java apps. string manipulation - string

I want to limit the no. of character that can be put on JTextField because on my database I have this column that has Sex,Status (which the no. of char. allowed is 1 only).
and Middle Initial (which the no. of char. allowed is 2 only).
This what I have in my mind :
(for Sex,Status column)
String text = jTextField2.getText();
int count = text.();
if (count>1) {
(delete the next character that will be input)
}
(for M.I. column)
String text = jTextField1.getText();
int count = text.();
if (count>2) {
(delete the next character that will be input)
}
is this possible? is there a command that will delete the next character, so the no. of char. is acceptable for my database?

Sure. Just use String#substring.
String middleInitial = "JKL";
middleInitial.substring(0, 2);
System.out.println(middleInitial); // => JK
Similarly, you can use substring(0, 1) for sex.
It might be better if sex is an enum, though.
public enum Sex {
MALE("m"), FEMALE("f");
final String symbol;
private Sex(String symbol) {
this.symbol = symbol;
}
}
Now you can use it like this:
String sex = "male";
Sex.valueOf(sex.toUpperCase());
Or directly
Sex.MALE;
Instead of a text field for sex, you might use a JComboBox so the user can only choose one of the two options. This way you're sure to have valid input.

Related

Get String Between 2 Strings with Arduino

I am looking for a way to get a String between 2 Strings using Arduino. This is the source String:
Hello, my name is John Doe# and my favourite number is 32#.
The output has to be:
String name = "John Doe"; //Between "name is " and "#"
String favouriteNumber = "32"; //Between "number is " and "#"
How can this be achieved with Arduino?
I am not able to find any information online about this. Those examples for C are not working anyway. I understand that using String is not recommended in Arduino, but I have to do it this way to make things simpler.
By the way, this method of using a '#' to indicate the end of the data is not an ideal way to do it as I would like the input to be more human readable and more natural. Would anyone please suggest another way to do this as well?
Thanks in advance!
Function midString find the substring that is between two other strings "start" and "finish". If such a string does not exist, it returns "". A test code is included too.
void setup() {
test();
}
void loop() {
delay(100);
}
String midString(String str, String start, String finish){
int locStart = str.indexOf(start);
if (locStart==-1) return "";
locStart += start.length();
int locFinish = str.indexOf(finish, locStart);
if (locFinish==-1) return "";
return str.substring(locStart, locFinish);
}
void test(){
Serial.begin(115200);
String str = "Get a substring of a String. The starting index is inclusive (the corresponding character is included in the substring), but the optional ending index is exclusive";
Serial.print(">");
Serial.print( midString( str, "substring", "String" ) );
Serial.println("<");
Serial.print(">");
Serial.print( midString( str, "substring", "." ) );
Serial.println("<");
Serial.print(">");
Serial.print( midString( str, "corresponding", "inclusive" ) );
Serial.println("<");
Serial.print(">");
Serial.print( midString( str, "object", "inclusive" ) );
Serial.println("<");
}
just searched for this and saw no answer so i cooked one up.
i prefer working with String as well because of code readability and simplicity.
for me its more important than squeezing every last drop of juice out of my arduino.
String name = GetStringBetweenStrings("Hello, my name is John Doe# and my favourite number is 32#." ,"name is ","#");
String GetStringBetweenStrings(String input, String firstdel, String enddel){
int posfrom = input.indexOf(firstdel) + firstdel.length();
int posto = input.indexOf(enddel);
return input.substring(posfrom, posto);
}
watch out for the first case its fine, but for the second one you would have to change the second filter sting to "#." so it doesn't use the first occurrence of the #

Add comma sequentially to string in C#

I have a string.
string str = "TTFTTFFTTTTF";
How can I break this string and add character ","?
result should be- TTF,TTF,FTT,TTF
You could use String.Join after you've grouped by 3-chars:
var groups = str.Select((c, ix) => new { Char = c, Index = ix })
.GroupBy(x => x.Index / 3)
.Select(g => String.Concat(g.Select(x => x.Char)));
string result = string.Join(",", groups);
Since you're new to programming. That's a LINQ query so you need to add using System.Linq to the top of your code file.
The Select extension method creates an anonymous type containing the char and the index of each char.
GroupBy groups them by the result of index / 3 which is an integer division that truncates decimal places. That's why you create groups of three.
String.Concat creates a string from the 3 characters.
String.Join concatenates them and inserts a comma delimiter between each.
Here is a really simple solution using StringBuilder
var stringBuilder = new StringBuilder();
for (int i = 0; i < str.Length; i += 3)
{
stringBuilder.AppendFormat("{0},", str.Substring(i, 3));
}
stringBuilder.Length -= 1;
str = stringBuilder.ToString();
I'm not sure if the following is better.
stringBuilder.Append(str.Substring(i, 3)).Append(',');
I would suggest to avoid LINQ in this case as it will perform a lot more operations and this is a fairly simple task.
You can use insert
Insert places one string into another. This forms a new string in your C# program. We use the string Insert method to place one string in the middle of another one—or at any other position.
Tip 1:
We can insert one string at any index into another. IndexOf can return a suitable index.
Tip 2:
Insert can be used to concatenate strings. But this is less efficient—concat, as with + is faster.
for(int i=3;i<=str.Length - 1;i+=4)
{
str=str.Insert(i,",");
}

D: how to remove last char in string?

