For the code below I'm reading input from stdin. Basically it's just some numbers delimited by spaces or line breaks. Specifically I'm trying to complete this challenge.
My goal is to create a list of numbers (without the first number) from the input. When I run the code below at hackerrank I get a list of a single number: (5)
Not sure what's going on, or how to fix. Would anyone know?
(map read-string (rest (line-seq (java.io.BufferedReader. *in*))))
line-seq gives one string for each line. read-string reads from a string, returning the first complete object found. Thus, you only get the first item on the line.
You could either us clojure.string/split to break up the string and use read-string on each part, or loop, accumulating the results of calling read on a PushbackReader made from the BufferedReader until you get no more input.
Since your input is
Input Format
The first line contains a single integer N.
The next line contains N integers: a0, a1,...aN-1 separated by space...
Sample Input
6
5 4 4 2 2 8
And you don't need to worry about validation / security, you can just
(let [n (read-string (read-line))
v (read-string (str "[" (read-line) "]"))]
(assert (== n (count v))) ;if you like
(comment solution here...))
Related
(def filename "dictionary2.txt")
(defn check_word [filename word]
(with-open [r (clojure.java.io/reader filename)]
(doseq [line (line-seq r)]
(if (compare line word)
(println word)))))
(check_word filename "wizard")
It prints "found" as many as the number of lines in the text file. Why is if statement always returning true? Word of "wizard" does exist in the dictionary file.
According to the documentation the compare function returns a negative number, zero, or a positive number depending on the order of its parameters. Numbers are considered truthy values, so they always make the then branch of a conditional expression execute. The only falsey values in Clojure are nil and false.
If you want to check that line equals word you can use just equality with (= line word).
I have a sequence of sequences and each sequence is similar to the following:
("9990999" "43" "ROADWAY" "MORRISON, VAN X DMD" "43 ROADWAY" "SOMETHINGTON" "XA" "00000" "501" "18050" "2500" "1180" "14370" "0")
clojure-csv won't help me here, because it -- as it should -- quotes fields with embedded commas. I want pipe-delimited output without quotes around each field, some of which contain embedded commas.
I have looked at a number of ways to remove the double quote characters including the following, but the quotes stay put.
(filter (fn [x] (not (= (str (first (str x))) (str (first (str \")))))) d1)
where d1 is the sequence above.
In addition to an answer, I am more interested in a pointer to documentation. I have been playing with this but to no avail.
As far as I understand you have a sequence of strings. Clojure provides a very specific toString implementation for sequences, you can see it here.
If you do (str d1) or simply type d1 in repl and press enter you'll see more or less what you typed: sequence of strings (String is printed as sequence of characters in double quotes).
Now if you want to concatenate all the string you can do this:
(apply str d1)
If you want to print it separated with commas you could do this:
(apply str (interpose "," d1))
To output is CSV format I would recommend to use clojure-csv.
Finally if you simply want to print the list but without the double quotes around strings you could do this:
(print d1)
Hope this helps.
EDIT1 (update due to changes in the question):
This can easily be achieved with:
(apply str (interpose "|" d1))
Please don't pay attention to double quotes around the entire result if you print it or spit it into a file you won't see them, this is just how Clojure prints string readably.
Alternatively if you have multiple sequences like that that you want to output at once you can still use clojure-csv but with different separator:
(ns csv-test.core
(:require [clojure-csv.core :as csv]))
(def d1 (list "9990999" "43" "ROADWAY" "MORRISON, VAN X DMD" "43 ROADWAY" "SOMETHINGTON" "XA" "00000" "501" "18050" "2500" "1180" "14370" "0"))
(print (csv/write-csv [d1] :delimiter "|"))
;;prints:
;;9990999|43|ROADWAY|MORRISON, VAN X DMD|43 ROADWAY|SOMETHINGTON|XA|00000|501|18050|2500|1180|14370|0
I'm working on writing a function in Clojure that will process a file character by character. I know that Java's BufferedReader class has the read() method that reads one character, but I'm new to Clojure and not sure how to use it. Currently, I'm just trying to do the file line-by-line, and then print each character.
(defn process_file [file_path]
(with-open [reader (BufferedReader. (FileReader. file_path))]
(let [seq (line-seq reader)]
(doseq [item seq]
(let [words (split item #"\s")]
(println words))))))
Given a file with this text input:
International donations are gratefully accepted, but we cannot make
any statements concerning tax treatment of donations received from
outside the United States. U.S. laws alone swamp our small staff.
