if loop on array

7 次查看（过去 30 天）
ali hassan
ali hassan 2022-2-7
评论： Image Analyst 2022-2-11
i want to use if loop on a array. let say if i have two arrays:
a=[1 2 3 4 5]
b=[2 3 4 5 6]
now if i want to use if loop such that when any value in array 'a' is 3 and corresponding value in array 'b' is 4,it should print 'ali'.
i tried the following code but it did'nt work.
a=[3 4 3 44 3];
b=[4 3 4 34 26];
if (any(a==3) & b==4)
sprintf('ali')
end

请先登录，再回答此问题。

采纳的回答

Image Analyst
Image Analyst 2022-2-10
Ran in:
Your parentheses are not right. Try it this way
a=[3 4 3 44 3];
b=[4 3 4 34 26];
if any(a==3 & b==4)
fprintf('ali\n')
end
ali
A=[1 2 5 66 7];
B=[6 4 4 77 8];
% if any element in array A is 5 and the corresponding B element is between 2 and 6
if any(A==5 & B>=2 & B<=6)
fprintf('ali')
else
fprintf('No Matches.\n')
end
ali
  5 个评论
显示 3更早的评论
ali hassan
ali hassan 2022-2-11
here we have a vector column 'FF_liter_ho_'.If i want to know that how many times non zero values were greater than 770 in a row.
so it happened only once in the column.
Image Analyst
Image Analyst 2022-2-11
Ran in:
@ali hassan if you have the Image Processing Toolbox, you can use regionprops():
ff_liter_ho = [nan, 75, 0, nan, nan, nan, nan, nan, nan, nan, 75, 37, nan, 0, 1, 2, 3, 4]
ff_liter_ho = 1×18
NaN 75 0 NaN NaN NaN NaN NaN NaN NaN 75 37 NaN 0 1 2 3 4
nonZeroIndexes = ff_liter_ho > 0
nonZeroIndexes = 1×18 logical array
0 1 0 0 0 0 0 0 0 0 1 1 0 0 1 1 1 1
props = regionprops(nonZeroIndexes, 'Area');
numRuns = length(props)
numRuns = 3
runLengths = [props.Area]
runLengths = 1×3
1 2 4

请先登录，再进行评论。

更多回答（1 个）

KSSV
KSSV 2022-2-7
a=[3 4 3 44 3];
b=[4 3 4 34 26];
if (any(a==3))
idx = a == 3 ;
if any(b(idx) == 4 )
fprintf('ali\n')
end
end
  4 个评论
显示 2更早的评论
ali hassan
ali hassan 2022-2-10
assume, i have an array
A=[1 2 5 66 7] and B=[6 4 4 77 8]
i want to put if condition such that if any element in array A is 5 and the corresponding is between 2 and 6, i want to know how many times the condition meets.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by