# Cookbook

An Octave cookbook. Each entry should go in a separate section and have the following subsection: problem, solution, discussion and maybe a see also.

## Contents

## Structures

### Retrieve a field value from all entries in a struct array

#### Problem

You have a struct array with multiple fields, and you want to acess the value from a specific field from all elements. For example, you want to return the age from all patients in the following case:

samples = struct ("patient", {"Bob", "Kevin", "Bob" , "Andrew"}, "age", [ 45 , 52 , 45 , 23 ], "protein", {"H2B", "CDK2" , "CDK2", "Tip60" }, "tube" , [ 3 , 5 , 2 , 18 ] );

#### Solution

Indexing the struct returns a comma separated list so use them to create a matrix.

[samples(:).age]

This however does not keep the original structure of the data, instead returning all values in a single column. To fix this, use `reshape()`

.

reshape ([samples(:).age], size (samples))

#### Discussion

Returning all values in a comma separated lists allows you to make anything out of them. If numbers are expected, create a matrix by enclosing them in square brackets. But if strings are to be expected, a cell array can also be easily generated with curly brackets

{samples(:).name}

You are also not limited to return all elements, you may use logical indexing from other fields to get values from the others:

[samples([samples(:).age] > 34).tube] ## return tube numbers from all samples from patients older than 34 [samples(strcmp({samples(:).protein}, "CDK2").tube] ## return all tube numbers for protein CDK2

## Input/output

## Mathematics

### Find if a number is even/odd

#### Problem

You have a number, or an array or matrix of them, and want to know if any of them is an odd or even number, i.e., their parity.

#### Solution

Check the remainder of a division by two. If the remainder is zero, the number is odd.

mod (value, 2) ## 1 if odd, zero if even

Since `mod()`

acceps a matrix, the following can be done:

any (mod (values, 2)) ## true if at least one number in values is even all (mod (values, 2)) ## true if all numbers in values are odd any (!logical (mod (values, 2))) ## true if at least one number in values is even all (!logical (mod (values, 2))) ## true if all numbers in values are even

#### Discussion

Since we are checking for the remainder of a division, the first choice would be to use `rem()`

. However, in the case of negative numbers `mod()`

will still return a positive number making it easier for comparisons. Another alternative is to use `bitand (X, 1)`

or `bitget (X, 1)`

but those are a bit slower.

Note that this solution applies to integers only. Non-integers such as 1/2 or 4.201 are neither even nor odd. If the source of the numbers are unknown, such as user input, some sort of checking should be applied for NaN, Inf, or non-integer values.

#### See also

Find if a number is an integer.