lichtblau
Jan 4 at 17:17

1

$begingroup$
Greetings to the boss: wrong design choice. It was always a pain to construct nicely aligned and consisent tables of real numbers. And what is even worse is that the documentation does not fit the behaviour of Mathematica anymore (at least in such type of details).I do remeber a time when the documentation coincided with the software (which was one reason I switched from Macsyma to Mathematica ...).
$endgroup$
– Rolf Mertig
Jan 4 at 21:59

1

$begingroup$
@RolfMertig I agree the alignment of numbers is not always easy. But I don't see what exact fractions have to do with formatting reals. On the topic of formatting reals, DecimalForm has seen some fixes..
$endgroup$
– Daniel Lichtblau
Jan 4 at 23:04

1

$begingroup$
@ΑλέξανδροςΖεγγ Did you run the code I posted?
$endgroup$
– Daniel Lichtblau
Jan 5 at 16:04

1

$begingroup$
Is there already a timeline for v12.0? If you have any say in this, please just make sure there's a release candidate sent to testers before a 12.0 final suddenly appears ...
$endgroup$
– Szabolcs
Jan 7 at 11:38

|
show 2 more comments

3

$begingroup$
@RolfMertig The decimal point disparity was a design choice (at the insistence of the boss), to only put in a decimal when the value does not coerce to an integer. As for using N, we opted not to coerce exact values, other than integers, into percentages. So that was also a design choice.
$endgroup$
– Daniel Lichtblau
Jan 4 at 17:17

1

$begingroup$
Greetings to the boss: wrong design choice. It was always a pain to construct nicely aligned and consisent tables of real numbers. And what is even worse is that the documentation does not fit the behaviour of Mathematica anymore (at least in such type of details).I do remeber a time when the documentation coincided with the software (which was one reason I switched from Macsyma to Mathematica ...).
$endgroup$
– Rolf Mertig
Jan 4 at 21:59

1

$begingroup$
@RolfMertig I agree the alignment of numbers is not always easy. But I don't see what exact fractions have to do with formatting reals. On the topic of formatting reals, DecimalForm has seen some fixes..
$endgroup$
– Daniel Lichtblau
Jan 4 at 23:04

1

$begingroup$
@ΑλέξανδροςΖεγγ Did you run the code I posted?
$endgroup$
– Daniel Lichtblau
Jan 5 at 16:04

1

$begingroup$
Is there already a timeline for v12.0? If you have any say in this, please just make sure there's a release candidate sent to testers before a 12.0 final suddenly appears ...
$endgroup$
– Szabolcs
Jan 7 at 11:38

@RolfMertig The decimal point disparity was a design choice (at the insistence of the boss), to only put in a decimal when the value does not coerce to an integer. As for using N, we opted not to coerce exact values, other than integers, into percentages. So that was also a design choice.

– Daniel Lichtblau
Jan 4 at 17:17

Greetings to the boss: wrong design choice. It was always a pain to construct nicely aligned and consisent tables of real numbers. And what is even worse is that the documentation does not fit the behaviour of Mathematica anymore (at least in such type of details).I do remeber a time when the documentation coincided with the software (which was one reason I switched from Macsyma to Mathematica ...).

– Rolf Mertig
Jan 4 at 21:59

@RolfMertig I agree the alignment of numbers is not always easy. But I don't see what exact fractions have to do with formatting reals. On the topic of formatting reals, DecimalForm has seen some fixes..

– Daniel Lichtblau
Jan 4 at 23:04

@ΑλέξανδροςΖεγγ Did you run the code I posted?

– Daniel Lichtblau
Jan 5 at 16:04

Is there already a timeline for v12.0? If you have any say in this, please just make sure there's a release candidate sent to testers before a 12.0 final suddenly appears ...

– Szabolcs
Jan 7 at 11:38

|
show 2 more comments

data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}, {5914, 6143,6182, 18000, 18173, 18344, 18454, 18506, 18800, 19216}}

There's no need to define additional variables,try

data[[2]] = 100 data[[2]]/data[[2, 1]]//Round[#,1]& (*//IntegerPart*)  ;

data

(*{{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}, 

{100,103, 104, 304, 307, 310, 312, 312, 317, 324}}*)

edited Jan 4 at 14:46

answered Jan 4 at 14:26

Ulrich Neumann

9,548617

$begingroup$
In many applications, Round would be more appropriate than IntegerPart
$endgroup$
– Bob Hanlon
Jan 4 at 14:35

$begingroup$
@ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
$endgroup$
– Ulrich Neumann
Jan 4 at 14:45

$begingroup$
@Ulrich If it is about speed, I'd prefer data[[2]] = Round[data[[2]] (100. /data[[2, 1]]), 1]. The converts to machine precision numbers and reduces the number of scalar-vector multiplications from 2 to 1.
$endgroup$
– Henrik Schumacher
Jan 4 at 15:01

$begingroup$
@ Henrik Thanks, my actual answer seems to be the power-version!
$endgroup$
– Ulrich Neumann
Jan 4 at 16:19

add a comment |

data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}, {5914, 6143,6182, 18000, 18173, 18344, 18454, 18506, 18800, 19216}}

There's no need to define additional variables,try

data[[2]] = 100 data[[2]]/data[[2, 1]]//Round[#,1]& (*//IntegerPart*)  ;

data

(*{{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}, 

{100,103, 104, 304, 307, 310, 312, 312, 317, 324}}*)

edited Jan 4 at 14:46

answered Jan 4 at 14:26

Ulrich Neumann

9,548617

$begingroup$
In many applications, Round would be more appropriate than IntegerPart
$endgroup$
– Bob Hanlon
Jan 4 at 14:35

