I am trying to connect redis via c# using using ServiceStack.Redis.
I have written below code to validate number based on the key specified.
argv[1] is key
argv[2] is number
string strScript = " local intCurrentVal = redis.call('GET', '' .. ARGV[1] .. ''); \n"
+ "if (tonumber(intCurrentVal) <= 0 ) then return 1 elseif ( (tonumber(intCurrentVal)) - (tonumber('' .. ARGV[2] .. '')) < 0 ) then return 0 end;"
+ "local intUpdatedVal = redis.call('SET', '' .. ARGV[1] .. '',( intCurrentVal - tonumber('' .. ARGV[2] .. '')));"
+ "local intCurr = redis.call('GET', '' .. ARGV[1] .. ''); return intCurr";
logical steps:
get the current value
check if current value should not be less then or equal to 0
check if current value - passed value should not be less then O
if current value - passed is not less then 0 then set the (current value - passed) as the current value
get the current value
Is it possible to optimize and tune the following lua script for performance. please help.
Original formatting is awful -- so is often performance.
local key = tostring(ARGV[1])
local number = tonumber(ARGV[2])
local current = tonumber(redis.call('GET', key))
if current <= 0 then
return 1
elseif current < number then
return 0
end
redis.call('SET', key, current - number)
return redis.call('GET', key)
Further optimization steps may include: localizing of global functions (like to string, tonumber, etc.), caching compiled chunk at LUA_REGISTRYINDEX table.
Related
Below is code to encode an integer value into an ASCII string. It is written in Python, and works fine from my testings.
def encode(value):
code = ''
while value%254 != value:
code = code + chr(value%254)
value = value/254
code = code + chr(value)
return code
def decode(code):
value = 0
length = len(code)
for i in range(0, length):
print code[i]
value = value * 254 + ord(code[length-1-i])
return value
code = encode(123456567)
print code
print decode(code)
However when I try the same implementation in Lua, the values encoded and decoded do not match up. Here is my Lua version:
function encode(value)
code = ''
while value%254 ~= value do
code = code .. string.char(value%254)
value = value/254
end
code = code .. string.char(value)
return code
end
function decode(code)
value = 0
code = string.reverse(code)
for i=1, #code do
local c = code:sub(i,i)
print(c)
value = value*254 + string.byte(c)
end
return value
end
code = encode(2555456)
print(decode(code))
Please note that I am trying to using mod 254 so that I can used 255 as a delimiter.
Use local whenever you are creating variables with similar names (for eg. code and value in your code).
When you use value = value / 254, you need to take only the integer part of the division and not the entire number.
Therefore:
function encode(value)
local code = ''
while value % 254 ~= value do
code = code .. string.char( value % 254 )
value = math.floor( value / 254 )
end
code = code .. string.char( value )
return code
end
function decode(code)
local value = 0
code = code:reverse()
for i = 1, #code do
local c = code:sub( i, i )
value = value * 254 + c:byte()
end
return value
end
Given a string s containing only lower case alphabets (a - z), find (i.e print) the characters that are repeated.
For ex, if string s = "aabcacdddec"
Output: a c d
3 approaches to this problem exists:
[brute force] Check every char of string (i.e s[i] with every other char and print if both are same)
Time complexity: O(n^2)
Space complexity: O(1)
[sort and then compare adjacent elements] After sorting (in O(n log(n) time), traverse the string and check if s[i] ans s[i + 1] are equal
Time complexity: O(n logn) + O(n) = O(n logn)
Space complexity: O(1)
[store the character count in an array] Create an array of size 26 (to keep track of a - z) and for every s[i], increment value stored at index = s[i] - 26 in the array. Finally traverse the array and print all elements (i.e 'a' + i) with value greater than 1
Time complexity: O(n)
Space complexity: O(1) but we have a separate array for storing the frequency of each element.
Is there a O(n) approach that DOES NOT use any array/hash table/map (etc)?
HINT: Use BIT Vectors
This is the element distinctness problem, so generally speaking - no there is no way to solve it in O(n) without extra space.
However, if you regard the alphabet as constant size (a-z characters only is pretty constant) you can either create a bitset of these characters, in O(1) space [ it is constant!] or check for each character in O(n) if it repeats more than once, it will be O(constant*n), which is still in O(n).
