# Difference between revisions of "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.

## 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 pacients in the following case:

``` cases(1).name         = "Bob";
cases(1).age          = 45;
cases(1).contaminated = true;

cases(2).name         = "Andrew";
cases(2).age          = 21;
cases(2).contaminated = true;

cases(2).name         = "Kevin";
cases(2).age          = 24;
cases(2).contaminated = false;
```

#### Solution

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

``` [cases(:).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 ([cases(:).age], size (cases))
```

#### 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

``` {cases(:).name}
```

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

``` [cases([cases(:).age] > 34).contaminated]         ## return contaminated state from all cases older than 34
[cases(strcmp({cases(:).protein}, "CDK2").tube]   ## return all tube numbers with protein CDK2
```

## 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.