Turn the following values into percentage [closed]
4
$begingroup$
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
edited Jan 5 at 14:52
J. M. is computer-less♦
96.3k10300461
asked Jan 4 at 14:06
user62202user62202
241
$endgroup$
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.
add a comment |
4
$begingroup$
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
edited Jan 5 at 14:52
J. M. is computer-less♦
96.3k10300461
asked Jan 4 at 14:06
user62202user62202
241
$endgroup$
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 thisdata = {{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 |
4
4
4
$begingroup$
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
edited Jan 5 at 14:52
J. M. is computer-less♦
96.3k10300461
asked Jan 4 at 14:06
user62202user62202
241
$endgroup$
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
list-manipulation numerics
edited Jan 5 at 14:52
J. M. is computer-less♦
96.3k10300461
asked Jan 4 at 14:06
user62202user62202
241
edited Jan 5 at 14:52
J. M. is computer-less♦
96.3k10300461
asked Jan 4 at 14:06
user62202user62202
241
edited Jan 5 at 14:52
J. M. is computer-less♦
96.3k10300461
edited Jan 5 at 14:52
J. M. is computer-less♦
96.3k10300461
edited Jan 5 at 14:52
J. M. is computer-less♦
96.3k10300461
96.3k10300461
asked Jan 4 at 14:06
user62202user62202
241
asked Jan 4 at 14:06
user62202user62202
241
asked Jan 4 at 14:06
user62202user62202
241
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 thisdata = {{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 thisdata = {{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
$begingroup$
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
$begingroup$
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
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
$begingroup$
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 |
3 Answers
3
active
oldest
votes
7
$begingroup$
(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]]]}
(...coming soon, in version 12)
answered Jan 4 at 16:16
Daniel LichtblauDaniel Lichtblau
46.6k275163
$endgroup$
$begingroup$
If I doNumberForm[100.,{4,1}]I get100.0. And forNumberForm[312.,{4,1}]312.0. It looks like thatPercentForm[1.]gives100%andPercentForm[3.12]results in312.%. Is this just the usual sloppiness or will this be fixed ? Also, why do you have to useN? Shouldn't this be done automatically, at least if the default ofPercentFormseems to be to print 1 digit to the right of the decimal point?
$endgroup$
– Rolf Mertig
Jan 4 at 16:59
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 usingN, 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,DecimalFormhas 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
|
show 2 more comments
5
$begingroup$
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}}*)
answered Jan 4 at 14:26
Ulrich NeumannUlrich Neumann
7,820516
$endgroup$
$begingroup$
In many applications,Roundwould be more appropriate thanIntegerPart
$endgroup$
– 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 preferdata[[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 |
1
$begingroup$
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
GerardF123GerardF123
1547
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
$begingroup$
(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]]]}
(...coming soon, in version 12)
answered Jan 4 at 16:16
Daniel LichtblauDaniel Lichtblau
46.6k275163
$endgroup$
$begingroup$
If I doNumberForm[100.,{4,1}]I get100.0. And forNumberForm[312.,{4,1}]312.0. It looks like thatPercentForm[1.]gives100%andPercentForm[3.12]results in312.%. Is this just the usual sloppiness or will this be fixed ? Also, why do you have to useN? Shouldn't this be done automatically, at least if the default ofPercentFormseems to be to print 1 digit to the right of the decimal point?
$endgroup$
– Rolf Mertig
Jan 4 at 16:59
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 usingN, 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,DecimalFormhas 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
|
show 2 more comments
7
$begingroup$
(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]]]}
(...coming soon, in version 12)
answered Jan 4 at 16:16
Daniel LichtblauDaniel Lichtblau
46.6k275163
$endgroup$
$begingroup$
If I doNumberForm[100.,{4,1}]I get100.0. And forNumberForm[312.,{4,1}]312.0. It looks like thatPercentForm[1.]gives100%andPercentForm[3.12]results in312.%. Is this just the usual sloppiness or will this be fixed ? Also, why do you have to useN? Shouldn't this be done automatically, at least if the default ofPercentFormseems to be to print 1 digit to the right of the decimal point?