I need to remove last char in string in my case it's comma (","):
foreach(line; fcontent.splitLines)
{
string row = line.split.map!(a=>format("'%s', ", a)).join;
writeln(row.chop.chop);
}
I have found only one way - to call chop two times. First remove \r\n and second remove last char.
Is there any better ways?
import std.array;
if (!row.empty)
row.popBack();
As it usually happens with string processing, it depends on how much Unicode do you care about.
If you only work with ASCII it is very simple:
import std.encoding;
// no "nice" ASCII literals, D really encourages Unicode
auto str1 = cast(AsciiString) "abcde";
str1 = str1[0 .. $-1]; // get slice of everything but last byte
auto str2 = cast(AsciiString) "abcde\n\r";
str2 = str2[0 .. $-3]; // same principle
In "last char" actually means unicode code point (http://unicode.org/glossary/#code_point) it gets a bit more complicated. Easy way is to just rely on D automatic decoding and algorithms:
import std.range, std.stdio;
auto range = "кириллица".retro.drop(1).retro();
writeln(range);
Here retro (http://dlang.org/phobos/std_range.html#.retro) is a lazy reverse iteration function. It takes any range (unicode string is a valid range) and returns wrapper that is capable of iterating it backwards.
drop (http://dlang.org/phobos/std_range.html#.drop) simply pops a single range element and ignores it. Calling retro again will reverse the iteration order back to normal, but now with the last element dropped.
Reason why it is different from ASCII version is because of nature of Unicode (specifically UTF-8 which D defaults to) - it does not allow random access to any code point. You actually need to decode them all one by one to get to any desired index. Fortunately, D takes care of all decoding for you hiding it behind convenient range interface.
For those who want even more Unicode correctness, it should be possible to operate on graphemes (http://unicode.org/glossary/#grapheme):
import std.range, std.uni, std.stdio;
auto range = "abcde".byGrapheme.retro.drop(1).retro();
writeln(range);
Sadly, looks like this specific pattern is not curently supported because of bug in Phobos. I have created an issue about it : https://issues.dlang.org/show_bug.cgi?id=14394
NOTE: Updated my answer to be a bit cleaner and removed the lambda function in 'map!' as it was a little ugly.
import std.algorithm, std.stdio;
import std.string;
void main(){
string fcontent = "I am a test\nFile\nwith some,\nCommas here and\nthere,\n";
auto data = fcontent
.splitLines
.map!(a => a.replaceLast(","))
.join("\n");
writefln("%s", data);
}
auto replaceLast(string line, string toReplace){
auto o = line.lastIndexOf(toReplace);
return o >= 0 ? line[0..o] : line;
}
module main;
import std.stdio : writeln;
import std.string : lineSplitter, join;
import std.algorithm : map, splitter, each;
enum fcontent = "some text\r\nnext line\r\n";
void main()
{
fcontent.lineSplitter.map!(a=>a.splitter(' ')
.map!(b=>"'" ~ b ~ "'")
.join(", "))
.each!writeln;
}
Take a look, I use this extension method to replace any last character or sub-string, for example:
string testStr = "Happy holiday!";<br>
Console.Write(testStr.ReplaceVeryLast("holiday!", "Easter!"));
public static class StringExtensions
{
public static string ReplaceVeryLast(this string sStr, string sSearch, string sReplace = "")
{
int pos = 0;
sStr = sStr.Trim();
do
{
pos = sStr.LastIndexOf(sSearch, StringComparison.CurrentCultureIgnoreCase);
if (pos >= 0 && pos + sSearch.Length == sStr.Length)
sStr = sStr.Substring(0, pos) + sReplace;
} while (pos == (sStr.Length - sSearch.Length + 1));
return sStr;
}
}

Binary search of an ArrayList object method to retrieve a string will not identify if strings are identical?

