So I want to search through a string to see if it contains the substring that I'm looking for. This is the algorithm I wrote up:
//Declares the String to be searched
String temp = "Hello World?";
//A String array is created to store the individual
//substrings in temp
String[] array = temp.split(" ");
//Iterates through String array and determines if the
//substring is present
for(String a : array)
{
if(a.equalsIgnoreCase("hello"))
{
System.out.println("Found");
break;
}
System.out.println("Not Found");
}
This algorithm works for "hello" but I don't know how to get it to work for "world" since it has a question mark attached to it.
Thanks for any help!
Take a look:
http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/String.html#contains(java.lang.CharSequence)
String.contains();
To get a containsIgnoreCase(), you'll have to make your searchword and your String toLowerCase().
Take a look at this answer:
How to check if a String contains another String in a case insensitive manner in Java?
return s1.toLowerCase().contains(s2.toLowerCase());
This will also be true for:
war of the worlds, because it will find world. If you don't want this behavior, youll have to change your method like #Bart Kiers said.
Split on the following instead:
"[\\s?.!,]"
which matches any space char, question mark, dot, exclamation or a comma (add more chars if you like).
Or do a temp = temp.toLowerCase() and then temp.contains("world").
You dont have to do this, it's already implemented:
IndexOf and others
You may want to use :
String string = "Hello World?";
boolean b = string.indexOf("Hello") > 0; // true
To ignore case, regular expressions must be used .
b = string.matches("(?i).*Hello.*");
One more variation to ignore case would be :
// To ignore case
b=string.toLowerCase().indexOf("Hello".toLowerCase()) > 0 // true
Related
I have a string which is :
1|name|lastname|email|tel \n
2|name|lastname|email|tel \n
I know that I have to use a loop to display all lines but the problem is that in my assignment
I can't use arrays or other classes than String and System.
Also I would like to sort names by ascending order without using sort method or arrays.
Do I have to use compareTo method to compare two names ?
If that's the case, how do I use compareTo method to sort names.
For example, if compareTo returns 1, that means that the name is greater than the other one. In that case how do I manage the return to sort name properly in the string ?
To display all substrings of the string as in the example, you can just go through all characters one by one and store them in a string. Whenever you hit a delimiter (e.g. | or \n), print the last string.
Here's a thread on iterating through characters of a string in Java:
What is the easiest/best/most correct way to iterate through the characters of a string in Java?
If you also need to sort the names in ascending order without an array, you will need to scan the input many times - sorting N strings takes at least N*log(N) steps. If this is a data structure question, PriorityQueue should do the trick for you - insert all substrings and then pop them out in a sorted fashion :)
building on the previous answer by StoneyKeys, since i do not have the privilege to comment, you can use a simple if statement that when the char is a delimiter, System.out.println() your previous scanned string. Then you can reset the string to an empty string in preparation for scanning the next string.
In java, there are special .equals() operators for strings and chars so when you won't be using == to check strings or char. Do look into that. To reset the value of string just assign it a new value. This is because the original variable points at a certain string ie "YHStan", by making it point at "", we are effectively "resetting" the string. ie scannedstr = "";
Please read the code and understand what each line of code does. The sample code and comments is only for your understanding, not a complete solution.
String str ="";
String value = "YH\nStan";
for (int i=0; i <value.length(); i++) {
char c = value.charAt(i);
String strc = Character.toString(c);
//check if its a delimiter, using a string or char .equals(), if it is print it out and reset the string
if (strc.equals("\n")) {
System.out.println(str);
str ="";
continue; // go to next iteration (you can instead use a else if to replace this)
}
//if its not delimiter append to str
str = str +strc;
//this is to show you how the str is changing as we go through the loop.
System.out.println(str);
}
System.out.println(str); //print out final string result
This gives a result of:
Y
YH
YH
S
St
Sta
Stan
Stan
I'm sure this is fairly simple, however I've tried googling the question but can't find an answer that fits my problem.
I'm playing around with string manipulation and one of the things I'm trying to do is get the first letter of each word. (And then place them all into a string)
I'm having trouble with registering each 'space' so that my If statement will be triggered. Here's what I have so far.
while (scanText.hasNext()) {
boolean isSpace = false;
if (scanText.hasNext(" ")) {isSpace = true;}
String s = scanText.next();
if (isSpace) {firstLetters += s + " ";}
}
Also, if there is a much better way to do this then please let me know
You can also split the original text by spaces, and collect the words.
String input = " Hello world aaa ";
String[] split = input.trim().split("\\s+"); // all types of whitespace; " +" to pick spaces only
// operate on "split" array containing words now: [Hello, world, aaa]
However using regexps here might be overkill.
