7.10: The find() function

The MATLAB built-in function find() is an efficient method to find the indices of data that satisfy some logical condition.

The syntax of the find function is:

indx = find( condition )

The output of the find() function is the set of indices for which the "condition" is true.

This is illustrated by these examples:

Example \(\PageIndex{1}\) find() with an == condition

a = 1:0.5:4 % a = [1.0 1.5, 2.0, 2.5, 3.0, 3.5, 4.0]
idx1 = find(a == 3.0) % Find the elements equal to 3 and return their indices
% idx1 = 5, becuz the 5th element of a is 3.0

% Check it with this code
a_idx1 = a(idx1) % = 3.0

Solution

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Example \(\PageIndex{2}\) find() with a < condition

a = 1:0.5:4 % Same vector as above
idx2 = find(a < 3.0)
% idx = 1 2 3 4
% These are the indices of the elements of a < 3

% Check it with this code
a(idx2) % = 1.0, 1.5, 2.0, 2.5

Solution

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Example \(\PageIndex{3}\) find() with a compound condition

a = 1:0.5:4 % Same vector as above
% Find the indices of the elements between 2.0 and 3.5, inclusive
% We need to use a compound conditional expression
idx3 = find( (2.0 <= a) & (a <= 3.5) )
% "&" = "and"
% That means that for a particular value of a, both conditional parts have to be true for
% for the compound conditional to be true.
% idx3 = 3 4 5 6
% These are the indices of the elements that satisfy both parts

% Check it with this code

a(idx3) % = 2.0, 2.5, 3.0, 3.5

Solution

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Example \(\PageIndex{4}\) The result of the compound condition itself

%% The condition by itself:
( (2.0 <= a) & (a <= 3.5) )
%ans =
% 1×7 logical array
% 0 0 1 1 1 1 0
% 0= False for the elements that do not satisfy both parts of the
% conditional
% 1=True for the elements that do satisfy both parts

Solution

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Example \(\PageIndex{5}\) What could go wrong?

%% What happens if we try this?
% A single condition, instead of a compound condition:
idx3b = find(2.0 <= a <= 3.5)

% The result is: idx3b = 1 2 3 4 5 6 7 (!?)
% This is clearly not what was desired.

% The condition by itself (2.0 <= a <= 3.5) returns:
% 1 1 1 1 1 1 1
% becuz every value of a is either >= 2 or <= 3.5
% You can only have a comparison between 1 number and 1 variable
% or 2 variables.
% You need to have each part of the compound be a complete condition by itself, as above, namely:

% ( (2.0 <= a) & (a <= 3.5) )

Solution

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Example \(\PageIndex{6}\) find() with 2 variables

a = 1: 0.5 :4 % As above
b = 2
idx4 = find(b == a)
% idx4 = 3
a(idx4) % 2

Solution

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Example \(\PageIndex{7}\) find() with no match.

When there are no elements that satisfy the condition,

then the find() command returns an empty vector.

a = 1:0.5:4
b = 20
idx4 = find(b == a)

% idx4 = empty

Solution

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Example \(\PageIndex{8}\)

%% Example 5: 2 vectors of the same length
a = 1:0.5:4
% a = [1.0 1.5, 2.0, 2.5, 3.0, 3.5, 4.0]
c = [ 0 1 2 3 4 5 6 ]
idx5 = find( a == c)
% idx5 = 3, because a(3) = b(3) = 2
% Also, even though both vectors have values of 1, 3, and 4
% these don't occur in the same place, so they are not matches.

Solution

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Exercise \(\PageIndex{1}\) Test Scores >= 90

% Use these test scores:
test_scores = [
73
72
88
94
77
92
88
65
86
78
97
65]
% Use the find() function to find and list the test scores >= 90

Answer

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