Fold operation (code/docs/test)

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Stephen Dolan committed 2012-12-28 15:05:34 +00:00
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@@ -841,7 +841,7 @@ sections:
input: '{}'
output: [42]
- title: Variables and Functions
- title: Advanced features
body: |
Variables are an absolute necessity in most programming languages, but
they're relegated to an "advanced feature" in jq.
@@ -858,6 +858,10 @@ sections:
(many jq functions such as `map` and `find` are in fact written
in jq).
Finally, jq has a `fold` operation, which is very powerful but a
bit tricky. Again, it's mostly used internally, to define some
useful bits of jq's standard library.
entries:
- title: Variables
body: |
@@ -962,6 +966,39 @@ sections:
- program: 'def addvalue(f): f as $x | map(. + $x); addvalue(.[0])'
input: '[[1,2],[10,20]]'
output: ['[[1,2,1,2], [10,20,1,2]]']
- title: Fold
body: |
The `fold` syntax in jq allows you to combine all of the
results of an expression by accumulating them into a single
answer. As an example, we'll pass `[3,2,1]` to this expression:
fold 0 as $sum (.[] | $sum + .)
The variable `$sum` is first given the value `0`. The body
of the fold (i.e. `.[] | $sum + .`) is evaluated. `.[]`
produces three results, `3`, `2`, and `1`. For the first
one, `$sum + .` gives `3`.
Having produced this answer, jq backtracks to find the next
result as per usual. However, this time, `$sum` is set to
the previous value of the body, so `$sum + .` gives
`5`. After the final backtracking, `$sum + .` gives
`6`. This final value is used as the value of the entire
`fold` expression, so the above filter returns `6`.
More formally, in order to evaluate `fold INIT as $VAR
(BODY)`, jq first sets `$VAR` to the value of `INIT`. It
then runs through `BODY`. Each time `BODY` produces a value,
`$VAR` is set to that value and jq backtracks to find the
next one. When `BODY` stops producing values, the final
value of `$VAR` is the result of the entire expression.
examples:
- program: 'fold 0 as $sum (.[] | $sum + .)'
input: '[10,2,5,3]'
output: ['20']
- title: Assignment