How do i add a for loop if command to a pre existing table

Hi im very new to Matlab with student version, no add on's I need to include a loop if command into a table, the following is my code
shop1=[4,5,3,6,0,5,5,6,4,5];
shop2=[5,3,1,1,3,5,3,6,3,3];
day=1:10;
profit1=3*shop1;
profit2=4*shop2;
disp(' day shop1 profit1')
disp(' ')
disp([day', shop1', profit1'])
disp(' ')
disp(' day shop2 profit2')
disp(' ')
disp([day', shop2', profit2'])
disp(' profit1 profit2')
disp([profit1', profit2'])
for d=1:10; % I need to insert these results as a third column of the profit1 profit2 table
if profit1(d)>profit2(d)
disp('Shop1 is leading')
elseif profit2(d)>profit1(d)
disp('Shop2 is leading')
elseif profit1(d)==profit2(d)
disp('It is a tie')
end
end
sales_equal=find(shop1==shop2);
disp('Days which sales were the same')
disp(sales_equal)
disp(' ')
disp('Number of days sales were the same')
disp(length(sales_equal))
any help is greatly appreciated thank you

 Akzeptierte Antwort

shop1 = [4,5,3,6,0,5,5,6,4,5];
shop2 = [5,3,1,1,3,5,3,6,3,3];
profit1 = 3 * shop1;
profit2 = 4 * shop2;
pool = {'Shop1 is leading', 'It is a tie', 'Shop2 is leading'};
compare = pool(sign(profit2 - profit1) + 2);
allData = cat(1, num2cell(profit1), num2cell(profit2), compare);
disp (' profit1 profit2 compare');
profit1 profit2 compare
fprintf(' %-13d%-13d%s\n', allData{:});
12 20 Shop2 is leading 15 12 Shop1 is leading 9 4 Shop1 is leading 18 4 Shop1 is leading 0 12 Shop2 is leading 15 20 Shop2 is leading 15 12 Shop1 is leading 18 24 Shop2 is leading 12 12 It is a tie 15 12 Shop1 is leading

Weitere Antworten (1)

dpb
dpb am 1 Okt. 2021
Begin learning to use MATLAB vector operations from the git-go -- naming variables sequentially with a numeric suffix is almost always a sure sign to use arrays instead. When don't then have to write duplicate code or use arcane, slow and very prone to error routes to avoid.
Use the builtin table for the display feature as well in the command window--something otoo...
shop=[4,5,3,6,0,5,5,6,4,5;
5,3,1,1,3,5,3,6,3,3].'; % Use an array oriented by column
profit=[3 4].*shop; % compute profits with automagic array expansion (note "dot" operator here)
tShopProfits=array2table([shop profit],'VariableNames',{'Shop 1','Shop 2','Profits 1','Profits 2'}); % build base table
tShopProfits=tShopProfits(:,[1 3 2 4]); % rearrange column order for convenient viewing
tShopProfits.Leader=string(compose('Shop %d',(tShopProfits.("Profits 1")<tShopProfits.("Profits 2"))+1)); % populate the winner column
isTie=tShopProfits.("Profits 1")==tShopProfits.("Profits 2"); % and fix up for ties
tShopProfits.Leader(isTie)=repmat("Tied",sum(isTie),1); % write the tied value where needed
The above yields--
>> disp(tShopProfits)
Shop 1 Profits 1 Shop 2 Profits 2 Leader
______ _________ ______ _________ ________
4.00 12.00 5.00 20.00 "Shop 2"
5.00 15.00 3.00 12.00 "Shop 1"
3.00 9.00 1.00 4.00 "Shop 1"
6.00 18.00 1.00 4.00 "Shop 1"
0.00 0.00 3.00 12.00 "Shop 2"
5.00 15.00 5.00 20.00 "Shop 2"
5.00 15.00 3.00 12.00 "Shop 1"
6.00 18.00 6.00 24.00 "Shop 2"
4.00 12.00 3.00 12.00 "Tied"
5.00 15.00 3.00 12.00 "Shop 1"
>>

3 Kommentare

dpb
dpb am 1 Okt. 2021
Bearbeitet: dpb am 1 Okt. 2021
OBTW, another "trick" to assign the Leader value from a lookup table with more than the two possibilities of T/F that the above required the second "fixup" line to handle ties --
LEADER=["Shop 1","Draw","Shop 2"]; % the values to assign by index (aka "lookup table")
tShopProfits.Leader=LEADER(sign(tShopProfits.("Profits 2")-tShopProfits.("Profits 1"))+2); % the "engine"
returns
>> tShopProfits =
10×5 table
Shop 1 Profits 1 Shop 2 Profits 2 Leader
______ _________ ______ _________ ________
4.00 12.00 5.00 20.00 "Shop 2"
5.00 15.00 3.00 12.00 "Shop 1"
3.00 9.00 1.00 4.00 "Shop 1"
6.00 18.00 1.00 4.00 "Shop 1"
0.00 0.00 3.00 12.00 "Shop 2"
5.00 15.00 5.00 20.00 "Shop 2"
5.00 15.00 3.00 12.00 "Shop 1"
6.00 18.00 6.00 24.00 "Shop 2"
4.00 12.00 3.00 12.00 "Draw"
5.00 15.00 3.00 12.00 "Shop 1"
>>
sign() returns [-1,0,1] so adding 2 converts to the index into the lookup table of [1:3]. A place where the ability in Fortran to define arrays with arbitrary starting indices would be extremely handy in MATLAB instead of the fixed and immutable base 1.
Thank you so much for your help I appreciate it
Glad to...if it does solve your problem, go ahead and Accept the answer if for no other reason than to indicate to others there is a solution.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Loops and Conditional Statements finden Sie in Hilfe-Center und File Exchange

Produkte

Version

R2021a

Gefragt:

am 1 Okt. 2021

Kommentiert:

dpb
am 1 Okt. 2021

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by