Pseudo code for 1st solution:
bit seen[] = new bit[SIZE_OF_ALPHABET] //contant!
bit printed[] = new bit[SIZE_OF_ALPHABET] //so is this!
for each i in seen.length: //init:
seen[i] = 0
printed[i] = 0
for each character c in string: //traverse the string:
i = intValue(c)
//already seen it and didn't print it? print it now!
if seen[i] == 1 and printed[i] == 0:
print c
printed[i] = 1
else:
seen[i] = 1
Pseudo code for 2nd solution:
for each character c from a-z: //constant number of repeats is O(1)
count = 0
for each character x in the string: //O(n)
if x==c:
count += 1
if count > 1
print count
Implementation in Java
public static void findDuplicate(String str) {
int checker = 0;
char c = 'a';
for (int i = 0; i < str.length(); ++i) {
int val = str.charAt(i) - c;
if ((checker & (1 << val)) > 0) {
System.out.println((char)(c+val));
}else{
checker |= (1 << val);
}
}
}
Uses as int as storage and performs bit wise operator to find the duplicates.
it is in O(n) .. explanation follows
Input as "abddc"
i==0
STEP #1 : val = 98 - 98 (0) str.charAt(0) is a and conversion char to int is 98 ( ascii of 'a')
STEP #2 : 1 << val equal to ( 1 << 0 ) equal to 1 finally 1 & 0 is 0
STEP #3 : checker = 0 | ( 1 << 0) equal to 0 | 1 equal to 1 checker is 1
i==1
STEP #1 : val = 99 - 98 (1) str.charAt(1) is b and conversion char to int is 99 ( ascii of 'b')
STEP #2 : 1 << val equal to ( 1 << 1 ) equal to 2 finally 1 & 2 is 0
STEP #3 : checker = 2 | ( 1 << 1) equal to 2 | 1 equal to 2 finally checker is 2
i==2
STEP #1 : val = 101 - 98 (3) str.charAt(2) is d and conversion char to int is 101 ( ascii of 'd')
STEP #2 : 1 << val equal to ( 1 << 3 ) equal to 8 finally 2 & 8 is 0
STEP #3 : checker = 2 | ( 1 << 3) equal to 2 | 8 equal to 8 checker is 8
i==3
STEP #1 : val = 101 - 98 (3) str.charAt(3) is d and conversion char to int is 101 ( ascii of 'd')
STEP #2 : 1 << val equal to ( 1 << 3 ) equal to 8 finally 8 & 8 is 8
Now print 'd' since the value > 0
You can also use the Bit Vector, depends upon the language it would space efficient. In java i would prefer to use int for this fixed ( just 26) constant case
The size of the character set is a constant, so you could scan the input 26 times. All you need is a counter to store the number of times you've seen the character corresponding to the current iteration. At the end of each iteration, print that character if your counter is greater than 1.
It's O(n) in runtime and O(1) in auxiliary space.
Implementation in C# (recursive solution)
static void getNonUniqueElements(string s, string nonUnique)
{
if (s.Count() > 0)
{
char ch = s[0];
s = s.Substring(1);
if (s.LastIndexOf(ch) > 0)
{
if (nonUnique.LastIndexOf(ch) < 0)
nonUnique += ch;
}
getNonUniqueElements(s, nonUnique);
}
else
{
Console.WriteLine(nonUnique);
return;
}
}
static void Main(string[] args)
{
getNonUniqueElements("aabcacdddec", "");
Console.ReadKey();
}
I want to convert string text to table and this text must be divided on characters. Every character must be in separate value of table, for example:
a="text"
--converting string (a) to table (b)
--show table (b)
b={'t','e','x','t'}
You could use string.gsub function
t={}
str="text"
str:gsub(".",function(c) table.insert(t,c) end)
Just index each symbol and put it at same position in table.
local str = "text"
local t = {}
for i = 1, #str do
t[i] = str:sub(i, i)
end
The builtin string library treats Lua strings as byte arrays.
An alternative that works on multibyte (Unicode) characters is the
unicode library that
originated in the Selene project.
Its main selling point is that it can be used as a drop-in replacement
for the string library, making most string operations “magically”
Unicode-capable.
If you prefer not to add third party dependencies your task can easily
be implemented using LPeg.
Here is an example splitter:
local lpeg = require "lpeg"
local C, Ct, R = lpeg.C, lpeg.Ct, lpeg.R
local lpegmatch = lpeg.match
local split_utf8 do
local utf8_x = R"\128\191"
local utf8_1 = R"\000\127"
local utf8_2 = R"\194\223" * utf8_x
local utf8_3 = R"\224\239" * utf8_x * utf8_x
local utf8_4 = R"\240\244" * utf8_x * utf8_x * utf8_x
local utf8 = utf8_1 + utf8_2 + utf8_3 + utf8_4
local split = Ct (C (utf8)^0) * -1
split_utf8 = function (str)
str = str and tostring (str)
if not str then return end
return lpegmatch (split, str)
end
end
This snippet defines the function split_utf8() that creates a table
of UTF8 characters (as Lua strings), but returns nil if the string
is not a valid UTF sequence.
