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# dsearch

search in ordered sets

### Calling Sequence

[ind, occ, info] = dsearch(X, s ) [ind, occ, info] = dsearch(X, s , ch )

### Arguments

- X
a mx-by-nx matrix of doubles, the entries to locate.

- s
a n-by-1 or 1-by-n matrix of doubles, the intervals (if ch="c") or the set (if ch="d"). The values in s must be in strictly increasing order: s(1) < s(2) < ... < s(n)

- ch
a 1-by-1 matrix of strings, the type of search (default ch="c"). Available values are ch="c" or ch="d".

- ind
a mx-by-nx matrix of doubles. The location of the X values in the intervals or the set defined by

`s`

.- occ
If ch="c", a (n-1)-by-1 or 1-by-(n-1) of doubles. If ch="d", a n-by-1 or 1-by-n of doubles. The number of entries from X in the intervals s.

- info
a 1-by-1 matrix of doubles. The number of X entries which are not in

`s`

.

### Description

This function locates the indices of the entries in X in the intervals or the set defined by s.

If `ch="c"`

(default), we consider the
intervals `I1 = [s(1), s(2)]`

and `Ik = (s(k), s(k+1)]`

for `k=2,...,n-1`

.
Notice that the bounds of `I1`

are closed,
while the left bounds of `I2`

, ..., `In`

are
opened.
For each entry `X(i)`

,
we search the interval `Ik`

which contains X(i), i.e. we search k such
that `s(k)<X(i)<=s(k+1)`

.

More precisely,

- ind(i)
is k such that

`Ik`

contains X(i) or 0 if X(i) is not in any interval`Ik`

.- occ(k)
is the number of components of

`X`

which are in`Ik`

- info
is the number of components of

`X`

which are not in any interval`Ik`

.

If `ch="d"`

, we consider the set {s(1),s(2),...,s(n)}.
For each X(i), search k such that X(i)=s(k).

More precisely,

- ind(i)
is k if X(i)=s(k) or 0 if X(i) is not in s.

- occ(k)
is the number of entries in X equal to s(k)

- info
is the number of entries in X which are not in the set {s(1),...,s(n)}.

The ch="c" option can be used to compute the empirical histogram of a function given a dataset. The ch="d" option can be used to compute the entries in X which are present in the set s.

### Examples

In the following example, we consider 3 intervals `I1=[5,11]`

,
`I2=(11,15]`

and `I3=(15,20]`

.
We are looking for the location of the entries of `X=[11 13 1 7 5 2 9]`

in these intervals.

[ind, occ, info] = dsearch([11 13 1 7 5 2 9], [5 11 15 20])

The previous example produces the following output.

-->[ind, occ, info] = dsearch([11 13 1 7 5 2 9], [5 11 15 20]) info = 2. occ = 4. 1. 0. ind = 1. 2. 0. 1. 1. 0. 1.

We now explain these results.

X(1)=11 is in the interval I1, hence ind(1)=1.

X(2)=13 is in the interval I2, hence ind(2)=2.

X(3)=1 is not in any interval, hence ind(3)=0.

X(4)=7 is in the interval I1, hence ind(4)=1.

X(5)=5 is in the interval I1, hence ind(5)=1.

X(6)=2 is not in any interval, hence ind(6)=0.

X(7)=9 is in the interval I1, hence ind(7)=1.

There are four X entries (5, 7, 9 and 11) in I1, hence occ(1)=4.

There is one X entry (i.e. 13) in I2, hence occ(2)=1.

There is no X entry in I3, hence occ(3)=0.

There are two X entries (i.e. 1, 2) which are not in any interval, hence info=2.

In the following example, we consider the set [5 11 15 20] and are searching the location of the X entries in this set.

[ind, occ, info] = dsearch([11 13 1 7 5 2 9], [5 11 15 20],"d" )

The previous example produces the following output.

-->[ind, occ, info] = dsearch([11 13 1 7 5 2 9], [5 11 15 20],"d" ) info = 5. occ = 1. 1. 0. 0. ind = 2. 0. 0. 0. 1. 0. 0.

The following is a detailed explanation for the previous results.