The program in it's entirety sorts an ArrayList of Student objects by integers highest average, last name, and also has the option to perform a search. My program works flawlessly except for my binary search, for which I can absolutely not determine the cause of failure. I have printed all the information as it comes up.
Here is the student class with the method that references the Student's first and last name (String).
public String getFirstName (){
return firstname;
}
public String getLastName(){
return lastname;
}
In addition, here is the code for the binary search. Yes, I know Collections has a method for this exact purpose, but for my class I need to write up the search myself.
private static void searchStudent(ArrayList<Student> a){
Scanner reader = new Scanner(System.in);
System.out.print("Please enter search term: ");
String term = reader.next();
//System.out.println(term + " " + term.length());
System.out.println("---SEARCH RESULTS:---");
for (int i = 0; i < a.size(); i++){
String fName = (a.get(i).getFirstName());
String lName = (a.get(i).getLastName());
//System.out.println(fName + " " + fName.length());
//System.out.println(lName + " " + lName.length());
if (term == fName){
System.out.println(a.get(i));
} else if (term == lName){
System.out.println(a.get(i));
}
}
}
In Java, you need to use .equals() to compare strings. E.g. instead of this:
if (term == fName){
you need to do this:
if (term.equals(fName)){
Otherwise, you are comparing references only.
Btw, this is not a binary search, it's a linear search. You can see one implementation of binary search e.g. here:
http://leepoint.net/notes-java/algorithms/searching/binarysearch.html
though to compare strings you would use .compareTo / .compareToIgnoreCase methods on the String class instead of < / > operators.

Sorting a string using another sorting order string [closed]