You can run this test code:
tests = {
en = [[Lua (/ˈluːə/ LOO-ə, from Portuguese: lua [ˈlu.(w)ɐ] meaning moon; ]]
.. [[explicitly not "LUA"[1]) is a lightweight multi-paradigm programming ]]
.. [[language designed as a scripting language with "extensible ]]
.. [[semantics" as a primary goal.]],
ru = [[Lua ([лу́а], порт. «луна») — интерпретируемый язык программирования, ]]
.. [[разработанный подразделением Tecgraf Католического университета ]]
.. [[Рио-де-Жанейро.]],
gr = [[Η Lua είναι μια ελαφρή προστακτική γλώσσα προγραμματισμού, που ]]
.. [[σχεδιάστηκε σαν γλώσσα σεναρίων με κύριο σκοπό τη δυνατότητα ]]
.. [[επέκτασης της σημασιολογίας της.]],
XX = ">\255< invalid"
}
-------------------------------------------------------------------------------
local limit = 14
for lang, str in next, tests do
io.write "\n"
io.write (string.format ("<%s %3d> ->", lang, #str))
local chars = split_utf8 (str)
if not chars then
io.write " INVALID!"
else
io.write (string.format (" <%3d>", #chars))
for i = 1, #chars > limit and limit or #chars do
io.write (string.format (" %q", chars [i]))
end
end
end
io.write "\n"
Btw., building a table with LPeg is significantly faster than calling
table.insert() repeatedly.
Here are stats for splitting the whole of Gogol’s Dead Souls (in
Russian, 1023814 bytes raw, 571395 characters UTF) on my machine:
library method time in ms
string table.insert() 380
string t [#t + 1] = c 310
string gmatch & for loop 280
slnunicode table.insert() 220
slnunicode t [#t + 1] = c 200
slnunicode gmatch & for loop 170
lpeg Ct (C (...)) 70
You can below code to achieve this easily.
t = {}
str = "text"
for i=1, string.len(str) do
t[i]= (string.sub(str,i,i))
end
for k , v in pairs(t) do
print(k,v)
end
-- 1 t
-- 2 e
-- 3 x
-- 4 t
Using string.sub
string.sub(s, i [, j])
Return a substring of the string passed. The substring starts at i. If the third argument j is not given, the substring will end at the end of the string. If the third argument is given, the substring ends at and includes j.
I have a script looping for 8 numbers, skipping negative and return the largest as follows:
biggest = 0
entry = 0
loop, 8
{
; MsgBox %A_Index%
if NegativeReadings%A_Index% not contains - ;filter out readings that are negative
{
; MsgBox % AttributeReadings%A_Index%
MsgBox %biggest%
; MsgBox % AttributeReadings%A_Index%
if (AttributeReadings[A_Index] > biggest)
{
biggest := AttributeReadings[A_Index]
entry = %A_Index%
}
}
}
MsgBox %entry%
When I feed in some a sample image with 100,100,150,100,50,100,110,75, the OCR returns the object array result correctly but the numeric comparison fails
I'm getting MsgBox %biggest% = 0,100,100,150,150,50,50,50 => %entry% = 8
Something wrong happens in between (50 > 150) I have little clue in dealing with data types in ahk, any help is welcomed
Ok, I figured it out on second shot tonight, the OCR is returning a string with tailing spaces, so it led to alphabetic comparison as mentioned by admin.
Trimmed it with regexp now it works just fine
Here is the code snippet for anyone who may come across this similar issue
#SingleInstance force
#Include OCR.ahk
; sleep 5000
OCR()
; global
{
AttributeReadings := Object()
loop, 8
{
NegativeReadings%A_Index% := GetOCR(730, (136 + (17 * (A_Index - 1))), 35, 17)
AttributeReadings[A_Index] := GetOCR(730, (136 + (17 * (A_Index - 1))), 35, 17, "numeric")
; MsgBox % AttributeReadings%A_Index%
}
biggest = 0
entry = 0
i = 0
loop, 8
{
; MsgBox %A_Index%
if NegativeReadings%A_Index% not contains - ;filter out readings that are negative
{
; MsgBox % AttributeReadings%A_Index%
; length := StrLen(biggest)
; MsgBox %biggest%, %length%
number := RegExReplace(AttributeReadings[A_Index], "([space])?", "")
MsgBox %number%
; MsgBox % AttributeReadings%A_Index%
if (number > biggest)
{
biggest := number
entry := i
}
}
i++
}
MsgBox %entry%
}
End::
OCR()
return
The code above basically reads in a list of numbers from an image and return the first largest non-negative. As to the negative filtering part, it's done in OCR since it's ok with my test cases, you may want to modify it depending what images you're working with
Is there built-in functionality for this?
