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Is there more Groovish way of adding an element to map of lists and initialize default list if not exists?
Or in other words what would be a Groovish way to code the below:
def mylist = [1,2,3,4]
def mymap = [:]
for (num in mylist){
if (num % 2 == 0){
pairity = "even"
} else {
pairity = "odd"
}
if (mymap.containsKey(pairity)){
println("Adding to Even")
mymap[pairity].add(num)
}
else {
println("adding to Odd")
mymap[pairity] = [num]
}
}
print(mymap.toString())
// adding to Odd
// adding to Odd
// Adding to Even
// Adding to Even
// [odd:[1, 3], even:[2, 4]]
You can use withDefault on a map to have automatically generate a value for a missing key on access.
[1,2,3,4].inject([:].withDefault{[]}){ m, i -> m[ i%2==0 ? 'odd' : 'even' ] << i; m }
// => [even:[1, 3], odd:[2, 4]]
You can simply groupby:
def mymap = mylist.groupBy { it % 2 == 0 ? 'even' : 'odd' }
That is effectively using the closure to partition the list on the condition.
The question is
int i = 6, x = 11;
switch(i % 3 ? 0 : 1)
{
case 0: x /=2; break;
case 1: x +=3;
case 2: x *=4;
}
cout << "x = " << x;
So the answer i should be getting is x=5, however, from the answer script it said it should be x= 56.
Why is that?
Both statements:
x +=3;
and
x *=4;
are executed. This is why (see here for conditional operator reference and here for the switch statement).
In the expression
i % 3 ? 0 : 1
i % 3 is the condition. Being an integer value, it's considered false if equal to zero, true otherwise. In your case i has value 6, thus i % 3 is 0, thus the condition is false and the value after the : is taken to be evaluated in the switch statement.
The value after the : is 1, so the code jumps to case 1 and executes x += 3. Now x is 11+3=14.
But there is no break statement before case 2, so the execution simply goes on with x *= 4 and x becomes 14*4=56.
Following is an example program in BASIC. Can someone tell me what this function returns if the marked condition is not true? I have to port the program to C++ and need to understand it. I have no BASIC knowledge - please bear with simple question.
FUNCTION CheckPoss (u)
tot = tot + 1
f = 0
SELECT CASE u
CASE 2
f = f + CheckIntersection(1, 3, 2, 1) 'A
CASE 3
f = f + CheckIntersection(2, 3, 3, 1) 'B
END SELECT
IF f = 0 THEN <============== This condition if true,
CheckPoss = 1 <============== then return value is 1
IF u = 9 THEN
PrintSolution
END IF
END IF
END FUNCTION
This is a good example of bad programming. First some unknown global variable is changed in this function. "tot = tot + 1"! Second line "F" is another unknown global variable is assigned "0". Or is this the only place this variable is used? In that case it is a variant implicitly declared here. Use a dim to declare it. It is legal in basic to do this. Globals should be passed as arguments to the function like this:
function CheckPoss(u as integer, tot as integer) as integer
dim f as integer
f=0
It is all about good practice so the input is clear and output is clear and all variable assignments should be through arguments passed to the function.
The return type is not declared either. Is this visual basic? or is it some older basic? Anyway the return type is a variant in case of visual basic. Older basic would be an integer type.
The output from this function will mostly likely be a zero if the condition is not met! That should also be clear in the code and it is not clear as it is, and I understand why you ask. I am amazed this piece of code comes from a working program.
Good luck with your project!
I don't know exactly that this function do.
On VB.net, the function follow the structure:
Public function CheckPoss(Byval u as integer)
... ' Just commands
return u ' Or other variable
end function
If not exist the 'return' command, the return of function will be 'null' character.
On C, The function will be:
int CheckPoss(int u){
tot++; // Increment tot variable (need be declared)
int f = 0;
switch(u){
case 2:
f += CheckIntersection(1, 3, 2, 1); // A
break;
case 3:
f += CheckIntersection(2, 3, 3, 1); // B
break;
}
if (f == 0){
if (u == 9){
PrintSolution();
}
return 1;
}
}
The return command need be the last command of this function. At case f != 0, the function must return trash (some value or character).