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I saw this in an interview question ,
Given a sorting order string, you are asked to sort the input string based on the given sorting order string.
for example if the sorting order string is dfbcae
and the Input string is abcdeeabc
the output should be dbbccaaee.
any ideas on how to do this , in an efficient way ?
The Counting Sort option is pretty cool, and fast when the string to be sorted is long compared to the sort order string.
create an array where each index corresponds to a letter in the alphabet, this is the count array
for each letter in the sort target, increment the index in the count array which corresponds to that letter
for each letter in the sort order string
add that letter to the end of the output string a number of times equal to it's count in the count array
Algorithmic complexity is O(n) where n is the length of the string to be sorted. As the Wikipedia article explains we're able to beat the lower bound on standard comparison based sorting because this isn't a comparison based sort.
Here's some pseudocode.
char[26] countArray;
foreach(char c in sortTarget)
{
countArray[c - 'a']++;
}
int head = 0;
foreach(char c in sortOrder)
{
while(countArray[c - 'a'] > 0)
{
sortTarget[head] = c;
head++;
countArray[c - 'a']--;
}
}
Note: this implementation requires that both strings contain only lowercase characters.
Here's a nice easy to understand algorithm that has decent algorithmic complexity.
For each character in the sort order string
scan string to be sorted, starting at first non-ordered character (you can keep track of this character with an index or pointer)
when you find an occurrence of the specified character, swap it with the first non-ordered character
increment the index for the first non-ordered character
This is O(n*m), where n is the length of the string to be sorted and m is the length of the sort order string. We're able to beat the lower bound on comparison based sorting because this algorithm doesn't really use comparisons. Like Counting Sort it relies on the fact that you have a predefined finite external ordering set.
Here's some psuedocode:
int head = 0;
foreach(char c in sortOrder)
{
for(int i = head; i < sortTarget.length; i++)
{
if(sortTarget[i] == c)
{
// swap i with head
char temp = sortTarget[head];
sortTarget[head] = sortTarget[i];
sortTarget[i] = temp;
head++;
}
}
}
In Python, you can just create an index and use that in a comparison expression:
order = 'dfbcae'
input = 'abcdeeabc'
index = dict([ (y,x) for (x,y) in enumerate(order) ])
output = sorted(input, cmp=lambda x,y: index[x] - index[y])
print 'input=',''.join(input)
print 'output=',''.join(output)
gives this output:
input= abcdeeabc
output= dbbccaaee
Use binary search to find all the "split points" between different letters, then use the length of each segment directly. This will be asymptotically faster then naive counting sort, but will be harder to implement:
Use an array of size 26*2 to store the begin and end of each letter;
Inspect the middle element, see if it is different from the element left to it. If so, then this is the begin for the middle element and end for the element before it;
Throw away the segment with identical begin and end (if there are any), recursively apply this algorithm.
Since there are at most 25 "split"s, you won't have to do the search for more than 25 segemnts, and for each segment it is O(logn). Since this is constant * O(logn), the algorithm is O(nlogn).
And of course, just use counting sort will be easier to implement:
Use an array of size 26 to record the number of different letters;
Scan the input string;
Output the string in the given sorting order.
This is O(n), n being the length of the string.
Interview questions are generally about thought process and don't usually care too much about language features, but I couldn't resist posting a VB.Net 4.0 version anyway.
"Efficient" can mean two different things. The first is "what's the fastest way to make a computer execute a task" and the second is "what's the fastest that we can get a task done". They might sound the same but the first can mean micro-optimizations like int vs short, running timers to compare execution times and spending a week tweaking every millisecond out of an algorithm. The second definition is about how much human time would it take to create the code that does the task (hopefully in a reasonable amount of time). If code A runs 20 times faster than code B but code B took 1/20th of the time to write, depending on the granularity of the timer (1ms vs 20ms, 1 week vs 20 weeks), each version could be considered "efficient".
Dim input = "abcdeeabc"
Dim sort = "dfbcae"
Dim SortChars = sort.ToList()
Dim output = New String((From c In input.ToList() Select c Order By SortChars.IndexOf(c)).ToArray())
Trace.WriteLine(output)
Here is my solution to the question
import java.util.*;
import java.io.*;
class SortString
{
public static void main(String arg[])throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
// System.out.println("Enter 1st String :");
// System.out.println("Enter 1st String :");
// String s1=br.readLine();
// System.out.println("Enter 2nd String :");
// String s2=br.readLine();
String s1="tracctor";
String s2="car";
String com="";
String uncom="";
for(int i=0;i<s2.length();i++)
{
if(s1.contains(""+s2.charAt(i)))
{
com=com+s2.charAt(i);
}
}
System.out.println("Com :"+com);
for(int i=0;i<s1.length();i++)
if(!com.contains(""+s1.charAt(i)))
uncom=uncom+s1.charAt(i);
System.out.println("Uncom "+uncom);
System.out.println("Combined "+(com+uncom));
HashMap<String,Integer> h1=new HashMap<String,Integer>();
for(int i=0;i<s1.length();i++)
{
String m=""+s1.charAt(i);
if(h1.containsKey(m))
{
int val=(int)h1.get(m);
val=val+1;
h1.put(m,val);
}
else
{
h1.put(m,new Integer(1));
}
}
StringBuilder x=new StringBuilder();
for(int i=0;i<com.length();i++)
{
if(h1.containsKey(""+com.charAt(i)))
{
int count=(int)h1.get(""+com.charAt(i));
while(count!=0)
{x.append(""+com.charAt(i));count--;}
}
}
x.append(uncom);
System.out.println("Sort "+x);
}
}
Here is my version which is O(n) in time. Instead of unordered_map, I could have just used a char array of constant size. i.,e. char char_count[256] (and done ++char_count[ch - 'a'] ) assuming the input strings has all ASCII small characters.
string SortOrder(const string& input, const string& sort_order) {
unordered_map<char, int> char_count;
for (auto ch : input) {
++char_count[ch];
}
string res = "";
for (auto ch : sort_order) {
unordered_map<char, int>::iterator it = char_count.find(ch);
if (it != char_count.end()) {
string s(it->second, it->first);
res += s;
}
}
return res;
}
private static String sort(String target, String reference) {
final Map<Character, Integer> referencesMap = new HashMap<Character, Integer>();
for (int i = 0; i < reference.length(); i++) {
char key = reference.charAt(i);
if (!referencesMap.containsKey(key)) {
referencesMap.put(key, i);
}
}
List<Character> chars = new ArrayList<Character>(target.length());
for (int i = 0; i < target.length(); i++) {
chars.add(target.charAt(i));
}
Collections.sort(chars, new Comparator<Character>() {
#Override
public int compare(Character o1, Character o2) {
return referencesMap.get(o1).compareTo(referencesMap.get(o2));
}
});
StringBuilder sb = new StringBuilder();
for (Character c : chars) {
sb.append(c);
}
return sb.toString();
}
In C# I would just use the IComparer Interface and leave it to Array.Sort
void Main()
{
// we defin the IComparer class to define Sort Order
var sortOrder = new SortOrder("dfbcae");
var testOrder = "abcdeeabc".ToCharArray();
// sort the array using Array.Sort
Array.Sort(testOrder, sortOrder);
Console.WriteLine(testOrder.ToString());
}
public class SortOrder : IComparer
{
string sortOrder;
public SortOrder(string sortOrder)
{
this.sortOrder = sortOrder;
}
public int Compare(object obj1, object obj2)
{
var obj1Index = sortOrder.IndexOf((char)obj1);
var obj2Index = sortOrder.IndexOf((char)obj2);
if(obj1Index == -1 || obj2Index == -1)
{
throw new Exception("character not found");
}
if(obj1Index > obj2Index)
{
return 1;
}
else if (obj1Index == obj2Index)
{
return 0;
}
else
{
return -1;
}
}
}

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