Reference · Maps

remove

remove(list | map | null, any) -> list | map

What it does

A new map without one key, or a new list without every item equal to a value. On a map, remove(m, k) drops the entry under k (turned into text, as all map keys are) and leaves the others in their order. On a list, remove(xs, x) drops each item that == x. The original is not changed.

Examples

let user = { name: "Ann", age: 41, tmp: true }let clean = remove(user, "tmp")print(clean, user)print(remove([1, 2, 1, 3], 1))

{ name: "Ann", age: 41 } { name: "Ann", age: 41, tmp: true }[2, 3]

Edge cases

• A key that is not there changes nothing and is not an error.
• Keys are text: remove(m, 5) drops "5".
• On a list, every equal item goes, not only the first; a value not in the list changes nothing.
• null as the map is an empty map and gives {}.
• On a map, an error value as the key is given back, and nothing is removed.

let m = { "5": "five", a: 1 }print(remove(m, "zz"), remove(m, 5))print(remove(["a", "b", "a"], "a"), remove([1, 2], 9))print(remove(null, "a"), remove(m, int("q")))

{ "5": "five", a: 1 } { a: 1 }["b"] [1, 2]{} err("not an int: \"q\"")

Errors

A string or a number as the first argument is a runtime error, not an error value; for text use replace.

print(remove("abc", "a"))

line 1: `remove` expects a map as argument 1, got str

Chaining

print({ a: 1, b: 2, c: 3 } | remove("b") | keys)print([3, 1, 3, 2] | remove(3) | sort)

["a", "c"][1, 2]

See also

set, merge, filter, has, replace

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