GNU Octave search a cell array of strings in linear time O(n):
(The 15 year old code in this answer was tested and correct on GNU Octave 3.8.2, 5.2.0 and 7.1.0)
The other answer has cellidx which was depreciated by octave, it still runs but they say to use ismember instead, like this:
%linear time string index search.
a = ["hello"; "unsorted"; "world"; "moobar"]
b = cellstr(a)
%b =
%{
% [1,1] = hello
% [2,1] = unsorted
% [3,1] = world
% [4,1] = moobar
%}
find(ismember(b, 'world')) %returns 3
ismember finds 'world' in index slot 3. This is a expensive linear time O(n) operation because it has to iterate through all elements whether or not it is found.
To achieve a logarathmic time O(log n) solution, then your list needs to come pre-sorted and then you can use binary search:
If your cell array is already sorted, you can do O(log-n) worst case:
function i = binsearch(array, val, low, high)
%binary search algorithm for numerics, Usage:
%myarray = [ 30, 40, 50.15 ]; %already sorted list
%binsearch(myarray, 30, 1, 3) %item 30 is in slot 1
if ( high < low )
i = 0;
else
mid = floor((low + high) / 2);
if ( array(mid) > val )
i = binsearch(array, val, low, mid-1);
elseif ( array(mid) < val )
i = binsearch(array, val, mid+1, high);
else
i = mid;
endif
endif
endfunction
function i = binsearch_str(array, val, low, high)
% binary search for strings, usage:
%myarray2 = [ "abc"; "def"; "ghi"]; #already sorted list
%binsearch_str(myarray2, "abc", 1, 3) #item abc is in slot 1
if ( high < low )
i = 0;
else
mid = floor((low + high) / 2);
if ( mystrcmp(array(mid, [1:end]), val) == 1 )
i = binsearch(array, val, low, mid-1);
elseif ( mystrcmp(array(mid, [1:end]), val) == -1 )
i = binsearch_str(array, val, mid+1, high);
else
i = mid;
endif
endif
endfunction
function ret = mystrcmp(a, b)
%this function is just an octave string compare, its behavior follows the
%strcmp(str1,str2)'s in C and java.lang.String.compareTo(...)'s in Java,
%that is:
% -returns 1 if string a > b
% -returns 0 if string a == b
% -return -1 if string a < b
% The gt() operator does not support cell array. If the single word
% is passed as an one-element cell array, converts it to a string.
a_as_string = a;
if iscellstr( a )
a_as_string = a{1}; %a was passed as a single-element cell array.
endif
% The gt() operator does not support cell array. If the single word
% is passed as an one-element cell array, converts it to a string.
b_as_string = b;
if iscellstr( b )
b_as_string = b{1}; %b was passed as a single-element cell array.
endif
% Space-pad the shortest word so as they can be used with gt() and lt() operators.
if length(a_as_string) > length( b_as_string )
b_as_string( length( b_as_string ) + 1 : length( a_as_string ) ) = " ";
elseif length(a_as_string) < length( b_as_string )
a_as_string( length( a_as_string ) + 1 : length( b_as_string ) ) = " ";
endif
letters_gt = gt(a_as_string, b_as_string); %list of boolean a > b
letters_lt = lt(a_as_string, b_as_string); %list of boolean a < b
ret = 0;
%octave makes us roll our own string compare because
%strings are arrays of numerics
len = length(letters_gt);
for i = 1:len
if letters_gt(i) > letters_lt(i)
ret = 1;
return
elseif letters_gt(i) < letters_lt(i)
ret = -1;
return
endif
end;
endfunction
%Assuming that myarray is already sorted, (it must be for binary
%search to finish in logarithmic time `O(log-n))` worst case, then do
myarray = [ 30, 40, 50.15 ]; %already sorted list
binsearch(myarray, 30, 1, 3) %item 30 is in slot 1
binsearch(myarray, 40, 1, 3) %item 40 is in slot 2
binsearch(myarray, 50, 1, 3) %50 does not exist so return 0
binsearch(myarray, 50.15, 1, 3) %50.15 is in slot 3
%same but for strings:
myarray2 = [ "abc"; "def"; "ghi"]; %already sorted list
binsearch_str(myarray2, "abc", 1, 3) %item abc is in slot 1
binsearch_str(myarray2, "def", 1, 3) %item def is in slot 2
binsearch_str(myarray2, "zzz", 1, 3) %zzz does not exist so return 0
binsearch_str(myarray2, "ghi", 1, 3) %item ghi is in slot 3
To sort your array if it isn't already:
Complexity of sorting depends on the kind of data you have and whatever sorting algorithm GNU octave language writers selected, it's somewhere between O(n*log(n)) and O(n*n).
