# Difference between revisions of "Cookbook"

Carandraug (talk | contribs) |
Carandraug (talk | contribs) (→Retrieve a field value from all entries in a struct array: strcmpi for indexing) |
||

Line 32: | Line 32: | ||

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

− | You are also not limited to return all elements, you may use logical indexing from other fields to get values from others | + | 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] | + | [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 | ||

== Input/output == | == Input/output == |

## Revision as of 14:29, 21 August 2012

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

## 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.