Reputation: 27
I'm a beginner at Clojure so I'll do my best to phrase this as well as I can,
I have a function that returns a list of nested lists after parsing a dataset of daily temperatures, each nested list corresponds to daily temps of a specific month e.g Feb 2014, Feb 2015 etc. and is padded out to 31 items using "-999" as filler to retain the dataset's structure.
raw dataset: https://www.metoffice.gov.uk/hadobs/hadcet/cetdl1772on.dat
(partition 31 (monthly-helper 2 (parse-into-list "CETdataDailyLong")))
=>((-15 7 15 -25 -5 -45 12 47 56 28 20 40 57 38 2 5 25 -3 0 7 7 -3 -10 -10 30 85 46 77 56 -999 -999)
(0 17 -28 -23 -30 5 -18 -3 -33 -23 -18 -3 -10 50 82 72 62 42 15 57 75 40 92 52 42 62 72 70 -999 -999 -999)
(-2 -12 4 28 12 0 44 27 -12 16 74 61 76 87 77 78 51 51 59 56 64 52 78 63 39 28 33 81 -999 -999 -999)
(97 58 75 103 33 46 88 101 56 47 66 36 52 47 58 42 42 37 63 77 76 43 55 85 58 57 55 66 -999 -999 -999)
(-59 19 28 55 47 30 52 49 42 50 45 25 34 70 40 54 24 13 25 54 85 29 27 38 25 73 44 50 40 -999 -999))
I'm trying to remove the -999 values from all nested lists in the list, I need to do this after partitioning the data to avoid having to partition the data arbitrarily by a number of days in each month. The closest I've got is below but it has no effect as it's only being applied to the top-level list instead of the values in each nested list, How would I need to modify this to get the result I'm looking for, Or to ask my original question; How do you apply a function to values at a specific nesting level?
(remove #(= -999 %)(partition 31 (monthly-helper 2 (parse-into-list "CETdataDailyLong"))))
Below is the minimal code with a chunk of the results from my partitioning function, I think it's very close but if you can show me what I'm missing I would really appreciate it, Thanks
(remove #(= -999 %)'(((-15 7 15 -25 -5 -45 12 47 56 28 20 40 57 38 2 5 25 -3 0 7 7 -3 -10 -10 30 85 46 77 56 -999 -999)
(0 17 -28 -23 -30 5 -18 -3 -33 -23 -18 -3 -10 50 82 72 62 42 15 57 75 40 92 52 42 62 72 70 -999 -999 -999)
(-2 -12 4 28 12 0 44 27 -12 16 74 61 76 87 77 78 51 51 59 56 64 52 78 63 39 28 33 81 -999 -999 -999)
(97 58 75 103 33 46 88 101 56 47 66 36 52 47 58 42 42 37 63 77 76 43 55 85 58 57 55 66 -999 -999 -999)
(-59 19 28 55 47 30 52 49 42 50 45 25 34 70 40 54 24 13 25 54 85 29 27 38 25 73 44 50 40 -999 -999))))
I've tried the below and loads of variations on it with map etc, but haven't got anywhere, Seeing a correct example would really help me understand where I'm going wrong.
(apply #(remove -999 %) (partition 31 (monthly-helper 2 (parse-into-list "CETdataDailyLong"))))
Exception: Wrong number of args (21) passed
Upvotes: 1
Views: 186
Reputation: 17859
i would start with an utility function, updating nested sequences at any level. it could look like this:
(defn update-nested [level f]
(cond (neg? level) identity
(zero? level) f
:else (partial map (update-nested (dec level) f))))
user> ((update-nested 0 (partial remove #{1})) [1 1 0 1])
;;=> (0)
user> ((update-nested 1 (partial remove #{1})) [[1 1 0 1] [0 0 1 0]])
;;=> ((0) (0 0 0))
user> ((update-nested 2 (partial remove #{1})) [[[1 1] [0 1]] [[0 0] [1 0]]])
;;=> ((() (0)) ((0 0) (0)))
user> ((update-nested 3 (partial remove #{1})) [[[[1 1] [0 1]]] [[[0 0] [1 0]]]])
;;=> (((() (0))) (((0 0) (0))))
user> ((update-nested 3 reverse) [[[[1 1] [0 1]]] [[[0 0] [1 0]]]])
;;=> ((((1 1) (1 0))) (((0 0) (0 1))))
Upvotes: 2
Reputation: 862
So iiuc, the:
What I see that you've tried:
remove
function w/ a predicate to remove if the value equals -999. The value in this case is '((-15 7 15 -25 -5 -45 12 ...))
