Tabulate

The list of $f$ applied to each index from 1 to $n$ (or a rectangular grid of them).

Tabulate(f, n)$[f(1), f(2), \ldots, f(n)]$.

Tabulate(f, n, m, …)the nested array of $f$ applied to every index tuple in the $n \times m \times \ldots$ grid, each 1-based.

Typed signatures

Type	Package	Applies
(generator: function, integer, integer?) -> list	compute-engine	always

Domain: Collections

Examples

Basic #

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In$\mathrm{Tabulate}(\operatorname{\_1}↦\operatorname{\_1}^2, 5)$Out$\bigl\lbrack1, 4, 9, 16, 25\bigr\rbrack$

A symbolic function shows the indices

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In$\mathrm{Tabulate}(f, 3)$Out$\bigl\lbrack f(1), f(2), f(3)\bigr\rbrack$

Scope #

The multi-dimensional form: a rectangular grid, indexed from 1

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In$\mathrm{Tabulate}((\operatorname{\_1}, \operatorname{\_2})↦\operatorname{\_1}\operatorname{\_2}, 2, 2)$Out$\bigl\lbrack\bigl\lbrack1, 2\bigr\rbrack, \bigl\lbrack2, 4\bigr\rbrack\bigr\rbrack$

A 2×3 grid

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In$\mathrm{Tabulate}(f, 2, 3)$Out$\bigl\lbrack\bigl\lbrack f(1, 1), f(1, 2), f(1, 3)\bigr\rbrack, \bigl\lbrack f(2, 1), f(2, 2), f(2, 3)\bigr\rbrack\bigr\rbrack$

live ↗

In$\mathrm{Tabulate}(\mathrm{Add}, 2, 2)$Out$\bigl\lbrack\bigl\lbrack2, 3\bigr\rbrack, \bigl\lbrack3, 4\bigr\rbrack\bigr\rbrack$

Three dimensions give a 2×2×2 array

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In$\mathrm{Tabulate}(f, 2, 2, 2)$Out$\bigl\lbrack\bigl\lbrack\bigl\lbrack f(1, 1, 1), f(1, 1, 2)\bigr\rbrack, \bigl\lbrack f(1, 2, 1), f(1, 2, 2)\bigr\rbrack\bigr\rbrack, \bigl\lbrack\bigl\lbrack f(2, 1, 1), f(2, 1, 2)\bigr\rbrack, \bigl\lbrack f(2, 2, 1), f(2, 2, 2)\bigr\rbrack\bigr\rbrack\bigr\rbrack$

Length 0 gives the empty list

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In$\mathrm{Tabulate}(f, 0)$Out$\bigl\lbrack\bigr\rbrack$

Applications #

The 3×3 Hilbert matrix, $1/(i+j-1)$

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In$\mathrm{Tabulate}((\operatorname{\_1}, \operatorname{\_2})↦\frac{1}{(\operatorname{\_1}+\operatorname{\_2})-1}, 3, 3)$Out$\bigl\lbrack\bigl\lbrack1, \frac{1}{2}, \frac{1}{3}\bigr\rbrack, \bigl\lbrack\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\bigr\rbrack, \bigl\lbrack\frac{1}{3}, \frac{1}{4}, \frac{1}{5}\bigr\rbrack\bigr\rbrack$

The 3×3 identity matrix

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In$\mathrm{Tabulate}((\operatorname{\_1}, \operatorname{\_2})↦\text{if }\operatorname{\_1}=\operatorname{\_2}\text{ then }1\text{ else }0, 3, 3)$Out$\bigl\lbrack\bigl\lbrack1, 0, 0\bigr\rbrack, \bigl\lbrack0, 1, 0\bigr\rbrack, \bigl\lbrack0, 0, 1\bigr\rbrack\bigr\rbrack$

Pascal's triangle as a lower-triangular matrix

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In$\mathrm{Tabulate}(\mathrm{Binomial}, 3, 3)$Out$\bigl\lbrack\bigl\lbrack1, 0, 0\bigr\rbrack, \bigl\lbrack2, 1, 0\bigr\rbrack, \bigl\lbrack3, 3, 1\bigr\rbrack\bigr\rbrack$

About

Examples	10
Checked against	Wolfram Language 10 agree