7.10: The find() function
- Page ID
- 87696
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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:
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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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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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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%% 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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%% 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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a = 1: 0.5 :4 % As above
b = 2
idx4 = find(b == a)
% idx4 = 3
a(idx4) % 2
Solution
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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 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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% 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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