My question is how can I capture the index of the vector row where a match occurred? In the code below, what am I doing wrong?
I have a vector of vectors
(def v1 [[44 2 3 4 5][1 6 7 5 10][11 12 13 14 15]])
a column index, and a comparison value
(def cmp-val1 11)
(def col-idx 0)
I want to return the row index where a comparison returned true. With col-idx = 0 and cmp-val1 = 11, I should see (first row-num) return 2, and it is returning 1.
(defn ret-match-row
"Return the index of the row, in which the cmp-val is found.
It is okay to increment 0."
[in-seq cmp-val col-idx]
(let [rn 0]
(let [row-num
(for [seq-row in-seq
:let [local-row-num (inc rn)]
:when (= cmp-val (nth seq-row col-idx nil))]
local-row-num)]
(first row-num))))
From lein repl:
bene-csv.core=> (ret-match-row v1 cmp-val1 col-idx)
1
=> (defn ret-match-row
[coll cmp idx]
(keep-indexed (fn [i v] (if (= cmp (get v idx)) i)) coll))
=> (ret-match-row v1 11 0)
(2)
A flexible answer comes from separating this into three distinct problems and composing them.
creating the data you seek
finding the data you want
presenting the way it should look.
first we number the rows by adding row numbers to them
(map vector v1 (range))
then filter out the rows not containing the number you want:
(filter (fn [[data index]] (some #{11} data)) (map vector v1 (range)))
> ([[11 12 13 14 15] 2])
here i used the trick that sets are functions that test their input for inclusion in the set which allows this to test for multiple values:
(filter (fn [[data index]] (some #{11 44} data)) (map vector v1 (range)))
> ([[44 2 3 4 5] 0] [[11 12 13 14 15] 2])
then since you only want to know where it matched and not what matched we filter that out:
(map second (filter (fn [[data index]] (some #{11 44} data)) (map vector v1 (range))))
> (0 2)
to wrap this into a nice function we write out the steps:
(defn with-row-numbers [col] (map vector col (range)))
(defn find-my-rows [rows goals]
(filter (fn [[data index]] (some (set goals) data)) rows))
(defn present-rows [rows] (map second rows))
and then compose them:
(defn ret-match-row [data rows]
(-> data
(with-row-numbers)
(find-my-rows rows)
(present-rows)))
(ret-match-row v1 [11])
(2)
sorry i couldn't help making it work with multiple values, its a habit.
(ret-match-row v1 [11 15 44])
> (0 2)
There may be other ways to do what your'e asking, but you can use a loop/recur to achieve the iteration you're after:
(defn ret-match-row [rows val col-idx]
(loop [[row & rows] rows
pos 0]
(cond
(not row)
nil
(= val (nth row col-idx))
pos
:not-found
(recur rows (inc pos)))))
(ret-match-row [[44 2 3 4 5]
[1 6 7 8 10]
[11 12 13 14 15]]
11
0)
;; => 2
You're also running into Clojure's immutability - the (inc rn) is not actually modifying rn. The loop / recur solution uses inc as well, but it passes the result of inc to the next iteration of the loop.
Clojure's for (list comprehension) form also will loop over all of the values in the sequence, resulting in a new sequence -- which is most likely not what you want. Even if you made the for loop do what you want, it will find all the matches, not just the first. The loop / recur example stops at the first match.
My take, using
clojure.contrib.seq find-first, indexed:
(defn ret-match-row [rows val col-idx]
(first
(find-first #(= val (nth (second %) col-idx))
(indexed rows))))
Related
Suppose I have a list of elements L, a function g, and a list of indices I.
Then, how can I map the function g only to the elements of the list L specified by the indices I?
For instance, if g is the squaring function, L is the list (1 2 3 4 5 6 7) and I is the set of indices (1 3 4), then I should obtain
(1 4 3 16 25 6 7), that is the list L in which I squared the elements in positions I.
(The first index is 0, like it is used in the nth function)
I can do it in some way or another, but I was wondering if there is a simple way to do it.
