can someone please help me understand what's going on here
lists:dropwhile(fun(X) -> X < 8 end, lists:seq(1,10)).
"\b\t\n" % ??? what is this ? why not [8,9,10]
lists:dropwhile(fun(X) -> X < 7 end, lists:seq(1,10)).
[7,8,9,10] % this is correct
Your results are actually correct in both cases. The unexpected string in the first case
is due to the fact that in Erlang strings are just lists of integers. Therefore, Erlang chooses to interpret your first list as a string, since it contains only printable ASCII codes. In the second case the list contains the code 7, which is not printable, so Erlang is forced to interpret it as an integer list.
You can always print the actual integer list by using
MyList = lists:dropwhile(fun(X) -> X < 8 end, lists:seq(1,10)),
io:format("~w", [MyList]).
Related
This code is an excerpt from this book.
count_characters(Str) ->
count_characters(Str, #{}).
count_characters([H|T], #{ H => N }=X) ->
count_characters(T, X#{ H := N+1 });
count_characters([H|T], X) ->
count_characters(T, X#{ H => 1 });
count_characters([], X) ->
X.
So,
1> count_characters("hello").
#{101=>1,104=>1,108=>2,111=>1}
What I understand from this is that, count_characters() takes an argument hello, and place it to the first function, i.e count_characters(Str).
What I don't understand is, how the string characters are converted into ascii value without using $, and got incremented. I am very new to erlang, and would really appreciate if you could help me understand the above. Thank you.
In erlang the string literal "hello" is just a more convenient way of writing the list [104,101,108,108,111]. The string format is syntactic sugar and nothing erlang knows about internally. An ascii string is internally string is internally stored as a list of 32-bit integers.
This also becomes confusing when printing lists where the values happen to be within the ascii range:
io:format("~p~n", [[65,66]]).
will print
"AB"
even if you didn't expect a string as a result.
As said previously, there is no string data type in Erlang, it uses the internal representation of an integer list, so
"hello" == [$h,$e,$l,$l,$o] == [104|[101|[108|[108|[111|[]]]]]]
Which are each a valid representation of an integer list.
To make the count of characters, the function use a new Erlang data type: a map. (available only since R17)
A map is a collection of key/value pairs, in your case the keys will be the characters, and the values the occurrence of each characters.
The function is called with an empty map:count_characters(Str, #{}).
Then it goes recursively through the list, and for each head H, 2 cases are posible:
The character H was already found, then the current map X will match with the pattern #{ H => N } telling us that we already found N times H, so we continue the recursion with the rest of the list and a new map where the value associated to H is now N+1: count_characters(T, X#{ H := N+1 }.
The character H is found for the first time, then we continue the recursion with the rest of the list and a new map where the key/value pair H/1 is added: count_characters(T, X#{ H => 1 }).
When the end of the list is reached, simply return the map: count_characters([], X) -> X.
Easy question here, probably, but searching did not find a similar question.
The # operator finds the length of a string, among other things, great. But with Lua being dynamically typed, thus no conversion operators, how does one type a number as a string in order to determine its length?
For example suppose I want to print the factorials from 1 to 9 in a formatted table.
i,F = 1,1
while i<10 do
print(i.."! == "..string.rep("0",10-#F)..F)
i=i+1
F=F*i
end
error: attempt to get length of global 'F' (a number value)
why not use tostring(F) to convert F to a string?
Alternatively,
length = math.floor(math.log10(number)+1)
Careful though, this will only work where n > 0!
There are probably a dozen ways to do this. The easy way is to use tostring as Dan mentions. You could also concatenate an empty string, e.g. F_str=""..F to get F_str as a string representation. But since you are trying to output a formatted string, use the string.format method to do all the hard work for you:
i,F = 1,1
while i<10 do
print(string.format("%01d! == %010d", i, F))
i=i+1
F=F*i
end
Isn't while tostring(F).len < 10 do useful?
Extremely just-started-yesterday new to F#.
What I want: To write code that parses the string "2 + 2" into (using as an example code from the tutorial project) Expr.Add(Expr.Num 2, Expr.Num 2) for evaluation. Some help to at least point me in the right direction or tell me it's too complex for my first F# project. (This is how I learn things: By bashing my head against stuff that's hard)
What I have: My best guess at code to extract the numbers. Probably horribly off base. Also, a lack of clue.
let script = "2 + 2";
let rec scriptParse xs =
match xs with
| [] -> (double)0
| y::ys -> (double)y
let split = (script.Split([|' '|]))
let f x = (split[x]) // "This code is not a function and cannot be applied."
let list = [ for x in 0..script.Length -> f x ]
let result = scriptParse
Thanks.