myarray = [ 9, 40, -3, 3.14, 20 ]; %not sorted list
myarray = sort(myarray)
myarray2 = [ "the"; "cat"; "sat"; "on"; "the"; "mat"]; %not sorted list
myarray2 = sortrows(myarray2)
Code buffs to make this backward compatible with GNU Octave 3. 5. and 7. goes to #Paulo Carvalho in the other answer here.
Yes check this: http://www.obihiro.ac.jp/~suzukim/masuda/octave/html3/octave_36.html#SEC75
a = ["hello"; "world"];
c = cellstr (a)
⇒ c =
{
[1,1] = hello
[2,1] = world
}
>>> cellidx(c, 'hello')
ans = 1
>>> cellidx(c, 'world')
ans = 2
The cellidx solution does not meet the OP's efficiency requirement, and is deprecated (as noted by help cellidx).
Håvard Geithus in a comment suggested using the lookup() function on a sorted cell array of strings, which is significantly more efficient than cellidx. It's still a binary search though, whereas most modern languages (and even many 20 year old ones) give us easy access to associative arrays, which would be a much better approach.
While Octave doesn't obviously have associated arrays, that's effectively what the interpreter is using for ocatve's variables, including structs, so you can make us of that, as described here:
http://math-blog.com/2011/05/09/associative-arrays-and-cellular-automata-in-octave/
Built-in Function: struct ("field", value, "field", value,...)
Built-in Function: isstruct (expr)
Built-in Function: rmfield (s, f)
Function File: [k1,..., v1] = setfield (s, k1, v1,...)
Function File: [t, p] = orderfields (s1, s2)
Built-in Function: fieldnames (struct)
Built-in Function: isfield (expr, name)
Function File: [v1,...] = getfield (s, key,...)
Function File: substruct (type, subs,...)
Converting Matlab to Octave is there a containers.Map equivalent? suggests using javaObject("java.util.Hashtable"). That would come with some setup overhead, but would be a performance win if you're using it a lot. It may even be viable to link in some library written in C or C++? Do think about whether this is a maintainable option though.
Caveat: I'm relatively new to Octave, and writing this up as I research it myself (which is how I wound up here). I haven't yet run tests on the efficiency of these techniques, and while I've got a fair knowledge of the underlying algorithms, I may be making unreasonable assumptions about what's actually efficient in Octave.
This is a version of mystrcmp() that works in Octave of recent version (7.1.0):
function ret = mystrcmp(a, b)
%this function is just an octave string compare, its behavior follows the
%strcmp(str1,str2)'s in C and java.lang.String.compareTo(...)'s in Java,
%that is:
% -returns 1 if string a > b
% -returns 0 if string a == b
% -return -1 if string a < b
% The gt() operator does not support cell array. If the single word
% is passed as an one-element cell array, converts it to a string.
a_as_string = a;
if iscellstr( a )
a_as_string = a{1}; %a was passed as a single-element cell array.
endif
% The gt() operator does not support cell array. If the single word
% is passed as an one-element cell array, converts it to a string.
b_as_string = b;
if iscellstr( b )
b_as_string = b{1}; %b was passed as a single-element cell array.
endif
% Space-pad the shortest word so as they can be used with gt() and lt() operators.
if length(a_as_string) > length( b_as_string )
b_as_string( length( b_as_string ) + 1 : length( a_as_string ) ) = " ";
elseif length(a_as_string) < length( b_as_string )
a_as_string( length( a_as_string ) + 1 : length( b_as_string ) ) = " ";
endif
letters_gt = gt(a_as_string, b_as_string); %list of boolean a > b
letters_lt = lt(a_as_string, b_as_string); %list of boolean a < b
ret = 0;
%octave makes us roll our own string compare because
%strings are arrays of numerics
len = length(letters_gt);
for i = 1:len
if letters_gt(i) > letters_lt(i)
ret = 1;
return
elseif letters_gt(i) < letters_lt(i)
ret = -1;
return
endif
end;
endfunction