SparseArray

A dense array built from position → value rules.

✓ checked against that system · ! checked, with disagreements (hover a chip)Wolfram LanguageSparseArray!↗

SparseArray(rules)a dense array whose dimensions are the largest index seen on each axis, elsewhere 0.

SparseArray(rules, dims)like the 1-argument form, with the dimensions given explicitly.

SparseArray(rules, dims, default)like the 2-argument form, with positions not covered by rules filled with default instead of 0.

Typed signatures

Type	Package	Applies
(any, any?, any?) -> collection	combinatorics	always

Domain: Collections

Details

• A dense list (no rules) is its own dense form: SparseArray({1, 2, 3}) is {1, 2, 3}.
• Each rule is $pos \to value$: $pos$ is a plain integer for a vector, or a list of integers for a matrix or higher-rank array.
• No distinct sparse storage type is kept — the result densifies immediately into an ordinary nested list, so Normal of it is unchanged, and it is not practical for arrays too large to materialise.

Examples

Basic #

live ↗

In$\mathrm{SparseArray}(\bigl\lbrack\mathrm{Rule}(1, x), \mathrm{Rule}(3, y)\bigr\rbrack, 4)$Out$\bigl\lbrack x, 0, y, 0\bigr\rbrack$

Scope #

With no dims given, they're the largest index seen per axis

live ↗

In$\mathrm{SparseArray}(\bigl\lbrack\mathrm{Rule}(\bigl\lbrack1, 1\bigr\rbrack, 5), \mathrm{Rule}(\bigl\lbrack2, 2\bigr\rbrack, 7)\bigr\rbrack)$Out$\bigl\lbrack\bigl\lbrack5, 0\bigr\rbrack, \bigl\lbrack0, 7\bigr\rbrack\bigr\rbrack$

A third argument fills the gaps instead of 0

live ↗

In$\mathrm{SparseArray}(\bigl\lbrack\mathrm{Rule}(2, 9)\bigr\rbrack, 3, -1)$Out$\bigl\lbrack-1, 9, -1\bigr\rbrack$

live ↗

In$\mathrm{SparseArray}(\bigl\lbrack\bigl\lbrack1, 0, 0\bigr\rbrack, \bigl\lbrack1, 2, 0\bigr\rbrack, \bigl\lbrack1, 2, 3\bigr\rbrack\bigr\rbrack)$Out$\bigl\lbrack\bigl\lbrack1, 0, 0\bigr\rbrack, \bigl\lbrack1, 2, 0\bigr\rbrack, \bigl\lbrack1, 2, 3\bigr\rbrack\bigr\rbrack$

Possible issues #

Normal is a no-op here — SparseArray already returns a plain dense list

live ↗

In$\operatorname{Normal}(\mathrm{SparseArray}(\bigl\lbrack\mathrm{Rule}(1, a)\bigr\rbrack, 2))$Out$\bigl\lbrack a, 0\bigr\rbrack$

About

Examples	5
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