Or, without a library, you can just make use of map-indexed:
(def I #{1 3 4})
(def L '(1 2 3 4 5 6 7))
(defn g [n] (* n n))
(map-indexed #(if (I %) (g %2) %2) L))
; or, with explicit parameters
(map-indexed (fn [i n] (if (I i) (g n) n)) L)
; Both produce a lazy-list of (1 4 3 16 25 6 7)
Of course, with better names, this would be a lot more readable.
I have I as a set here instead of a list so lookups can be done efficiently. If you have it as a list originally, you can convert it to a set using set.
Also note, this produces a lazy-list. If you want a different structure, you can use vec for example to force it into a vector afterward.
(require '[com.rpl.specter :as s])
(use '[plumbing.core])
(s/transform [s/INDEXED-VALS (fn->> first (contains? #{1 3 4})) s/LAST]
(fn [x] (* x x))
'(1 2 3 4 5 6 7))
I would say, you can do it with a simple map call:
(defn g [x]
(* x x))
(def L '(1 2 3 4 5 6 7))
(def I '(1 3 4))
(map #(g (nth L %)) I)
;; => (4 16 25)
The mindset here is, for each indexes of I, I lookup the associated value in L and I compute g function over it.
Another option is to loop over the desired indexes to change, using assoc to replace items in a vector:
(ns tst.demo.core
(:use tupelo.core tupelo.test) )
(defn update-idx
[data txfn idxs]
(reduce
(fn [result idx]
(let [val-orig (get result idx)
val-new (txfn val-orig)]
(assoc result idx val-new)))
(vec data) ; coerce data to vector
idxs))
(dotest
(let [plus100 (fn [x] (+ 100 x))]
(is= (update-idx (range 10) plus100 [2 3 5 7]))
[0 1 102 103 4 105 6 107 8 9]))
I have a list of numbers 2 4 3 7 4 9 8 5 12 24 8.
I need to find numbers which are repeated more than once in clojure.
I have used frequencies function to find. But the result is
{2 1,4 2,3 1,7 1,9 1,8 2,5 1,12 1,24 1}
I intially thought of considering them as key value and then take each key value once and see if val is > 1. if value is greater than 1 then I need to inc 1.
But I am unable to work this out.
Can anyone please help me??
Is there anyway I can make this into [[2 1][4 2][3 1][7 1][9 1][8 2][5 1][12 1][24 1]] and consider each vector recursively or any better idea you can think of.
Thank you.
The function below will just continue on where you have stuck:
(defn find-duplicates [numbers]
(->> numbers
(frequencies)
(filter (fn [[k v]] (> v 1)))
(keys)))
It will filter map entries that have values greater than 1 and then extract their keys.
(find-duplicates [2 4 3 7 4 9 8 5 12 24 8])
;; => (4 8)
If you want the repeated items:
(defn repeated [coll]
(->> coll
frequencies
(remove #(= 1 (val %)))
keys))
(repeated [2 4 3 7 4 9 8 5 12 24 8])
;(4 8)
If you just want to count them:
(defn repeat-count [coll]
(->> coll
frequencies
(remove #(= 1 (val %)))
count))
(repeat-count [2 4 3 7 4 9 8 5 12 24 8])
;2
You can do it lazily, so that it will work on an endless sequence:
(defn repeated [coll]
((fn ff [seen xs]
(lazy-seq
(when-let [[y & ys] (seq xs)]
(case (seen y)
::several (ff seen ys)
::once (cons y (ff (assoc seen y ::several) ys))
(ff (assoc seen y ::once) ys)))))
{} coll))
(repeated [2 4 3 7 4 9 8 5 12 24 8])
;(4 8)
This is similar to core distinct.
... and finally, for brevity, ...
(defn repeated [coll]
(for [[k v] (frequencies coll) :when (not= v 1)] k))
I stole the use of keys from Piotrek Byzdyl's answer. It is only supposed to apply to a map. but works perfectly well here on a sequence of map-entries.
(->> s (group-by identity) (filter (comp next val)) keys)
You are on the right track.
If you seq over hash-map, e. g. via map, you get the kv tuple structure you described and can destructure an individual tuple in the element transformation function:
(->> s
(frequencies)
(map (fn [[number times]]
(cond-> number ; take number
(> times 1) (inc))))) ; inc if (times > 1), otherwise return number
You can use this approach.
