How to find the position of a number in an array?

2,825 次查看(过去 30 天)
If I have a vector, a = [7 8 8 2 5 6], how do I compute the positions of the value 8? I expect 2 and 3 or (1,2) and (1,3).

采纳的回答

Walter Roberson
Walter Roberson 2024-11-14,0:00
编辑:MathWorks Support Team 2024-11-14,6:28
You can use the “find” function to return the positions corresponding to an array element value. For example: a = [7 8 8 2 5 6]; linearIndices = find(a==8) linearIndices = 2 3 To get the row and column indices separately, use: [row,col] = find(a==8) row = 1 1 col = 2 3 If you only need the position of one occurrence, you could use the syntax “find(a==8,1)”. You can also specify a direction if you specifically want the first or last occurrence, such as “find(a==8,1,’first’). For more information on these options, see “find”.
  2 个评论
Arvind Andrew Das
Arvind Andrew Das 2022-7-19
i did the find function, but it gives me the ans vector with all of the values of the array and the results are in true or false for my condition( 0 & 1 ).
Walter Roberson
Walter Roberson 2022-7-20
find() can never return 0. Perhaps you are looking at ans for a different operation?

请先登录,再进行评论。

更多回答(4 个)

Bhagyesh Shiyani
Bhagyesh Shiyani 2019-12-5
what if i want both 8 positions, any code?
  2 个评论
Florian Reinbold
Florian Reinbold 2020-1-15
Hi Bhagyesh
i would suggest this one:
[val, idx] = find(a==8);
It seems to make a great job! :)
Cheers
Florian
Walter Roberson
Walter Roberson 2020-1-15
This will not return value and index, it will return row and column numbers.

请先登录,再进行评论。


Sorne Duong
Sorne Duong 2021-7-21
a = 1, 3, 6, 9, 10, 15
We know the fourth value is 9, but how to find the fourth value in MATLAB?
  4 个评论

请先登录,再进行评论。


Nilesh Kumar Bibhuti
编辑:Walter Roberson 2021-10-15
BRO , THIS WILL GIVE U THE DESIRED OUTPUT . HAPPY CODING :)
#include <stdio.h>
#define MAX_SIZE 100 // Maximum array size
int main()
{
int arr[MAX_SIZE] ,brr[MAX_SIZE] ;
int size, i, toSearch, found ,k=0;
/* Input size of array */
printf("Enter size of array: ");
scanf("%d", &size);
/* Input elements of array */
printf("Enter elements in array: ");
for (i = 0; i < size; i++)
{
scanf("%d", &arr[i]);
}
printf("\nEnter element to search: ");
scanf("%d", &toSearch);
for (i = 0; i < size; i++)
{
if (arr[i] == toSearch)
{
brr[k] = i+1 ;
printf("%d ",brr[k]);
k++ ;
}
}
return 0;
}
  1 个评论
Walter Roberson
Walter Roberson 2021-10-15
In C, int size should really be size_t size and your scanf() should be using %lu instead of %d . i and k should also be size_t
You should be checking the return status of each scanf() call .
Also
int main()
should be
int main(void)
unless you are using K&R C from before C was standardized.
The user is expecting the positions to be returned, rather than displayed.
You probably shouldn't be assuming integer for the array, but that would be an acceptable limitation if specifically documented.
You do not use or initialize found.

请先登录,再进行评论。


Korosh Agha Mohammad Ghasemi
移动:Voss 2024-6-25
To find the positions of the value 8 in the vector a = [7 8 8 2 5 6], you can use the find function in MATLAB. Here's how you can do it:
a = [7 8 8 2 5 6];
% Find the positions of the value 8
positions = find(a == 8);
disp('Positions of value 8:');
disp(positions);
This code will output the indices of the elements in a that are equal to 8. For the given vector, it will display 2 and 3.
If you need the positions in a 1-based index format like (1,2) and (1,3) (assuming a is a row vector), you can directly use the indices obtained from the find function since MATLAB indexing starts from 1.
Here's the complete example:
a = [7 8 8 2 5 6];
% Find the positions of the value 8
positions = find(a == 8);
% Display the positions in (row, column) format
for i = 1:length(positions)
fprintf('(1,%d)\n', positions(i));
end
This script will output:
(1,2)
(1,3)
This clearly indicates the positions of the value 8 in the vector a.

类别

Help CenterFile Exchange 中查找有关 Loops and Conditional Statements 的更多信息

标签

尚未输入任何标签。

Community Treasure Hunt

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

Start Hunting!

Translated by