Unexpected result from reduce function
I would like to get the smallest element from a vector. For this I use combine the reduce
and min
functions. However, when providing my own implementation of min
I get unexpected results:
user=> (reduce (fn [x y] (< x y) x y) [1 2 3 2 1 0 1 2])
2
user=> (reduce min [1 2 3 2 1 0 1 2 3])
0
The reduce with standard min
returns 0 as expected. However, when I p开发者_如何学Pythonrovide my own implementation it returns 2. What am I doing wrong?
You're missing an if
:
(reduce (fn [x y] (if (< x y) x y)) ...)
^-- note the if
works fine. :-)
You are missing if
around the function's body. What happens now is:
user> (use 'clojure.contrib.trace)
nil
user> (defn foo [x y] (< x y) x y)
#'user/foo
user> (dotrace [foo] (reduce foo [1 2 3 2 1 0 1 2]))
TRACE t2229: (foo 1 2)
TRACE t2229: => 2
TRACE t2230: (foo 2 3)
TRACE t2230: => 3
TRACE t2231: (foo 3 2)
TRACE t2231: => 2
TRACE t2232: (foo 2 1)
TRACE t2232: => 1
TRACE t2233: (foo 1 0)
TRACE t2233: => 0
TRACE t2234: (foo 0 1)
TRACE t2234: => 1
TRACE t2235: (foo 1 2)
TRACE t2235: => 2
2
In other words the function you pass in always returns y, so in the last iteration 2 is returned, since 2 is the last number of the sequence you reduced.
Also note that min
already is based on reduce
:
(defn min
"Returns the least of the nums."
([x] x)
([x y] (if (< x y) x y))
([x y & more]
(reduce min (min x y) more)))
A fn used with reduce probably needs to handle 3 arities - 0, 1, and 2 arguments.
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