My suggestion is:
int CheckPoss(int u){
tot++; // I think that this must count how times you call this function
int f;
if(u == 2){
f = CheckIntersection(1, 3, 2, 1); // A
}else if(u == 3){
f = CheckIntersection(2, 3, 3, 1); // B
}else{
f = 1; // Case else
}
if (f == 0){
if (u == 9)
PrintSolution();
return 1;
}
}
I have the following code, which works, but I'm wondering if there is a "groovier" way of doing this:
/**
* 10 digit - #-######-##-#
* 13 digit - ###-#-######-##-#
* */
private formatISBN(String isbn) {
if (isbn?.length() == 10) {
def part1 = isbn.substring(0, 1)
def part2 = isbn.substring(1, 7)
def part3 = isbn.substring(7, 9)
def part4 = isbn.substring(9, 10)
return "${part1}-${part2}-${part3}-${part4}"
} else if (isbn?.length() == 13) {
def part1 = isbn.substring(0, 3)
def part2 = isbn.substring(3, 4)
def part3 = isbn.substring(4, 10)
def part4 = isbn.substring(10, 12)
def part5 = isbn.substring(12, 13)
return "${part1}-${part2}-${part3}-${part4}-${part5}"
} else {
return isbn
}
}
You could first use the [] string operator to get the substrings instead of substring and drop the intermediate variables. For example in the case for length == 10:
"${isbn[0]}-${isbn[1..6]}-${isbn[7..8]}-${isbn[9]}"
Now, there is a bit of repetition there. You can get instead first get all the isbn segments and then .join them with '-':
[isbn[0], isbn[1..6], isbn[7..8], isbn[9]].join('-')
And, even further, instead of referencing isbn every time, you can make a list of the ranges you want to get and then get them all the same time using collect:
[0, 1..6, 7..8, 9].collect { isbn[it] }.join('-')
If you're going for code golfing, you can also do:
('-'+isbn)[1, 0, 2..7, 0, 8..9, 0, 10]
I'll leave it to you to figure out how that works, but i guess it's probably not a good idea to leave that on production code, unless you want to surprise future maintainers hehe.
Also, notice that the format when length == 13 is the same as for length == 10 but with a different prefix, you can then reuse the same function in that case. The whole function (with a couple of tests) would be:
/**
* 10 digit - #-######-##-#
* 13 digit - ###-#-######-##-#
**/
def formatIsbn(isbn) {
switch (isbn?.length()) {
case 10: return [0, 1..6, 7..8, 9].collect { isbn[it] }.join('-')
case 13: return isbn.take(3) + '-' + formatIsbn(isbn.drop(3))
default: return isbn
}
}
assert formatIsbn('abcdefghij') == 'a-bcdefg-hi-j'
assert formatIsbn('abcdefghijklm') == 'abc-d-efghij-kl-m'
Now, i think there are some bad smells in that code. Can isbn be null? At least to me, this doesn't look like a function that needs to bother about the nullity of its argument, or at least that's not clear by reading its name (it should be called something like formatIsbnOrNull instead if both ISBN strings and null values are accepted). If null values are not valid, then let it blow up with a NullPointerException when accessing isbn.length() so the caller know they have passed a wrong argument, instead of silently returning the same null.
The same goes for the return ISBN at the end. Is it expected for that function to receive a string that's neither 10 nor 13 characters long? If not, better throw new IllegalArgumentException() and let the caller know they have called it wrongly.
Finally, i'm not sure if this is the most "readable" solution. Another possible solution is having a string for the format, like '###-#-######-##-#' and then replace the #s by the isbn characters. I think it might be more self-documenting:
def formatIsbn(isbn) {
def format = [
10: '#-######-##-#',
13: '###-#-######-##-#'
][isbn.length()]
def n = 0
format.replaceAll(/#/) { isbn[n++] }
}
Consider adding the method to the String class, as shown here. Note that this answer is a spin on a clever suggestion in epidemian's answer (re: collect).
