Turn the following values into percentage












2














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!






list-manipulation numerics data







share|improve this question









New contributor




user62202 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • 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
    3 hours ago












  • 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
    3 hours ago
















2














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!






list-manipulation numerics data







share|improve this question









New contributor




user62202 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • 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
    3 hours ago












  • 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
    3 hours ago














2












2








2







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!






list-manipulation numerics data







share|improve this question









New contributor




user62202 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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!






list-manipulation numerics data




list-manipulation numerics data






share|improve this question









New contributor




user62202 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




user62202 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 3 hours ago









Szabolcs

158k13432926




158k13432926






New contributor




user62202 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 3 hours ago









user62202

111




111




New contributor




user62202 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





user62202 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user62202 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • 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
    3 hours ago












  • 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
    3 hours ago


















  • 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
    3 hours ago












  • 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
    3 hours ago
















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
3 hours ago






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
3 hours ago














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
3 hours ago




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
3 hours ago










3 Answers
3






active

oldest

votes


















4














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}}*)





share|improve this answer























  • In many applications, Round would be more appropriate than IntegerPart
    – Bob Hanlon
    3 hours ago










  • @ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
    – Ulrich Neumann
    3 hours ago












  • @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
    2 hours ago










  • @ Henrik Thanks, my actual answer seems to be the power-version!
    – Ulrich Neumann
    1 hour ago



















2














(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)






share|improve this answer





















  • If I do NumberForm[100.,{4,1}] I get 100.0. And for NumberForm[312.,{4,1}] 312.0. It looks like that PercentForm[1.] gives 100% and PercentForm[3.12] results in 312.%. Is this just the usual sloppiness or will this be fixed ? Also, why do you have to use N? Shouldn't this be done automatically, at least if the default of PercentForm seems to be to print 1 digit to the right of the decimal point?
    – Rolf Mertig
    50 mins ago












  • @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
    31 mins ago



















1














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}}






share|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "387"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });






    user62202 is a new contributor. Be nice, and check out our Code of Conduct.










    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f188823%2fturn-the-following-values-into-percentage%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    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}}*)





    share|improve this answer























    • In many applications, Round would be more appropriate than IntegerPart
      – Bob Hanlon
      3 hours ago










    • @ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
      – Ulrich Neumann
      3 hours ago












    • @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
      2 hours ago










    • @ Henrik Thanks, my actual answer seems to be the power-version!
      – Ulrich Neumann
      1 hour ago
















    4














    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}}*)





    share|improve this answer























    • In many applications, Round would be more appropriate than IntegerPart
      – Bob Hanlon
      3 hours ago










    • @ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
      – Ulrich Neumann
      3 hours ago












    • @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
      2 hours ago










    • @ Henrik Thanks, my actual answer seems to be the power-version!
      – Ulrich Neumann
      1 hour ago














    4












    4








    4






    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}}*)





    share|improve this answer














    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}}*)






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 3 hours ago

























    answered 3 hours ago









    Ulrich Neumann

    7,580516




    7,580516












    • In many applications, Round would be more appropriate than IntegerPart
      – Bob Hanlon
      3 hours ago










    • @ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
      – Ulrich Neumann
      3 hours ago












    • @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
      2 hours ago










    • @ Henrik Thanks, my actual answer seems to be the power-version!
      – Ulrich Neumann
      1 hour ago


















    • In many applications, Round would be more appropriate than IntegerPart
      – Bob Hanlon
      3 hours ago










    • @ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
      – Ulrich Neumann
      3 hours ago












    • @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
      2 hours ago










    • @ Henrik Thanks, my actual answer seems to be the power-version!
      – Ulrich Neumann
      1 hour ago
















    In many applications, Round would be more appropriate than IntegerPart
    – Bob Hanlon
    3 hours ago




    In many applications, Round would be more appropriate than IntegerPart
    – Bob Hanlon
    3 hours ago












    @ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
    – Ulrich Neumann
    3 hours ago






    @ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
    – Ulrich Neumann
    3 hours ago














    @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
    2 hours ago




    @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
    2 hours ago












    @ Henrik Thanks, my actual answer seems to be the power-version!
    – Ulrich Neumann
    1 hour ago




    @ Henrik Thanks, my actual answer seems to be the power-version!
    – Ulrich Neumann
    1 hour ago











    2














    (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)






    share|improve this answer





















    • If I do NumberForm[100.,{4,1}] I get 100.0. And for NumberForm[312.,{4,1}] 312.0. It looks like that PercentForm[1.] gives 100% and PercentForm[3.12] results in 312.%. Is this just the usual sloppiness or will this be fixed ? Also, why do you have to use N? Shouldn't this be done automatically, at least if the default of PercentForm seems to be to print 1 digit to the right of the decimal point?
      – Rolf Mertig
      50 mins ago












    • @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
      31 mins ago
















    2














    (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)






    share|improve this answer





















    • If I do NumberForm[100.,{4,1}] I get 100.0. And for NumberForm[312.,{4,1}] 312.0. It looks like that PercentForm[1.] gives 100% and PercentForm[3.12] results in 312.%. Is this just the usual sloppiness or will this be fixed ? Also, why do you have to use N? Shouldn't this be done automatically, at least if the default of PercentForm seems to be to print 1 digit to the right of the decimal point?
      – Rolf Mertig
      50 mins ago












    • @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
      31 mins ago














    2












    2








    2






    (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)






    share|improve this answer












    (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)