Assuming that scanText is a Scanner object, you could use something like stated on the documentation:
Scanner s = new Scanner(input).useDelimiter("\\s+"); //regex for spaces
https://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html
I have a serious problem with extracting terms from each string line. To be more specific, I have one csv formatted file which is actually not csv format (it saves all terms into line[0] only)
So, here's just example string line among thousands of string lines:
(split() doesn't work.!!! )
test.csv
"31451 CID005319044 15939353 C8H14O3S2 beta-lipoic acid C1C[S#](=O)S[C##H]1CCCCC(=O)O "
"12232 COD05374044 23439353 C924O3S2 saponin CCCC(=O)O "
"9048 CTD042032 23241 C3HO4O3S2 Berberine [C##H]1CCCCC(=O)O "
I want to extract "beta-lipoic acid" ,"saponin" and "Berberine" only which is located in 5th position.
You can see there are big spaces between terms, so that's why I said 5th position.
In this case, how can I extract terms located in 5th position for each line?
One more thing: the length of whitespace between each of the six terms is not always equal. the length could be one, two, three, four, or five, or something like that.
Because the length of whitespace is random, I can not use the .split() function.
For example, in the first line I would get "beta-lipoic" instead "beta-lipoic acid.**
Here is a solution for your problem using the string split and index of,
import java.util.ArrayList;
public class StringSplit {
public static void main(String[] args) {
String[] seperatedStr = null;
int fourthStrIndex = 0;
String modifiedStr = null, finalStr = null;
ArrayList<String> strList = new ArrayList<String>();
strList.add("31451 CID005319044 15939353 C8H14O3S2 beta-lipoic acid C1C[S#](=O)S[C##H]1CCCCC(=O)O ");
strList.add("12232 COD05374044 23439353 C924O3S2 saponin CCCC(=O)O ");
strList.add("9048 CTD042032 23241 C3HO4O3S2 Berberine [C##H]1CCCCC(=O)O ");
for (String item: strList) {
seperatedStr = item.split("\\s+");
fourthStrIndex = item.indexOf(seperatedStr[3]) + seperatedStr[3].length();
modifiedStr = item.substring(fourthStrIndex, item.length());
finalStr = modifiedStr.substring(0, modifiedStr.indexOf(seperatedStr[seperatedStr.length - 1]));
System.out.println(finalStr.trim());
}
}
}
Output:
beta-lipoic acid
saponin
Berberine
Option 1 : Use spring.split and check for multiple consecutive spaces. Like the code below:
String s[] = str.split("\\s\\s+");
for (String string : s) {
System.out.println(string);
}
Option 2 : Implement your own string split logic by browsing through all the characters. Sample code below (This code is just to give an idea. I didnot test this code.)
public static List<String> getData(String str) {
List<String> list = new ArrayList<>();
String s="";
int count=0;
for(char c : str.toCharArray()){
System.out.println(c);
if (c==' '){
count++;
}else {
s = s+c;
}
if(count>1&&!s.equalsIgnoreCase("")){
list.add(s);
count=0;
s="";
}
}
return list;
}
This would be a relatively easy fix if it weren't for beta-lipoic acid...
Assuming that only spaces/tabs/other whitespace separate terms, you could split on whitespace.
Pattern whitespace = Pattern.compile("\\s+");
String[] terms = whitespace.split(line); // Not 100% sure of syntax here...
// Your desired term should be index 4 of the terms array
While this would work for the majority of your terms, this would also result in you losing the "acid" in "beta-lipoic acid"...
Another hacky solution would be to add in a check for the 6th spot in the array produced by the above code and see if it matches English letters. If so, you can be reasonably confident that the 6th spot is actually part of the same term as the 5th spot, so you can then concatenate those together. This falls apart pretty quickly though if you have terms with >= 3 words. So something like
Pattern possibleEnglishWord = Pattern.compile([[a-zA-Z]*); // Can add dashes and such as needed
if (possibleEnglishWord.matches(line[5])) {
// return line[4].append(line[5]) or something like that
}
Another thing you can try is to replace all groups of spaces with a single space, and then remove everything that isn't made up of just english letters/dashes
line = whitespace.matcher(line).replaceAll("");
Pattern notEnglishWord = Pattern.compile("^[a-zA-Z]*"); // The syntax on this is almost certainly wrong
notEnglishWord.matcher(line).replaceAll("");
Then hopefully the only thing that is left would be the term you're looking for.
Hopefully this helps, but I do admit it's rather convoluted. One of the issues is that it appears that non-term words may have only one space between them, which would fool Option 1 as presented by Hirak... If that weren't the case that option should work.
