Turn the following values into percentage [on hold]
4
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 13 hours ago
J. M. is computer-less♦
96k10300460
asked yesterday
user62202
211
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user62202 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Szabolcs, Thies Heidecke, Bob Hanlon, m_goldberg, bbgodfrey 3 hours ago
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
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 13 hours ago
J. M. is computer-less♦
96k10300460
asked yesterday
user62202
211
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.
put on hold as off-topic by Szabolcs, Thies Heidecke, Bob Hanlon, m_goldberg, bbgodfrey 3 hours ago
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.
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;
– Xminer
yesterday
1
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
yesterday
add a comment |
4
4
4
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 13 hours ago
J. M. is computer-less♦
96k10300460
asked yesterday
user62202
211
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
list-manipulation numerics
edited 13 hours ago
J. M. is computer-less♦
96k10300460
asked yesterday
user62202
211
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.
edited 13 hours ago
J. M. is computer-less♦
96k10300460
asked yesterday
user62202
211
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.
edited 13 hours ago
J. M. is computer-less♦
96k10300460
edited 13 hours ago
J. M. is computer-less♦
96k10300460
edited 13 hours ago
J. M. is computer-less♦
96k10300460
96k10300460
asked yesterday
user62202
211
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 yesterday
user62202
211
asked yesterday
user62202
211
211
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.
put on hold as off-topic by Szabolcs, Thies Heidecke, Bob Hanlon, m_goldberg, bbgodfrey 3 hours ago
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.
put on hold as off-topic by Szabolcs, Thies Heidecke, Bob Hanlon, m_goldberg, bbgodfrey 3 hours ago
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.
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;
– Xminer
yesterday
1
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
yesterday
add a comment |
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;
– Xminer
yesterday
1
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
yesterday
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
yesterday
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
yesterday
1
1
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
yesterday
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
yesterday
add a comment |
3 Answers
3
active
oldest
votes
6
(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 yesterday
Daniel Lichtblau
46.5k275162
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?
– Rolf Mertig
yesterday
2
@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.
– Daniel Lichtblau
yesterday
1
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
yesterday
1
@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..
– Daniel Lichtblau
yesterday
What is coming soon in version 12?
– Αλέξανδρος Ζεγγ
19 hours ago
|
show 1 more comment
5
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 yesterday
Ulrich Neumann
7,610516
In many applications,Roundwould be more appropriate thanIntegerPart
– Bob Hanlon
yesterday
@ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
– Ulrich Neumann
yesterday
@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.
– Henrik Schumacher
yesterday
@ Henrik Thanks, my actual answer seems to be the power-version!
– Ulrich Neumann
yesterday
add a comment |
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}}
answered yesterday
GerardF123
1447
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
6
(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 yesterday
Daniel Lichtblau
46.5k275162
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?
– Rolf Mertig
yesterday
2
@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.
– Daniel Lichtblau
yesterday
1
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
yesterday
1
@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..
– Daniel Lichtblau
yesterday
What is coming soon in version 12?
– Αλέξανδρος Ζεγγ
19 hours ago
|
show 1 more comment
6
(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 yesterday
Daniel Lichtblau
46.5k275162
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?
– Rolf Mertig
yesterday
2
@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.
– Daniel Lichtblau
yesterday
1
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
yesterday
1
@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..
– Daniel Lichtblau
yesterday
What is coming soon in version 12?
– Αλέξανδρος Ζεγγ
19 hours ago
|
show 1 more comment
6
6
6
(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 yesterday
Daniel Lichtblau
46.5k275162
(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 yesterday
Daniel Lichtblau
46.5k275162
answered yesterday
Daniel Lichtblau
46.5k275162
answered yesterday
Daniel Lichtblau
46.5k275162
answered yesterday
Daniel Lichtblau
46.5k275162
46.5k275162
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?
– Rolf Mertig
yesterday
2
@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.
– Daniel Lichtblau
yesterday
1
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
yesterday
1
@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..
– Daniel Lichtblau
yesterday
What is coming soon in version 12?
– Αλέξανδρος Ζεγγ
19 hours ago
|
show 1 more comment
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?
– Rolf Mertig
yesterday
2
@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.
– Daniel Lichtblau
yesterday
1
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
yesterday
1
@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..
– Daniel Lichtblau
yesterday
What is coming soon in version 12?
– Αλέξανδρος Ζεγγ
19 hours 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
yesterday
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
yesterday
2
2
@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
yesterday
@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
yesterday
1
1
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
yesterday
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
yesterday
1
1
@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
yesterday
@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
yesterday
What is coming soon in version 12?
– Αλέξανδρος Ζεγγ
19 hours ago
What is coming soon in version 12?
– Αλέξανδρος Ζεγγ
19 hours ago
|
show 1 more comment
5
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 yesterday
Ulrich Neumann
7,610516
In many applications,Roundwould be more appropriate thanIntegerPart
– Bob Hanlon
yesterday
@ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
– Ulrich Neumann
yesterday
@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.
– Henrik Schumacher
yesterday
@ Henrik Thanks, my actual answer seems to be the power-version!
– Ulrich Neumann
yesterday
add a comment |
5
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 yesterday
Ulrich Neumann
7,610516
In many applications,Roundwould be more appropriate thanIntegerPart
– Bob Hanlon
yesterday
@ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
– Ulrich Neumann
yesterday
@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.
– Henrik Schumacher
yesterday
@ Henrik Thanks, my actual answer seems to be the power-version!
– Ulrich Neumann
yesterday
add a comment |
5
5
5
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 yesterday
Ulrich Neumann
7,610516
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 yesterday
Ulrich Neumann
7,610516
edited yesterday
answered yesterday
Ulrich Neumann
7,610516
answered yesterday
Ulrich Neumann
7,610516
answered yesterday
Ulrich Neumann
7,610516
7,610516
In many applications,Roundwould be more appropriate thanIntegerPart
– Bob Hanlon
yesterday
@ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
– Ulrich Neumann
yesterday
@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.
– Henrik Schumacher
yesterday
@ Henrik Thanks, my actual answer seems to be the power-version!
– Ulrich Neumann
yesterday
add a comment |
In many applications,Roundwould be more appropriate thanIntegerPart
– Bob Hanlon
yesterday
@ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
– Ulrich Neumann
yesterday
@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.
– Henrik Schumacher
yesterday
@ Henrik Thanks, my actual answer seems to be the power-version!
– Ulrich Neumann
yesterday
In many applications,
Round would be more appropriate than IntegerPart– Bob Hanlon
yesterday
In many applications,
Round would be more appropriate than IntegerPart– Bob Hanlon
yesterday
@ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
– Ulrich Neumann
yesterday
@ Bob Hanlon Thanks, you're right (but it is a liitle bit slower ;-) )
– Ulrich Neumann
yesterday
@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
yesterday
@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
yesterday
@ Henrik Thanks, my actual answer seems to be the power-version!
– Ulrich Neumann
yesterday
@ Henrik Thanks, my actual answer seems to be the power-version!
– Ulrich Neumann
yesterday
add a comment |
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}}
answered yesterday
GerardF123
1447
add a comment |
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}}
answered yesterday
GerardF123
1447
add a comment |
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}}
answered yesterday
GerardF123
1447
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 yesterday
GerardF123
1447
answered yesterday
GerardF123
1447
answered yesterday
GerardF123
1447
answered yesterday
GerardF123
1447
1447
add a comment |
add a comment |
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
yesterday
1
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
yesterday