My output looks like this:
[International donations are gratefully accepted, but we cannot make]
[any statements concerning tax treatment of donations received from]
[outside the United States. U.S. laws alone swamp our small staff.]
Though I would expect it to look like:
["international" "donations" "are" .... ]
So my question is, how can I convert the function above to read character by character? Or even, how to make it work as I expect it to? Also, any tips for making my Clojure code better would be greatly appreciated.
(with-open [reader (clojure.java.io/reader "path/to/file")] ...
I prefer this way to get a reader in clojure. And, by character by character, do you mean in file access level, like read, which allow you control how many bytes to read?
Edit
As #deterb pointed out, let's check the source code of line-seq
(defn line-seq
"Returns the lines of text from rdr as a lazy sequence of strings.
rdr must implement java.io.BufferedReader."
{:added "1.0"
:static true}
[^java.io.BufferedReader rdr]
(when-let [line (.readLine rdr)]
(cons line (lazy-seq (line-seq rdr)))))
I faked a char-seq
(defn char-seq
[^java.io.Reader rdr]
(let [chr (.read rdr)]
(if (>= chr 0)
(cons chr (lazy-seq (char-seq rdr))))))
I know this char-seq reads all chars into memory[1], but I think it shows that you can directly call .read on BufferedReader. So, you can write your code like this:
(let [chr (.read rdr)]
(if (>= chr 0)
;do your work here
))
How do you think?
[1] According to #dimagog's comment, char-seq not read all char into memory thanks to lazy-seq
I'm not familiar with Java or the read() method, so I won't be able to help you out with implementing it.
One first thought is maybe to simplify by using slurp, which will return a string of the text of the entire file with just (slurp filename). However, this would get the whole file, which maybe you don't want.
Once you have a string of the entire file text, you can process any string character by character by simply treating it as though it were a sequence of characters. For example:
=> (doseq [c "abcd"]
(prntln c))
a
b
c
d
=> nil
Or:
=> (remove #{\c} "abcd")
=> (\a \b \d)
You could use map or reduce or any sort of sequence manipulating function. Note that after manipulating it like a sequence, it will now return as a sequence, but you could easily wrap the outer part in (reduce str ...) to return it back to a string at the end--explicitly:
=> (reduce str (remove #{\c} "abcd"))
=> "abd"
As for your problem with your specific code, I think the problem lies with what words is: a vector of strings. When you print each words you are printing a vector. If at the end you replaced the line (println words) with (doseq [w words] (println w))), then it should work great.
Also, based on what you say you want your output to look like (a vector of all the different words in the file), you wouldn't want to only do (println w) at the base of your expression, because this will print values and return nil. You would simply want w. Also, you would want to replace your doseqs with fors--again, to avoid return nil.
Also, on improving your code, it looks generally great to me, but--and this is going with all the first change I suggest above (but not the others, because I don't want to draw it all out explicitly)--you could shorten it with a fun little trick:
(doseq [item seq]
(let [words (split item #"\s")]
(doseq [w words]
(println w))))
;//Could be rewritten as...
(doseq [item s
:let [words (split item #"\s")]
w words]
(println w))
You're pretty close - keep in mind that Strings are a sequence. (concat "abc" "def") results in the sequence (\a \b \c \d \e \f).
mapcat is another really useful function for this - it will lazily concatenate the results of applying the mapping fn to the sequence. This means that mapcating the result of converting all of the line strings to a seq will be the lazy sequence of characters you're after.
I did this as (mapcat seq (line-seq reader)).
For other advice:
For creating the reader, I would recommend using the clojure.java.io/reader function instead of directly creating the classes.
Consider breaking apart the reading the file and the processing (in this case printing) of the strings from each other. While it is important to keep the full file parsing inside the withopen clause, being able to test the actual processing code outside of the file reading code is quite useful.
When navigating multiple (potentially nested) sequences consider using for. for does a nice job handling nested for loop type cases.
(take 100 (for [line (repeat "abc") char (seq line)] (prn char)))
Use prn for debugging output. It gives you real output, as compared to user output (which hides certain details which users don't normally care about).
can you suggest me the shortest and easiest way for extracting substring from string sequence? I'm getting this collection from using enlive framework, which takes content from certain web page, and here is what I am getting as result:
("background-image:url('http://s3.mangareader.net/cover/gantz/gantz-r0.jpg')"
"background-image:url('http://s3.mangareader.net/cover/deadman-wonderland/deadman-wonderland-r0.jpg')"
"background-image:url('http://s3.mangareader.net/cover/12-prince/12-prince-r1.jpg')" )
What I would like is to get some help in extracting the URL from the each string in the sequence.i tried something with partition function, but with no success. Can anyone propose a regex, or any other approach for this problem?