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 1 hour ago









    Daniel Lichtblau

    46.5k275162




    46.5k275162












    • If I do NumberForm[100.,{4,1}] I get 100.0. And for NumberForm[312.,{4,1}] 312.0. It looks like that PercentForm[1.] gives 100% and PercentForm[3.12] results in 312.%. Is this just the usual sloppiness or will this be fixed ? Also, why do you have to use N? Shouldn't this be done automatically, at least if the default of PercentForm seems to be to print 1 digit to the right of the decimal point?
      – Rolf Mertig
      50 mins ago












    • @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
      31 mins ago


















    • If I do NumberForm[100.,{4,1}] I get 100.0. And for NumberForm[312.,{4,1}] 312.0. It looks like that PercentForm[1.] gives 100% and PercentForm[3.12] results in 312.%. Is this just the usual sloppiness or will this be fixed ? Also, why do you have to use N? Shouldn't this be done automatically, at least if the default of PercentForm seems to be to print 1 digit to the right of the decimal point?
      – Rolf Mertig
      50 mins ago












    • @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
      31 mins ago
















    If I do NumberForm[100.,{4,1}] I get 100.0. And for NumberForm[312.,{4,1}] 312.0. It looks like that PercentForm[1.] gives 100% and PercentForm[3.12] results in 312.%. Is this just the usual sloppiness or will this be fixed ? Also, why do you have to use N? Shouldn't this be done automatically, at least if the default of PercentForm seems to be to print 1 digit to the right of the decimal point?
    – Rolf Mertig
    50 mins ago






    If I do NumberForm[100.,{4,1}] I get 100.0. And for NumberForm[312.,{4,1}] 312.0. It looks like that PercentForm[1.] gives 100% and PercentForm[3.12] results in 312.%. Is this just the usual sloppiness or will this be fixed ? Also, why do you have to use N? Shouldn't this be done automatically, at least if the default of PercentForm seems to be to print 1 digit to the right of the decimal point?
    – Rolf Mertig
    50 mins ago














    @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
    31 mins ago




    @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
    31 mins ago











    1














    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}}






    share|improve this answer


























      1














      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}}






      share|improve this answer
























        1












        1








        1






        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}}






        share|improve this answer












        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}}







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 2 hours ago









        GerardF123

        1447




        1447






















            user62202 is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            user62202 is a new contributor. Be nice, and check out our Code of Conduct.













            user62202 is a new contributor. Be nice, and check out our Code of Conduct.












            user62202 is a new contributor. Be nice, and check out our Code of Conduct.
















            Thanks for contributing an answer to Mathematica Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f188823%2fturn-the-following-values-into-percentage%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Understanding the information contained in the Deep Space Network XML data?

            Image
            up vote 1 down vote favorite NASA's Deep Space Network activity can be viewed conveniently at https://eyes.nasa.gov/dsn/dsn.html but if you would like data in a more raw XML form, it can be accessed as described in this answer Below is a small Python script that reads the XML into a list of dictionaries, and saves it to disk as a JSON. The dictionaries are organized at the higher level by dish, then by spacecraft per dish. Question: What are the most-likely meanings of all the data field contained in the Deep Space Network XML data? This is the example url from the other answer, used in the script below: https://eyes.nasa.gov/dsn/data/1365107113.xml def dictify(r,root=True): # https://stackoverflow.com/questions/2148119/how-to-convert-an-xml-string-to-a-dictionary-in-python/30923963#30923963 if root: ret
            Read more

            Ross-on-Wye

            Image
            Ross-on-Wye Welsh: Rhosan ar Wy Town centre, looking north from Market House Ross-on-Wye Ross-on-Wye shown within Herefordshire Population 10,700 (2011) [1] OS grid reference SO597241 Unitary authority Herefordshire Ceremonial county Herefordshire Region West Midlands Country England Sovereign state United Kingdom Post town ROSS-ON-WYE Postcode district HR9 Dialling code 01989 Police West Mercia Fire Hereford and Worcester Ambulance West Midlands EU Parliament West Midlands UK Parliament Hereford and South Herefordshire List of places UK England Herefordshire 51°54′50″N 2°35′13″W  /  51.914°N 2.587°W  / 51.914; -2.587 Coordinates: 51°54′50″N 2°35′13″W  /  51.914°N 2.587°W  / 51.914; -2.587 Ross-on-Wye (Welsh: Rhosan ar Wy [2] ) is a small market town with a population of 10,700 (according to the 2011 census), in south eastern Herefordshire, England, located on the River Wye, and on
            Read more

            Eastern Orthodox Church

            Image
            Not to be confused with Oriental Orthodoxy, Eastern Christianity, or Eastern Catholic Churches. "Eastern Orthodoxy", "Orthodox Church", "Orthodox Catholic Church", and "Orthodox Christian Church" redirect here. For other uses, see Orthodox (disambiguation). This article may be too long to read and navigate comfortably . The readable prose size is 132 kilobytes. Please consider splitting content into sub-articles, condensing it, or adding or removing subheadings. ( January 2018 ) Part of a series on the Eastern Orthodox Church Mosaic of Christ Pantocrator, Hagia Sophia Overview Structure Theology (History of theology) Liturgy Church history Holy Mysteries View of salvation View of Mary View of icons Background Crucifixion / Resurrection / Ascension of Jesus Christianity Christian Church Apostolic succession Four Marks of the Church Orthodoxy Organization Autocephaly Patriarchate
            Read more