I'm using Math.Net (http://numerics.mathdotnet.com/) to work with matrices.
I need a method that returns a matrix as a string.
So if my matrix looks like this:
{{1.0, 2}
{3 , 4}}
I need my return string to equal
"1 2 3 4"
Here is my code
var M = Matrix<double>.Build;
var mMatrix = M.DenseOfArray(new[,] {{ 1.0, 2 },
{ 3 , 4}});
StringBuilder builder = new StringBuilder();
foreach (var m in unitMatrix.Enumerate())
{
builder.Append(m + " ");
}
return builder.ToString();
This returns "1 3 2 4"
How do I make it return "1 2 3 4"?
You can enumerate row by row using mMatrix.EnumerateRows(), and then enumerate through all the values of each row. For example, you could write this as:
String.Join(" ", mMatrix.EnumerateRows().SelectMany(x => x.Enumerate()))
or if it is ok to build up an intermediate array:
String.Join(" ", mMatrix.ToRowWiseArray())
Alternatively you could use the existing string formatting functions, even though they are a bit weird to use, e.g.:
mMatrix.ToMatrixString(int.MaxValue,0,int.MaxValue,0,"","",""," "," ", x => x.ToString())
Related
I have a string that contains different ranges and I need to find their value
var str = "some text x = 1..14, y = 2..4 some text"
I used the substringBefore() and substringAfter() methodes to get the x and y but I can't find a way to get the values because the numbers could be one or two digits or even negative numbers.
One approach is to use a regex, e.g.:
val str = "some text x = 1..14, y = 2..4 some text"
val match = Regex("x = (-?\\d+[.][.]-?\\d+).* y = (-?\\d+[.][.]-?\\d+)")
.find(str)
if (match != null)
println("x=${match.groupValues[1]}, y=${match.groupValues[2]}")
// prints: x=1..14, y=2..4
\\d matches a single digit, so \\d+ matches one or more digits; -? matches an optional minus sign; [.] matches a dot; and (…) marks a group that you can then retrieve from the groupValues property. (groupValues[0] is the whole match, so the individual values start from index 1.)
You could easily add extra parens to pull out each number separately, instead of whole ranges.
(You may or may not find this as readable or maintainable as string-manipulation approaches…)
Is this solution fit for you?
val str = "some text x = 1..14, y = 2..4 some text"
val result = str.replace(",", "").split(" ")
var x = ""; var y = ""
for (i in 0..result.count()-1) {
if (result[i] == "x") {
x = result[i+2]
} else if (result[i] == "y") {
y = result[i+2]
}
}
println(x)
println(y)
Using KotlinSpirit library
val rangeParser = object : Grammar<IntRange>() {
private var first: Int = -1
private var last: Int = -1
override val result: IntRange
get() = first..last
override fun defineRule(): Rule<*> {
return int {
first = it
} + ".." + int {
last = it
}
}
}.toRule().compile()
val str = "some text x = 1..14, y = 2..4 some text"
val ranges = rangeParser.findAll(str)
https://github.com/tiksem/KotlinSpirit
I have a list of strings that contain names and a series of numbers, ex: "John James Hollywood 1 2 3".
I want to parse out the names using a loop so that I can scan each part of the text and store it in a new string variable. Then I tried to use a conditional to put any numbers in a different variable. I will calculate the average of the numbers and then print the name and the average of the numbers, so the output looks like this:
John James Hollywood: 2
I don't know whether to use the input string stream or what my first steps are.
public static String parseNameNumbers(String text)
{
Scanner scnr = new Scanner(text);
Scanner inSS = null;
int numSum = 0;
int countNum = 0;
String newText = new String();
String sep = "";
while(scnr.hasNext()) {
if(scnr.hasNextInt()) {
numSum = numSum + scnr.nextInt();
countNum++;
}
else {
newText = newText + sep + scnr.next();
sep = " ";
}
}
return newText + ": " + (numSum/countNum);
}
I'm trying to write an algorithm that will generate all strings of length nm, with exactly n of each number 1, 2, ... m,
For instance all strings of length 6, with exactly two 1's, two 2's and two 3's e.g. 112233, 121233,
I managed to do this with just 1's and 2's using a recursive method, but can't seem to get something that works when I introduce 3's.
When m = 2, the algorithm I have is:
generateAllStrings(int len, int K, String str)
{
if(len == 0)
{
output(str);
}
if(K > 0)
{
generateAllStrings(len - 1, K - 1, str + '2');
}
if(len > K)
{
generateAllStrings(len - 1, K, str + '1');
}
}
I've tried inserting similar conditions for the third number but the algorithm doesn't give a correct output. After that I wouldn't even know how to generalise for 4 numbers and above.
Is recursion the right thing to do? Any help would be appreciated.