(def c [2 4 3 7 4 9 8 5 12 24 8])
(->> c
sort
(partition-by identity)
(filter #(> (count %) 1))
(map first))
I have the following variable
(def a [[1 2] [3 4] [5 6]])
and want to return
[[1 3 5][2 4 6]]
and if input is
[[1 2] [3 4] [5 6] [7 8 9]] then the required result is
[[1 3 5 7] [2 4 6 8] [9]]
How to do it in clojure?
(persistent!
(reduce
(fn [acc e]
(reduce-kv
(fn [acc2 i2 e2]
(assoc! acc2 i2 ((fnil conj []) (get acc2 i2) e2)))
acc
e))
(transient [])
[[1 2 3] [:a :b] [\a] [111] [200 300 400 500]]))
;;=> [[1 :a \a 111 200] [2 :b 300] [3 400] [500]]
An empty vector can be updated via the update-in fn at the 0th index, a non-empty vector can be, additionally, updated at the index immediately following the last value.
The reduction here is about passing the outer accumulator to the inner reducing function, updating it accordingly, and then returning it back to the outer reducing function, which in turn will pass again to the inner rf for processing the next element.
EDIT: Updated to fastest version.
I like ifett's implementation, though it seems weird to use reduce-kv to build a vector that could be easily build with map/mapv.
So, here is how I would've done it:
(defn transpose [v]
(mapv (fn [ind]
(mapv #(get % ind)
(filter #(contains? % ind) v)))
(->> (map count v)
(apply max)
range)))
(->> (range)
(map (fn [i]
(->> a
(filter #(contains? % i))
(map #(nth % i)))))
(take-while seq))
Notice that this algorithm creates a lazy seq of lazy seqs so you that you will only pay for the transformations you really consume. If you insist on creating vectors instead, wrap the forms in vec at the necessary places - or if you are using Clojurescript or don't mind a Clojure 1.7 alpha use transducers to create vectors eagerly without paying for laziness or immutability:
(into []
(comp
(map (fn [i]
(into [] (comp (filter #(contains? % i))
(map #(nth % i)))
a)))
(take-while seq))
(range))
I find this easy to understand:
(defn nth-column [matrix n]
(for [row matrix] (nth row n)))
(defn transpose [matrix]
(for [column (range (count (first matrix)))]
(nth-column matrix column)))
(transpose a)
=> ((1 3 5) (2 4 6))
nth-column is a list comprehension generating a sequence from the nth element of each sequence (of rows).
Then transpose-matrix is simply iterating over the columns creating a sequence element for each, consisting of (nth-column matrix column) i.e. the sequence of elements for that column.
(map
(partial filter identity) ;;remove nil in each sub-list
(take-while
#(some identity %) ;;stop on all nil sub-list
(for [i (range)]
(map #(get % i) a)))) ;; get returns nil on missing values
Use get to have nil on missing values, iterate (for) on an infinite range, stop on all nil sub-list, remove nil from sub-lists. Add vector constructor before first map and in it's function (first argument) if you really need vectors.
EDIT: please leave a comment if you think this is not useful. We can all learn from mistakes.
What is the best way of implementing map function together with an updatable state between applications of function to each element of sequence? To illustrate the issue let's suppose that we have a following problem:
I have a vector of the numbers. I want a new sequence where each element is multiplied by 2 and then added number of 10's in the sequence up to and including the current element. For example from:
[20 30 40 10 20 10 30]
I want to generate:
[40 60 80 21 41 22 62]
Without adding the count of 10 the solution can be formulated using a high level of abstraction:
(map #(* 2 %) [20 30 40 10 20 10 30])
Having count to update forced me to "go to basic" and the solution I came up with is:
(defn my-update-state [x state]
(if (= x 10) (+ state 1) state)
)
(defn my-combine-with-state [x state]
(+ x state))
(defn map-and-update-state [vec fun state update-state combine-with-state]
(when-not (empty? vec)
(let [[f & other] vec
g (fun f)
new-state (update-state f state)]
(cons (combine-with-state g new-state) (map-and-update-state other fun new-state update-state combine-with-state))
)))
(map-and-update-state [20 30 40 50 10 20 10 30 ] #(* 2 %) 0 my-update-state my-combine-with-state )
My question: is it the appropriate/canonical way to solve the problem or I overlooked some important concepts/functions.