Note:
This code augments String with asIsbn().
The range [0..2] does not need the call to asIsbn(), but the symmetry of using collect twice is irresistable.
Groovy returns the last expression in if/else, so 'return' is not necessary
/**
* 10 digit - #-######-##-#
* 13 digit - ###-#-######-##-#
**/
String.metaClass.asIsbn = { ->
if (delegate.length() == 10) {
[0, 1..6, 7..8, 9].collect { delegate[it] }.join('-')
} else if (delegate.length() == 13) {
[0..2, 3..12].collect { delegate[it].asIsbn() }.join('-')
} else {
delegate
}
}
assert "abcdefghij".asIsbn() == 'a-bcdefg-hi-j'
assert "abcdefghijklm".asIsbn() == 'abc-d-efghij-kl-m'
assert "def".asIsbn() == "def"
String s = null
assert s?.asIsbn() == null
I would try using Regex... I think it's pretty much readable if you know how to use regex, and it's javascript inspired syntax in groovy is pretty cool also.
One more thing: it's pretty clear, looking at the capture groups, what your string looks like for the desired formatting.
private formatISBN(String isbn) {
if (isbn?.length() == 10) {
m = isbn =~ /(\d{1})(\d{6})(\d{2})(\d{1})/
return "${m[0][1]}-${m[0][2]}-${m[0][3]}-${m[0][4]}"
} else if (isbn?.length() == 13) {
m = isbn =~ /(\d{3})(\d{1})(\d{6})(\d{2})(\d{1})/
return "${m[0][1]}-${m[0][2]}-${m[0][3]}-${m[0][4]}-${m[0][5]}"
} else {
return isbn
}
}
Btw, #epidemian suggestion using backreferences is great! I think the code would look like:
private formatISBN(String isbn) {
if (isbn?.length() == 10) {
return isbn.replaceAll(/(\d{1})(\d{6})(\d{2})(\d{1})/, '$1-$2-$3-$4')
} else if (isbn?.length() == 13) {
return isbn.replaceAll(/(\d{3})(\d{1})(\d{6})(\d{2})(\d{1})/, '$1-$2-$3-$4-$5')
} else {
return isbn
}
}
Dunno if I like this any better. I'd make the position map a static final, too.
private isbnify(String isbn) {
def dashesAt = [ 10: [[0,1], [1,7], [7,9], [9,10]],
13: [[0,3], [3,4], [4,10], [10,12], [12,13]]]
def dashes = dashesAt[isbn?.length()]
(dashes == null) ? isbn
: dashes.collect { isbn.substring(*it) }.join('-')
}
Ranges make for a bit less clutter, IMO:
private isbnify3(String isbn) {
def dashesAt = [ 10: [0, 1..6, 7..8, 9],
13: [0..2, 3, 4..9, 10..11, 12]]
def dashes = dashesAt[isbn?.length()]
dashes == null ? isbn : dashes.collect { isbn[it] }.join("-")
}
With an inject-with-two-accumulators it should be easy to do a list-of-dash-positions version, too.
This should be a comment to #everton, but I don't have the 50 reputation needed to do that yet. So this answer is really just a suggested variation on #everton's answer.
One less regex by making the first 3 digits optional. The downside is having to remove a leading '-' if the ISBN is 10 characters. (I also prefer \d over \d{1}.)
private formatISBN(String isbn) {
String result = isbn.replaceAll(/^(\d{3})?(\d)(\d{6})(\d{2})(\d)$/,
'$1-$2-$3-$4-$5')
if (result) {
return result.startsWith('-') ? result[1..-1] : result
} else {
return isbn // return value unchanged, pattern didn't match
}
}
println formatISBN('1234567890')
println formatISBN('9991234567890')
println formatISBN('123456789') // test an ISBN that's too short
println formatISBN('12345678901234') // test an ISBN that's too long
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