MAP_APPLY

SinceVersion dev

Description

Applies a lambda expression to every entry in a map and constructs a new map from the transformed keys and values. The lambda receives the key and value of each entry and returns a pair of expressions. The first expression is used as the new key and the second expression as the new value.

Syntax

MAP_APPLY((<key>, <value>) -> (<new_key>, <new_value>), <map>)
MAP_APPLY((<key>, <value>) -> STRUCT(<new_key>, <new_value>), <map>)

Parameters

Parameter	Description
<key>	The lambda parameter representing the key of a map entry.
<value>	The lambda parameter representing the value of a map entry.
<new_key>	An expression that produces the key in the returned map. Its type must be supported by MAP keys.
<new_value>	An expression that produces the value in the returned map.
<map>	A MAP<K, V> expression.

Return Value

Returns a MAP<K2, V2>, where K2 and V2 are the types of the first and second fields returned by the lambda.

• If <map> is empty, returns an empty map.
• If <map> is NULL, returns NULL.
• NULL fields returned by the lambda are retained as NULL keys or values.
• If multiple entries are transformed to the same key, the value from the last occurrence is retained.
• The lambda must return a non-nullable STRUCT with exactly two fields. Tuple syntax (<new_key>, <new_value>) is equivalent to STRUCT(<new_key>, <new_value>); a nullable struct or a result with a different number of fields causes an analysis error.

Examples

SELECT map_apply((k, v) -> (k + 10, v * 2), map(1, 10, 2, 20)) AS result;

+------------------+
| result           |
+------------------+
| {11:20, 12:40}   |
+------------------+

SELECT map_apply((k, v) -> (1, v), map(1, 10, 2, 20)) AS result;

+--------+
| result |
+--------+
| {1:20} |
+--------+

SELECT map_apply((k, v) -> (k, v), CAST(NULL AS MAP<INT, INT>)) AS result;

+--------+
| result |
+--------+
| NULL   |
+--------+