PS:
The original problem is walking AST (abstract syntax tree) and generating new AST together with updating symbol table, so when proposing the solution to the problem above please keep it in mind.
I do not worry about blowing up stack, so replacement with loop+recur is not
my concern here.
Is using global Vars or Refs instead of passing state as an argument a definite no-no?
You can use reduce to accumulate a pair of the number of 10s seen so far and the current vector of results.:
(defn map-update [v]
(letfn [(update [[ntens v] i]
(let [ntens (+ ntens (if (= 10 i) 1 0))]
[ntens (conj v (+ ntens (* 2 i)))]))]
(second (reduce update [0 []] v))))
To count # of 10 you can do
(defn count-10[col]
(reductions + (map #(if (= % 10) 1 0) col)))
Example:
user=> (count-10 [1 2 10 20 30 10 1])
(0 0 1 1 1 2 2)
And then a simple map for the final result
(map + col col (count-10 col)))
Reduce and reductions are good ways to traverse a sequence keeping a state. If you feel your code is not clear you can always use recursion with loop/recur or lazy-seq like this
(defn twice-plus-ntens
([coll] (twice-plus-ntens coll 0))
([coll ntens]
(lazy-seq
(when-let [s (seq coll)]
(let [item (first s)
new-ntens (if (= 10 item) (inc ntens) ntens)]
(cons (+ (* 2 item) new-ntens)
(twice-plus-ntens (rest s) new-ntens)))))))
have a look at map source code evaluating this at your repl
(source map)
I've skipped chunked optimization and multiple collection support.
You can make it a higher-order function this way
(defn map-update
([mf uf coll] (map-update mf uf (uf) coll))
([mf uf val coll]
(lazy-seq
(when-let [s (seq coll)]
(let [item (first s)
new-status (uf item val)]
(cons (mf item new-status)
(map-update mf uf new-status (rest s))))))))
(defn update-f
([] 0)
([item status]
(if (= item 10) (inc status) status)))
(defn map-f [item status]
(+ (* 2 item) status))
(map-update map-f update-f in)
The most appropriate way is to use function with state
(into
[]
(map
(let [mem (atom 0)]
(fn [val]
(when (== val 10) (swap! mem inc))
(+ #mem (* val 2)))))
[20 30 40 10 20 10 30])
also see
memoize
standard function
I have used the following expression to retrieve the index of the smallest number in a vector. However, I would like to avoid the use of .indexOf (for efficiency reasons and maybe numeric precision, although I guess the numbers are implicitly converted to strings).
(.indexOf [1 2 3 4 0 5]
(reduce #(if (< %1 %2) %1 %2) [1 2 3 4 0 5] ))
Would it be possible to do it differently using reduce?
user=> (first (apply min-key second (map-indexed vector [1 2 4 0 5])))
3
I'd suggest using loop/recur if you want to do this efficiently, perhaps something like the following:
(defn min-index [v]
(let [length (count v)]
(loop [minimum (v 0)
min-index 0
i 1]
(if (< i length)
(let [value (v i)]
(if (< value minimum)
(recur value i (inc i))
(recur minimum min-index (inc i))))
min-index))))
The idea is to iterate across the whole vector, keeping track of the minimum and the index of the minimum value found so far at each point.
You can also use reduce:
(def v [1 2 3 4 0 5])
(second (reduce (fn [[curr-min min-idx curr-idx] val]
(if (< val curr-min)
[val curr-idx (inc curr-idx)]
[curr-min min-idx (inc curr-idx)])) [(first v) 0 0] v)) ;; => 4
The result of reduce is actually a three-element vector consisting of the minimum value, its index, and an index tracker (which is not important), respectively. And it traverses the collection once.
The initial value provided to reduce is basically the first element of the collection.
I know the question is old, but it is here for posterity's sake.
Following up on #Alex Taggart's answer, using thread-last macro:
user=> (->> [1 2 4 0 5]
(map-indexed vector) ; [[0 1] [1 2] [2 4] [3 0] [4 5]]
(apply min-key second) ; [3 0]
first)
3