X(1)=11 is in the set

`s`

at position #2, hence ind(1)=2.X(2)=13 is not in the set

`s`

, hence ind(2)=0.X(3)=1 is not in the set

`s`

, hence ind(3)=0.X(4)=7 is not in the set

`s`

, hence ind(4)=0.X(5)=5 is in the set

`s`

at position #1, hence ind(5)=1.X(6)=2 is not in the set

`s`

, hence ind(6)=0.X(7)=9 is not in the set

`s`

, hence ind(7)=0.There is one entry X (i.e. 5) equal to 5, hence occ(1)=1.

There is one entry X (i.e. 11) equal to 1, hence occ(2)=1.

There are no entries matching

`s(3)`

, hence occ(3)=0.There are no entries matching

`s(4)`

, hence occ(4)=0.There are five X entries (i.e. 13, 1, 7, 2, 9) which are not in the set

`s`

, hence info=5.

The values in `s`

must be in increasing order, whatever the
value of the `ch`

option.
If this is not true, an error is generated, as in the following session.

-->dsearch([11 13 1 7 5 2 9], [2 1]) !--error 999 dsearch : the array s (arg 2) is not well ordered -->dsearch([11 13 1 7 5 2 9], [2 1],"d") !--error 999 dsearch : the array s (arg 2) is not well ordered

### Advanced Examples

In the following example, we compare the empirical histogram of uniform random numbers in [0,1) with the uniform distribution function. To perform this comparison, we use the default search algorithm based on intervals (ch="c"). We generate X as a collection of m=50 000 uniform random numbers in the range [0,1). We consider the n=10 values equally equally spaced values in the [0,1] range and consider the associated intervals. Then we count the number of entries in X which fall in the intervals: this is the empirical histogram of the uniform distribution function. The expectation for occ/m is equal to 1/(n-1).

m = 50000 ; n = 10; X = grand(m,1,"def"); s = linspace(0,1,n)'; [ind, occ] = dsearch(X, s); e = 1/(n-1)*ones(1,n-1); scf() ; plot(s(1:n-1), occ/m,"bo"); plot(s(1:n-1), e,"r-"); legend(["Experiment","Expectation"]); xtitle("Uniform random numbers","X","P(X)");

In the following example, we compare the histogram of binomially distributed random numbers with the binomial probability distribution function B(N,p), with N=8 and p=0.5. To perform this comparison, we use the discrete search algorithm based on a set (ch="d").

N = 8 ; p = 0.5; m = 50000; X = grand(m,1,"bin",N,p); s = (0:N)'; [ind, occ] = dsearch(X, s, "d"); Pexp = occ/m; Pexa = binomial(p,N); scf() ; plot(s,Pexp,"bo"); plot(s,Pexa',"r-"); xtitle("Binomial distribution B(8,0.5)","X","P(X)"); legend(["Experiment","Expectation"]);

In the following example, we use piecewise Hermite polynomials to interpolate a dataset.

// define Hermite base functions function y=Ll(t, k, x) // Lagrange left on Ik y=(t-x(k+1))./(x(k)-x(k+1)) endfunction function y=Lr(t, k, x) // Lagrange right on Ik y=(t-x(k))./(x(k+1)-x(k)) endfunction function y=Hl(t, k, x) y=(1-2*(t-x(k))./(x(k)-x(k+1))).*Ll(t,k,x).^2 endfunction function y=Hr(t, k, x) y=(1-2*(t-x(k+1))./(x(k+1)-x(k))).*Lr(t,k,x).^2 endfunction function y=Kl(t, k, x) y=(t-x(k)).*Ll(t,k,x).^2 endfunction function y=Kr(t, k, x) y=(t-x(k+1)).*Lr(t,k,x).^2 endfunction x = [0 ; 0.2 ; 0.35 ; 0.5 ; 0.65 ; 0.8 ; 1]; y = [0 ; 0.1 ;-0.1 ; 0 ; 0.4 ;-0.1 ; 0]; d = [1 ; 0 ; 0 ; 1 ; 0 ; 0 ; -1]; X = linspace(0, 1, 200)'; ind = dsearch(X, x); // plot the curve Y = y(ind).*Hl(X,ind) + y(ind+1).*Hr(X,ind) + d(ind).*Kl(X,ind) + d(ind+1).*Kr(X,ind); scf(); plot(X,Y,"k-"); plot(x,y,"bo") xtitle("Hermite piecewise polynomial"); legend(["Polynomial","Data"]); // NOTE : it can be verified by adding these ones : // YY = interp(X,x,y,d); plot2d(X,YY,3,"000")

### See Also

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