One option would be to list off all distinct permutations of the string 111...1222...2...nnn....n. There are nice algorithms for enumerating all distinct permutations of a string in time proportional to the length of the string, and they'd probably be a good way to go about solving this problem.
To use a simple recursive algorithm, give each recursion the permutation so far (variable perm), and the number of occurances of each digit that is still available (array count).
Run the code snippet to generate all unique permutations for n=2 and m=4 (set: 11223344).
function permutations(n, m) {
var perm = "", count = []; // start with empty permutation
for (var i = 0; i < m; i++) count[i] = n; // set available number for each digit = n
permute(perm, count); // start recursion with "" and [n,n,n...]
function permute(perm, count) {
var done = true;
for (var i = 0; i < count.length; i++) { // iterate over all digits
if (count[i] > 0) { // more instances of digit i available
var c = count.slice(); // create hard copy of count array
--c[i]; // decrement count of digit i
permute(perm + (i + 1), c); // add digit to permutation and recurse
done = false; // digits left over: not the last step
}
}
if (done) document.write(perm + "<BR>"); // no digits left: complete permutation
}
}
permutations(2, 4);
You can easily do this using DFS (or BFS alternatively). We can define an graph such that each node contains one string and a node is connected to any node that holds a string with a pair of int swaped in comparison to the original string. This graph is connected, thus we can easily generate a set of all nodes; which will contain all strings that are searched:
set generated_strings
list nodes
nodes.add(generateInitialString(N , M))
generated_strings.add(generateInitialString(N , M))
while(!nodes.empty())
string tmp = nodes.remove(0)
for (int i in [0 , N * M))
for (int j in distinct([0 , N * M) , i))
string new = swap(tmp , i , j)
if (!generated_strings.contains(new))
nodes.add(new)
generated_strings.add(new)
//generated_strings now contains all strings that can possibly be generated.
Hey guys I have string "69 - 13" How to detect "-" in the string and how to sum the numbers in the string 69+13=82 ?
There are various method to do that (componentsSeparatedByString, NSScanner, ...).
Here is one using only Swift library functions:
let str = "69 - 13"
// split string into components:
let comps = split(str, { $0 == "-" || $0 == " " }, maxSplit: Int.max, allowEmptySlices: false)
// convert strings to numbers (use zero if the conversion fails):
let nums = map(comps) { $0.toInt() ?? 0 }
// compute the sum:
let sum = reduce(nums, 0) { $0 + $1 }
println(sum)
Here is an updated implementation in Swift 4 that relies on higher order functions to perform the operation:
let string = "69+13"
let number = string.components(separatedBy: CharacterSet.decimalDigits.inverted)
.compactMap({ Int($0) })
.reduce(0, +)
print(number) // 82
The components(separatedBy: CharacterSet.decimalDigits.inverted) removes all non-digit values and creates an array for each group of values (in this case 69 and 13)
Int($0) converts your string value into an Int
compactMap gets rid of any nil values, ensuring that only valid values are left
reduce then sums up the values that remain in your array
The function f in the following code simply attempts to print out it's arguments and how many it receives. However, it expands array parameters (but not arraylists) as illustrated on the line f(x) // 3. Is there anyway to get f not to expand array parameters, or alternatively at the very least detect that it has happened, and perhaps correct for it. The reason for this is because my "real" f function isn't as trivial and instead passes it's parameters to a given function g, which often isn't a variable parameter function which instead expects an array directly as an argument, and the expansion by f mucks that up.
def f = {
Object... args ->
print "There are: ";
print args.size();
println " arguments and they are: ";
args.each { println it };
println "DONE"
}
def x = new int[2];
x[0] = 1;
x[1] = 2;
f(1,2); // 1
f([1,2]); // 2
f(x); // 3
I doubt there is any clean solution to this, as it behaves as Java varargs. You may test the size of the array inside the closure, or, as in Java, use a method overload:
public class Arraying {
public static void main(String[] args) {
// prints "2"
System.out.println( concat( new Object[] { "a", "b" } ) );
// prints "a". Commenting the second concat(), prints "1"
System.out.println( concat( "a" ) );
// prints "3"
System.out.println( concat( "a", "b", "c" ) );
}
static String concat(Object... args) {
return String.valueOf(args.length);
}
static String concat(Object obj) { return obj.toString(); }
}
If you comment the concat(Object obj) method, all three methods will match the concat(Object... args).
You can use a label for the argument as follow:
def f = {
Object... args ->
print "There are: ";
print args.size();
println " arguments and they are: ";
args.each { println it };
println "DONE"
}
def x = new int[2];
x[0] = 1;
x[1] = 2;
f(1,2); // 1
f([1,2]); // 2
f(a:x); // <--- used label 'a', or anything else
then the output is:
There are: 2 arguments and they are:
1
2
DONE
There are: 1 arguments and they are:
[1, 2]
DONE
There are: 1 arguments and they are:
[a:[1, 2]]
DONE