The immediate issue that you're running into is that split is an array of strings. To access an element of this array, the syntax is split.[x], not split[x] (which would apply split to the singleton list [x], assuming it were a function).
Here are a few other issues:
Your definition of list is probably wrong: x ranges up to the length of script, not the length of the array split. If you want to convert an array or other sequence to a list you can just use List.ofSeq or Seq.toList instead of an explicit list comprehension [...].
Your "casts" to double are a bit odd - that's not the right syntax for performing conversions in F#, although it will work in this case. double is a function, so the parentheses are unnecessary and what you are doing is really calling double 0 and double y. You should just use 0.0 for the first case, and in the second case, it's unclear what you are converting from.
In general, it would probably be better to do a bit more design up front to decide what your overall strategy will be, since it's not clear to me that you'll be able to piece together a working parser based on your current approach. There are several well known techniques for writing a parser - are you trying to use a particular approach?
I have a binary M such that 34= will always be present and the rest may vary between any number of digits but will always be an integer.
M = [<<"34=21">>]
When I run this command I get an answer like
hd([X || <<"34=", X/binary >> <- M])
Answer -> <<"21">>
How can I get this to be an integer with the most care taken to make it as efficient as possible?
[<<"34=",X/binary>>] = M,
list_to_integer(binary_to_list(X)).
That yields the integer 21
As of R16B, the BIF binary_to_integer/1 can be used:
OTP-10300
Added four new bifs, erlang:binary_to_integer/1,2,
erlang:integer_to_binary/1, erlang:binary_to_float/1 and
erlang:float_to_binary/1,2. These bifs work similarly to how
their list counterparts work, except they operate on
binaries. In most cases converting from and to binaries is
faster than converting from and to lists.
These bifs are auto-imported into erlang source files and can
therefore be used without the erlang prefix.
So that would look like:
[<<"34=",X/binary>>] = M,
binary_to_integer(X).
A string representation of a number can be converted by N-48. For multi-digit numbers you can fold over the binary, multiplying by the power of the position of the digit:
-spec to_int(binary()) -> integer().
to_int(Bin) when is_binary(Bin) ->
to_int(Bin, {size(Bin), 0}).
to_int(_, {0, Acc}) ->
erlang:trunc(Acc);
to_int(<<N/integer, Tail/binary>>, {Pos, Acc}) when N >= 48, N =< 57 ->
to_int(Tail, {Pos-1, Acc + ((N-48) * math:pow(10, Pos-1))}).
The performance of this is around 100 times slower than using the list_to_integer(binary_to_list(X)) option.
I need a base converter function for Lua. I need to convert from base 10 to base 2,3,4,5,6,7,8,9,10,11...36 how can i to this?
In the string to number direction, the function tonumber() takes an optional second argument that specifies the base to use, which may range from 2 to 36 with the obvious meaning for digits in bases greater than 10.
In the number to string direction, this can be done slightly more efficiently than Nikolaus's answer by something like this:
local floor,insert = math.floor, table.insert
function basen(n,b)
n = floor(n)
if not b or b == 10 then return tostring(n) end
local digits = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
local t = {}
local sign = ""
if n < 0 then
sign = "-"
n = -n
end
repeat
local d = (n % b) + 1
n = floor(n / b)
insert(t, 1, digits:sub(d,d))
until n == 0
return sign .. table.concat(t,"")
end
This creates fewer garbage strings to collect by using table.concat() instead of repeated calls to the string concatenation operator ... Although it makes little practical difference for strings this small, this idiom should be learned because otherwise building a buffer in a loop with the concatenation operator will actually tend to O(n2) performance while table.concat() has been designed to do substantially better.
There is an unanswered question as to whether it is more efficient to push the digits on a stack in the table t with calls to table.insert(t,1,digit), or to append them to the end with t[#t+1]=digit, followed by a call to string.reverse() to put the digits in the right order. I'll leave the benchmarking to the student. Note that although the code I pasted here does run and appears to get correct answers, there may other opportunities to tune it further.
For example, the common case of base 10 is culled off and handled with the built in tostring() function. But similar culls can be done for bases 8 and 16 which have conversion specifiers for string.format() ("%o" and "%x", respectively).
Also, neither Nikolaus's solution nor mine handle non-integers particularly well. I emphasize that here by forcing the value n to an integer with math.floor() at the beginning.
Correctly converting a general floating point value to any base (even base 10) is fraught with subtleties, which I leave as an exercise to the reader.
you can use a loop to convert an integer into a string containting the required base. for bases below 10 use the following code, if you need a base larger than that you need to add a line that mapps the result of x % base to a character (usign an array for example)
x = 1234
r = ""
base = 8
while x > 0 do
r = "" .. (x % base ) .. r
x = math.floor(x / base)
end
print( r );