$endgroup$
– Rolf Mertig
Jan 4 at 16:59
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 usingN, 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,DecimalFormhas 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
|
show 2 more comments
7
7
7
$begingroup$
(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]]]}
(...coming soon, in version 12)
answered Jan 4 at 16:16
Daniel LichtblauDaniel Lichtblau
46.6k275163
$endgroup$
(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]]]}
(...coming soon, in version 12)
answered Jan 4 at 16:16
Daniel LichtblauDaniel Lichtblau
46.6k275163
answered Jan 4 at 16:16
Daniel LichtblauDaniel Lichtblau
46.6k275163
answered Jan 4 at 16:16
Daniel LichtblauDaniel Lichtblau
46.6k275163
answered Jan 4 at 16:16
Daniel LichtblauDaniel Lichtblau
46.6k275163
46.6k275163
$begingroup$
If I doNumberForm[100.,{4,1}]I get100.0. And forNumberForm[312.,{4,1}]312.0. It looks like thatPercentForm[1.]gives100%andPercentForm[3.12]results in312.%. Is this just the usual sloppiness or will this be fixed ? Also, why do you have to useN? Shouldn't this be done automatically, at least if the default ofPercentFormseems to be to print 1 digit to the right of the decimal point?
$endgroup$
– Rolf Mertig
Jan 4 at 16:59
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 usingN, 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,DecimalFormhas 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
|
show 2 more comments
$begingroup$
If I doNumberForm[100.,{4,1}]I get100.0. And forNumberForm[312.,{4,1}]312.0. It looks like thatPercentForm[1.]gives100%andPercentForm[3.12]results in312.%. Is this just the usual sloppiness or will this be fixed ? Also, why do you have to useN? Shouldn't this be done automatically, at least if the default ofPercentFormseems to be to print 1 digit to the right of the decimal point?
$endgroup$
– Rolf Mertig
Jan 4 at 16:59
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 usingN, 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,DecimalFormhas 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
$begingroup$
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?$endgroup$
– Rolf Mertig
Jan 4 at 16:59
$begingroup$
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?$endgroup$
– Rolf Mertig
Jan 4 at 16:59
3
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
$begingroup$
@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
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
$begingroup$
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
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
$begingroup$
@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
1
$begingroup$
@ΑλέξανδροςΖεγγ Did you run the code I posted?
$endgroup$
– Daniel Lichtblau
Jan 5 at 16:04
$begingroup$
@ΑλέξανδροςΖεγγ Did you run the code I posted?
$endgroup$
– Daniel Lichtblau
Jan 5 at 16:04
|
show 2 more comments
5
$begingroup$
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}}*)
answered Jan 4 at 14:26
Ulrich NeumannUlrich Neumann
7,820516
$endgroup$
$begingroup$
In many applications,Roundwould be more appropriate thanIntegerPart
$endgroup$
– 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 preferdata[[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 |
5
$begingroup$
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}}*)
answered Jan 4 at 14:26
Ulrich NeumannUlrich Neumann
7,820516
$endgroup$
$begingroup$
In many applications,Roundwould be more appropriate thanIntegerPart
$endgroup$
– 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 preferdata[[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 |
5
5
5
$begingroup$
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}}*)
answered Jan 4 at 14:26
Ulrich NeumannUlrich Neumann
7,820516
$endgroup$
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}}*)
answered Jan 4 at 14:26
Ulrich NeumannUlrich Neumann
7,820516
edited Jan 4 at 14:46
answered Jan 4 at 14:26
Ulrich NeumannUlrich Neumann
7,820516
answered Jan 4 at 14:26
Ulrich NeumannUlrich Neumann
7,820516
answered Jan 4 at 14:26
Ulrich NeumannUlrich Neumann
7,820516
7,820516
$begingroup$
In many applications,Roundwould be more appropriate thanIntegerPart
$endgroup$
– 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 preferdata[[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,Roundwould be more appropriate thanIntegerPart
$endgroup$
– 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 preferdata[[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
$begingroup$
In many applications,
Round would be more appropriate than IntegerPart$endgroup$
– Bob Hanlon
Jan 4 at 14:35
$begingroup$
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
$begingroup$
@ 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
$begingroup$
@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
$begingroup$
@ Henrik Thanks, my actual answer seems to be the power-version!
$endgroup$
– Ulrich Neumann
Jan 4 at 16:19
add a comment |
1
$begingroup$
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
GerardF123GerardF123
1547
$endgroup$
add a comment |
1
$begingroup$
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
GerardF123GerardF123
1547
$endgroup$
add a comment |
1
1
1
$begingroup$
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
GerardF123GerardF123
1547
$endgroup$
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
GerardF123GerardF123
1547
answered Jan 4 at 15:20
GerardF123GerardF123
1547
answered Jan 4 at 15:20
GerardF123GerardF123
1547
answered Jan 4 at 15:20
GerardF123GerardF123
1547
1547
add a comment |
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