Storing Outputs from a Nested Loop

1 回表示 (過去 30 日間)
A M
A M 2017 年 1 月 30 日
コメント済み: Iddo Weiner 2017 年 2 月 1 日
I have the following problem, wherein, i want to store outputs from the nested loops. I am just presenting a simple scenario to depict the situation:
for a=[1:1:10]
for b=[1:1:10]
c=a*b
end
end
c in this case would store the last value i.e. 100. How to develop an array which stores all the values of c from 1 to 100 as c(1,1)=1, c(1,2)=2 and so on.

採用された回答

Iddo Weiner
Iddo Weiner 2017 年 1 月 31 日
By initiating an empty array and then filling it with the looping indices:
c = nan(10,10);
for a = 1:10
for b= 1:10
c(a,b)=a*b;
end
end
% visualize c
imagesc(c)
  8 件のコメント
Niels
Niels 2017 年 1 月 31 日
Matlab uses (as you should also do) smart shortcuts for its functions and variables, such as numel for Number of Elements
Iddo Weiner
Iddo Weiner 2017 年 2 月 1 日
I think you are supposed to get a 10*10 matrix in the example you posted. Yes, nan() initiates an empty array. You can also use zeros() to initiate a matrix of zeros. BTW - this code works also without the imitation line, i.e.:
for a = 1:10
for b= 1:10
c(a,b)=a*b;
end
end
BUT this is highly un-recommended, because the shape of c changes constantly - which is a very inefficient use of memory. When you pre-allocate then c's shape is constant and the whole code runs much quicker (of course in the example case the array is very small and there won't be a significant difference).
Hope this helps, good luck

サインインしてコメントする。

その他の回答 (1 件)

Niels
Niels 2017 年 1 月 31 日
some changes:
aMax=10; % equal to #of rows
bMax=10; % equal to #of colums
c=zeros(aMax,bMax);
for a=1:aMax % let a run from 1 to aMax, stepwidth:=1
for b=1:bMax
c(a,b)=a*b; % hope you know that c is not running fom 1 to 100
end
end
if you want c to cointain to values 1:(aMax*bMax) (100 in current state) change
c(a,b)=(a-1)*aMax+b;

カテゴリ

Help Center および File ExchangeLoops and Conditional Statements についてさらに検索

Community Treasure Hunt

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

Start Hunting!

Translated by