Oh by the way, if you do end up doing this, put the Pattern declarations outside of any loops. They only need to be created once.
Given a string, I need to print out all permutations of the string. How should I do that? I have tried
for(int i = 0; i<word.length();i++)
{
for(int j='a';j<='z';j++){
word = word.charAt(i)+""+(char)j;
System.out.println(word);
}
}
Is there a good way about doing this?
I'm not 100% sure that I understand what you are trying to do. I'm going to go by your original wording of the question and your comment to #ErstwhileIII's answer, which make me think that it's not really "permutations" (i.e. rearrangement of the letters in the word) that you are looking for, but rather possible single-letter modifications (not sure what a better word for this would be either), like this:
Take a word like "hello" and print a list of all "versions" you can get by adding one "typo" to it:
hello -> aello, bello, cello, ..., zello, hallo, hbllo, hcllo, ..., hzllo, healo, heblo, ...
If that's indeed what you're looking for, the following code will do that for you pretty efficiently:
public void process(String word) {
// Convert word to array of letters
char[] letters = word.toCharArray();
// Run through all positions in the word
for (int pos=0; pos<letters.length; pos++) {
// Run through all letters for the current position
for (char letter='a'; letter<='z'; letter++) {
// Replace the letter
letters[pos] = letter;
// Re-create a string and print it out
System.out.println(new String(letters));
}
// Set the current letter back to what it was
letters[pos] = word.charAt(pos);
}
}
OH .. to print out all permutations of a string, consider your algorithm first. What is the definition of "all permutations" .. for example:
String "a" would have answer a only
String "ab" would have answer: ab, ba
String "abc" would have answer: abc acb, bca, bac, cba, cab
Reflect on the algorithm you would use (write it down in english) .. then translate to Java code
While not the most efficient, a recursive solution might be easiest to use (i.e. for a string of length n, go through each of the characters and follow that with the permutations of the string with that character removed).
EDIT: Ok... you changed your request. Permutations is a whole other story. I think this will help: Generating all permutations of a given string
Not sure what you are trying to do... Example 1 is to get the alphabet one letter next to another. Example 2 is to print whatever you gave us there as an example.
//Example 1
String word=""; //empty string
for(int i = 65; i<=90;i++){ //65-90 are the Ascii numbers for capital letters
word+=(char)i; //cast int to char
}
System.out.println(word);
//Example 2
String word="";
for (int i=65;i<=90;i++){
word+=(char)i+"rse";
if(i!=90){ //you don't want this at the end of your sentence i suppose :)
word+=", ";
}
}
System.out.println(word);
I have this code which searches a string array and returns the result if the input string matches the 1st characters of a string:
for (int i = 0; i < countryCode.length; i++) {
if (textlength <= countryCode[i].length()) {
if (etsearch
.getText()
.toString()
.equalsIgnoreCase(
(String) countryCode[i].subSequence(0,
textlength))) {
text_sort.add(countryCode[i]);
image_sort.add(flag[i]);
condition_sort.add(condition[i]);
}
}
}
But i want to get those string also where the input string matches not only in the first characters but also any where in the string? How to do this?
You have three way to search if an string contain substring or not:
String string = "Test, I am Adam";
// Anywhere in string
b = string.indexOf("I am") > 0; // true if contains
// Anywhere in string
b = string.matches("(?i).*i am.*"); // true if contains but ignore case
// Anywhere in string
b = string.contains("AA") ; // true if contains but ignore case
I have not enough 'reputation points' to reply in the comments, but there is an error in the accepted answer. indexOf() returns -1 when it cannot find the substring, so it should be:
b = string.indexOf("I am") >= 0;
Check out the contains(CharSequence) method
Try this-
etsearch.getText().toString().contains((String) countryCode[i]);
You could use contains. As in:
myString.contains("xxx");
See also: http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/String.html, specifically the contains area.
Use the following:
public boolean contains (CharSequence cs)
Since: API Level 1
Determines if this String contains the sequence of characters in the CharSequence passed.
Parameters
cs the character sequence to search for.
Returns
true if the sequence of characters are contained in this string, otherwise false
Actually, the default Arrayadapter class in android only seems to search from the beginning of whole words but by using the custom Arrayadapter class, you can search for parts of the arbitrary string. I have created the whole custom class and released the library. It is very easy to use. You can check the implementation and usage from here or by clicking on the link provided below.
https://github.com/mohitjha727/Advanced-Search-Filter
You can refer to this simple example-
We have names " Mohit " and "Rohan" and if We put " M" only then Mohit shows up in search result, but when we put "oh" then both Mohit and Rohan show up as they have the common letter 'oh'.
Thank You.