which does not equal -999, so you end up w/ what you started with.apply
takes a function and a single sequence of args. You passed in 21 lists to apply
.With all this probably understood, I think the easiest solution is a nested for
loop. A for loop returns a list of your values, optionally modified by a function. Each value is a list, so you need to go deeper w/ another for loop.
; Remove -999's, three levels deep, with for.
(defn remove-999s [s-of-s]
; All data
(for [year s-of-s]
; For all years
(for [month year]
; For all months
; (filter #(not (= % -999)) month) would also work
(remove #(= % -999) month))))
(remove-999s '(((-15 7 15 -25 -5 -45 12 47 56 28 20 40 57 38 2 5 25 -3 0 7 7 -3 -10 -10 30 85 46 77 56 -999 -999) (0 17 -28 -23 -30 5 -18 -3 -33 -23 -18 -3 -10 50 82 72 62 42 15 57 75 40 92 52 42 62 72 70 -999 -999 -999) (-2 -12 4 28 12 0 44 27 -12 16 74 61 76 87 77 78 51 51 59 56 64 52 78 63 39 28 33 81 -999 -999 -999) (97 58 75 103 33 46 88 101 56 47 66 36 52 47 58 42 42 37 63 77 76 43 55 85 58 57 55 66 -999 -999 -999)(-59 19 28 55 47 30 52 49 42 50 45 25 34 70 40 54 24 13 25 54 85 29 27 38 25 73 44 50 40 -999 -999))))
Here's the result, without the -999's.
; (((-15 7 15 -25 -5 -45 12 47 56 28 20 40 57 38 2 5 25 -3 0 7 7 -3 -10 -10 30 85 46 77 56)
; (0 17 -28 -23 -30 5 -18 -3 -33 -23 -18 -3 -10 50 82 72 62 42 15 57 75 40 92 52 42 62 72 70)
; (-2 -12 4 28 12 0 44 27 -12 16 74 61 76 87 77 78 51 51 59 56 64 52 78 63 39 28 33 81)
; (97 58 75 103 33 46 88 101 56 47 66 36 52 47 58 42 42 37 63 77 76 43 55 85 58 57 55 66)
; (-59 19 28 55 47 30 52 49 42 50 45 25 34 70 40 54 24 13 25 54 85 29 27 38 25 73 44 50 40))) [End of data]
Because Clojure doesn't allow nested #'s, and nesting fn
's gets gross, if you want to use maps like Biped suggests, you'll probably want to use it with letfn
or defn
. Here's how I did it:
; Remove -999's, three levels deep, with maps.
(defn remove-999s [s-of-s]
(letfn [(is-999 [v] (= v -999))
( map-month [s] (remove is-999 s))
( map-year [s] (map map-month s)) ]
(map map-year s-of-s))) ; Gives the same results.
After writing this, I realized that for
is like a weird map
, so either can be used.
Another alternative's loop
and recur
or otherwise classic recursion.
Upvotes: 2
Reputation: 16035
(require '[com.rpl.specter :as s])
(def data '(your list here))
(s/setval (s/walker #(= % -999)) s/NONE data)
Upvotes: 1
Reputation: 1281
Your first exhibit is a list-of-lists. And your desired output is also a list-of-lists -- but different lists. Therefore, you want map
instead of apply
.
Upvotes: 1