Replace

expr rewritten by lhs -> rhs, but ONLY if the whole expression matches — unlike ReplaceAll, this never looks inside subexpressions.

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

Replace(expr, lhs -> rhs)rhs (with wildcards substituted) if expr matches lhs at the top level, else expr unchanged

Replace(expr, {rule1, rule2, …})like the 2-argument form, trying each rule in order and using the first that matches

Replace(expr, {{rules…}, {rules…}})each set of rules on its own, the answers as a list

Typed signatures

Type	Package	Applies
(any, expression<Rule> | list<expression<Rule>> | list<list<expression<Rule>>>) -> any	combinatorics	always

Domain: Collections

Details

• TOP LEVEL ONLY: this is the difference from compute-engine's own ReplaceAll (Wolfram's /.), which recurses into every subexpression. Replace is Wolfram's own /. restricted to depth 0 — closer to Wolfram's ReplaceAll would actually be ReplaceAll here.
• Built on BoxedExpression's own match/subs rather than compute-engine's replace method, which (probed) silently no-ops on a literal, non-wildcard match target reached through a boxed Rule.

Examples

Basic #

live ↗

In$\mathrm{Replace}(x+1, \mathrm{Rule}(\operatorname{\_a}+1, 100\operatorname{\_a}))$Out$100x$

Only the whole expression is tried, so $x^2$ inside the sum is not replaced

live ↗

In$\mathrm{Replace}(1+x^2, \mathrm{Rule}(x^2, a+b))$Out$x^2+1$

live ↗

In$\mathrm{Replace}(x^2, \mathrm{Rule}(x^2, a+b))$Out$a+b$

live ↗

In$\mathrm{Replace}(x, \bigl\lbrack\bigl\lbrack\mathrm{Rule}(x, a)\bigr\rbrack, \bigl\lbrack\mathrm{Rule}(x, b)\bigr\rbrack\bigr\rbrack)$Out$\bigl\lbrack a, b\bigr\rbrack$

Scope #

live ↗

In$\mathrm{Replace}(b, \bigl\lbrack\mathrm{Rule}(a, 1), \mathrm{Rule}(b, 2)\bigr\rbrack)$Out$2$

No rule matches, so the expression is unchanged

live ↗

In$\mathrm{Replace}(x, \bigl\lbrack\mathrm{Rule}(y, 2), \mathrm{Rule}(z, 3)\bigr\rbrack)$Out$x$

Properties #

An empty rule list does nothing

live ↗

In$\mathrm{Replace}(x, \bigl\lbrack\bigr\rbrack)$Out$x$

live ↗

In$\mathrm{Replace}(x, \bigl\lbrack\bigl\lbrack\mathrm{Rule}(x, 1)\bigr\rbrack, \bigl\lbrack\mathrm{Rule}(y, 2)\bigr\rbrack\bigr\rbrack)$Out$\bigl\lbrack1, x\bigr\rbrack$

Possible issues #

unlike ReplaceAll, a match one level down doesn't count

live ↗

In$\mathrm{Replace}(\bigl\lbrack x+1\bigr\rbrack, \mathrm{Rule}(\operatorname{\_a}+1, 100\operatorname{\_a}))$Out$\bigl\lbrack x+1\bigr\rbrack$

About

Examples	9
Checked against	Wolfram Language 7 agree
Crosswalk	Wolfram Language

See also: MatchQ, ReplacePart