$begingroup$
@ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
$endgroup$
– Ulrich Neumann
Jan 4 at 14:45

$begingroup$
@Ulrich If it is about speed, I'd prefer data[[2]] = Round[data[[2]] (100. /data[[2, 1]]), 1]. The converts to machine precision numbers and reduces the number of scalar-vector multiplications from 2 to 1.
$endgroup$
– Henrik Schumacher
Jan 4 at 15:01

$begingroup$
@ Henrik Thanks, my actual answer seems to be the power-version!
$endgroup$
– Ulrich Neumann
Jan 4 at 16:19

add a comment |

data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}, {5914, 6143,6182, 18000, 18173, 18344, 18454, 18506, 18800, 19216}}

There's no need to define additional variables,try

data[[2]] = 100 data[[2]]/data[[2, 1]]//Round[#,1]& (*//IntegerPart*)  ;

data

(*{{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}, 

{100,103, 104, 304, 307, 310, 312, 312, 317, 324}}*)

edited Jan 4 at 14:46

answered Jan 4 at 14:26

Ulrich Neumann

9,548617

data = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}, {5914, 6143,6182, 18000, 18173, 18344, 18454, 18506, 18800, 19216}}

There's no need to define additional variables,try

data[[2]] = 100 data[[2]]/data[[2, 1]]//Round[#,1]& (*//IntegerPart*)  ;

data

(*{{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017}, 

{100,103, 104, 304, 307, 310, 312, 312, 317, 324}}*)

edited Jan 4 at 14:46

answered Jan 4 at 14:26

Ulrich Neumann

9,548617

edited Jan 4 at 14:46

answered Jan 4 at 14:26

Ulrich Neumann

9,548617

answered Jan 4 at 14:26

Ulrich Neumann

9,548617

answered Jan 4 at 14:26

Ulrich Neumann

9,548617

$begingroup$
In many applications, Round would be more appropriate than IntegerPart
$endgroup$
– Bob Hanlon
Jan 4 at 14:35

$begingroup$
@ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
$endgroup$
– Ulrich Neumann
Jan 4 at 14:45

$begingroup$
@Ulrich If it is about speed, I'd prefer data[[2]] = Round[data[[2]] (100. /data[[2, 1]]), 1]. The converts to machine precision numbers and reduces the number of scalar-vector multiplications from 2 to 1.
$endgroup$
– Henrik Schumacher
Jan 4 at 15:01

$begingroup$
@ Henrik Thanks, my actual answer seems to be the power-version!
$endgroup$
– Ulrich Neumann
Jan 4 at 16:19

add a comment |

$begingroup$
In many applications, Round would be more appropriate than IntegerPart
$endgroup$
– Bob Hanlon
Jan 4 at 14:35

$begingroup$
@ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
$endgroup$
– Ulrich Neumann
Jan 4 at 14:45

$begingroup$
@Ulrich If it is about speed, I'd prefer data[[2]] = Round[data[[2]] (100. /data[[2, 1]]), 1]. The converts to machine precision numbers and reduces the number of scalar-vector multiplications from 2 to 1.
$endgroup$
– Henrik Schumacher
Jan 4 at 15:01

$begingroup$
@ Henrik Thanks, my actual answer seems to be the power-version!
$endgroup$
– Ulrich Neumann
Jan 4 at 16:19

In many applications, Round would be more appropriate than IntegerPart

– Bob Hanlon
Jan 4 at 14:35

@ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )

– Ulrich Neumann
Jan 4 at 14:45

@Ulrich If it is about speed, I'd prefer data[[2]] = Round[data[[2]] (100. /data[[2, 1]]), 1]. The converts to machine precision numbers and reduces the number of scalar-vector multiplications from 2 to 1.

– Henrik Schumacher
Jan 4 at 15:01

@ Henrik Thanks, my actual answer seems to be the power-version!

– Ulrich Neumann
Jan 4 at 16:19

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Similar to the above:

k = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 

   2017} , {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 

   18800, 19216}}



k[[2]]=Flatten[Floor[100*Rest[k]/Flatten[Rest[k]][[1]]]]

Gives

{{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017},
{100, 103, 104, 304, 307, 310, 312, 312, 317, 324}}

answered Jan 4 at 15:20

GerardF123

1547

add a comment |

Similar to the above:

k = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 

   2017} , {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 

   18800, 19216}}



k[[2]]=Flatten[Floor[100*Rest[k]/Flatten[Rest[k]][[1]]]]

Gives

{{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017},
{100, 103, 104, 304, 307, 310, 312, 312, 317, 324}}

answered Jan 4 at 15:20

GerardF123

1547

add a comment |

Similar to the above:

k = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 

   2017} , {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 

   18800, 19216}}



k[[2]]=Flatten[Floor[100*Rest[k]/Flatten[Rest[k]][[1]]]]

Gives

{{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017},
{100, 103, 104, 304, 307, 310, 312, 312, 317, 324}}

answered Jan 4 at 15:20

GerardF123

1547

Similar to the above:

k = {{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 

   2017} , {5914, 6143, 6182, 18000, 18173, 18344, 18454, 18506, 

   18800, 19216}}



k[[2]]=Flatten[Floor[100*Rest[k]/Flatten[Rest[k]][[1]]]]

Gives

{{2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017},
{100, 103, 104, 304, 307, 310, 312, 312, 317, 324}}

answered Jan 4 at 15:20

GerardF123

1547

answered Jan 4 at 15:20

GerardF123

1547

answered Jan 4 at 15:20

GerardF123

1547

answered Jan 4 at 15:20

GerardF123

1547

add a comment |

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