Thanks
re-seq to the resque!
(map #(re-seq #"http.*jpg" %) d)
(("http://s3.mangareader.net/cover/gantz/gantz-r0.jpg")
("http://s3.mangareader.net/cover/deadman-wonderland/deadman-wonderland-r0.jpg")
("http://s3.mangareader.net/cover/12-prince/12-prince-r1.jpg"))
user>
re-find is even better:
user> (map #(re-find #"http.*jpg" %) d)
("http://s3.mangareader.net/cover/gantz/gantz-r0.jpg"
"http://s3.mangareader.net/cover/deadman-wonderland/deadman-wonderland-r0.jpg"
"http://s3.mangareader.net/cover/12-prince/12-prince-r1.jpg")
because it doesn't add an extra layer of seq.
Would something simple like this work for you?
(defn extract-url [s]
(subs s (inc (.indexOf s "'")) (.lastIndexOf s "'")))
This function will return a string containing all the characters between the first and last single quotes.
Assuming your sequence of strings is named ss, then:
(map extract-url ss)
;=> ("http://s3.mangareader.net/cover/gantz/gantz-r0.jpg"
; "http://s3.mangareader.net/cover/deadman-wonderland/deadman-wonderland-r0.jpg"
; "http://s3.mangareader.net/cover/12-prince/12-prince-r1.jpg")
This is definitely not a generic solution, but it fits the input you have provided.
In Emacs I would like to write some regexp that does the following:
First, return a list of all dictionary words that can be formed within "hex space". By this I mean:
#000000 - #ffffff
so #00baba would be a word (that can be looked up in the dictionary)
so would #baba00
and #abba00
and #0faded
...where trailing and leading 0's are considered irrelevant. How would I write this? Is my question clear enough?
Second, I would like to generate a list of words that can be made using numbers as letters:
0 = o
1 = i
3 = e
4 = a
...and so on. How would I write this?
First, load your dictionary. I'll assume that you're using /var/share/dict/words, which is nearly always installed by default when you're running Linux. It lists one word per line, which is a very handy format for this sort of thing.
Next run M-x keep-lines. It'll ask you for a regular expression and then delete any line that doesn't match it. Use the regex ^[a-f]\{,6\}$ and it will filter out anything that can't be part of a color.
Specifically, the ^ makes the regex start at the beginning of the line, the [a-f] matches any one character that is between a and f (inclusive), the {,6} lets it match between 0 and 6 instances of the previous item (in this case the character class [a-f] and finally the $ tells it that the next thing must be the end of the line.
This will return a list of all instances of #000000 - #ffffff in the buffer, although this pattern may not be restrictive enough for your purposes.
(let ((my-colour-list nil))
(save-excursion
(goto-char (point-min))
(while (re-search-forward "#[0-9a-fA-F]\\{6\\}" nil t)
(add-to-list 'my-colour-list (match-string-no-properties 0)))
my-colour-list))
I'm not actually certain that this is what you were asking for. What do you mean by "dictionary"?
A form that will return you a hash table with all the elements you specify in it could be this:
(let ((hash-table (make-hash-table :test 'equal)))
(dotimes (i (exp 256 3))
(puthash (concat "#" (format "%06x" i)) t hash-table))
hash-table)
I'm not sure how Emacs will manage that size of elements (16 million). As you don't want the 0, you can generate the space without that format, and removing trailing 0's. I don't know what do you want to do with the rest of the numbers. You can write the function step by step like this then:
(defun find-hex-space ()
(let (return-list)
(dotimes (i (exp 256 3))
(let* ((hex-number (strip-zeros (format "%x" i)))
(found-word (gethash hex-number *dictionary*)))
(if found-word (push found-word return-list))))
return-list))
Function strip-zeros is easy to write, and here I suppose your words are in a hash called *dictionary*. strip-zeros could be something like this:
(defun strip-zeros (string)
(let ((sm (string-match "^0*\\(.*?\\)0*$" string)))
(if sm (match-string 1 string) string)))
I don't quite understand your second question. The words would be also using the hex space? Would you then consider only the words formed by numbers, or would also include the letters in the word?