Can you split a string in Java without storing what has been split into variables? (Assignment requirement :()
I have tried things which worked on other programming languages however nothing I try seems to work:
(Attempting to see if the second item in a space delimited string (x) is +)
if ((x.split.(" ")).(1) = "+") {
// Do something
}
if ((x.split.(1).(" ")) = "+") {
// Do something
}
Well, what is returned is of type String[]. So if you know that there will be two items, you can reference it as an array..
if(x.split(" ")[1].equals("+"))
Extra Reading
You should look at String Comparison.
String.split returns an array, so this is how it could be done. Note the use of '.equals()'. In Java the == operator checks if the pointer value is the same.
if (x.split.(" ")[1].equals("+")) {
// Do something
}
(And of course this could throw an out of bounds exception if the split wouldn't make an array of size >= 2)
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 have a strange problem when adding a value to a String array which is later involved in an array sort using a hash map. I have a filename XFR900a, and the XFR900 part is added to the array using the following code;
private ArrayList<String> Types = new ArrayList<String>();
...
Types.add(name.substring(0,(name.length() - 1));
System.out.println(name.substring(0,(name.length() - 1));
I even print the line which gives "XFR900", however the array sort later on behaves differently when I use the following code instead;
Types.add("XFR900");
System.out.println(name.substring(0,(name.length() - 1));
which is simply the substring part done manually, very confusing.
Are there any good alternatives to substring, as there must be some odd non ascii character in there?
Phil
UPDATE
Thanks for your comments everyone. Here is some of the code that later compares the string;
for (int i=0;i< matchedArray.size();i++){
//run through the arrays
if (last == matchedArray.get(i)) {
//add arrays to a data array
ArrayList data = new ArrayList();
data.add(matchedArray1.get(i));
data.add(matchedArray2.get(i));
data.add(matchedArray3.get(i));
data.add(matchedArray4.get(i));
data.add(matchedArray5.get(i));
//put into hash map
map.put(matchedArray.get(i), data);
}
else {
//TODO
System.out.println("DO NOT MATCH :" + last + "-" + matchedArray.get(i));
As you can see I have added a test System.out.println("DO NOT MATCH" ... and below is some the output;
DO NOT MATCH :FR99-XFR900
DO NOT MATCH :XFR900-XFR900
I only run the substring on the XFR900a filename. The problem is that for the test line to be printed last != matchedArray.get(i) however they are then the same when printed out to the display.
Phil
You should never use the == operator to compare the content of strings. == checks if it is the same object. Write last.equals(matchedArray.get(i)) instead. The equals() method checks if to object are equal, not if they are the same. In case of String it checks if the two strings consists of the same characters. This might eliminate your strange behaviour.
PS: The behaviour of == on string is a little unpredictable because the java virtual machine does some optimization. If two strings are equal it is possible that the jvm uses the same object for both. This is possible because String objects are immutable anyway. This would explain the difference in behaviour if you write down the substring manually. In the one case the jvm optimizes, in the other it doesn't.
Use .equals() rather than == because they are strings!
if (last.equals(matchedArray.get(i))) {}
Never use == operator if you wanted to check the value since operator will check the Object reference equality, use equals operator which check on the value not the reference i.e. for (int i=0;i< matchedArray.size();i++){
//run through the arrays
if (last.equals(matchedArray.get(i))) { // Line edited
//add arrays to a data array
ArrayList data = new ArrayList();
data.add(matchedArray1.get(i));
data.add(matchedArray2.get(i));
data.add(matchedArray3.get(i));
data.add(matchedArray4.get(i));
data.add(matchedArray5.get(i));
//put into hash map
map.put(matchedArray.get(i), data);
}
else {
//TODO
System.out.println("DO NOT MATCH :" + last + "-" + matchedArray.get(i));
I have an array in containing numbers that represent cable sizes (1, 1.5, 2.5, etc), stored as strings.
In my program, the array is loaded into a spinner, which is working fine.
However, when the item is selected and stored in a variable, I want to check what string was selected, and set another numerical variable to 2.5 so I can do a calculation later in the program.
I tried the following:
if (conductorSize = "1" ) {conCsa = 1;}
else if (conductorSize = "1.5") {conCsa = 1.5;}
conductorSize being the variable holding the selected string, and conCsa being the variable
set to a numerical variable for calculation.
The compiler says that I cannot convert a string to boolean. What's happening?
If you are doing string comparisons, use .equals()
Example taken from here:
String s = "something", t = "maybe something else";
if (s == t) // Legal, but usually WRONG.
if (s.equals(t)) // RIGHT <<<<<<<<<<<<< Use this.
if (s > t) // ILLEGAL
if (s.compareTo(t) > 0) // CORRECT>
As Ed S. points out you are using the assignment operator. However since you are comparing a String you need to use the equals method.
if ("1".equals(conductorSize)) {conCsa = 1;}
else if ("1.5".equals(conductorSize)) {conCsa = 1.5;}
Alternatively, you could just create a new float from your String:
float conCsa;
try {
conCsa = Float.parseFloat(conductorSize);
}catch(NumberFormatException e){
conCsa = 0.0f; //set to a default value
}
It looks like what you're trying to do might better be expressed in this way:
conCsa = Double.parseDouble(conductorSize);
In general you need to use the .equals() method.
If performance is extremely important and you are comparing against string literals, take a look at String.intern(). It'll allow you to do super-fast == comparisons and avoid a full character-by-character scan as in .equals().
Performance would have to be really, really important though, to justify such a non-standard approach.
When you have cable sizes which are constants, you need to use Enums , which will help you in reducing no of if condition comparisons.
I am reading in a csv file in Java and, depending on the format of the string on a given line, I have to do something different with it. The three different formats contained in the csv file are (using random numbers):
833
"79, 869"
"56-57, 568"
If it is just a single number (833), I want to add it to my ArrayList. If it is two numbers separated by a comma and surrounded by quotations ("79, 869)", I want to parse out the first of the two numbers (79) and add it to the ArrayList. If it is three numbers surrounded by quotations (where the first two numbers are separated by a dash, and the third by a comma ["56-57, 568"], then I want to parse out the third number (568) and add it to the ArrayList.
I am having trouble using str.contains() to determine if the string on a given line contains a dash or not. Can anyone offer me some help? Here is what I have so far:
private static void getFile(String filePath) throws java.io.IOException {
BufferedReader reader = new BufferedReader(new FileReader(filePath));
String str;
while ((str = reader.readLine()) != null) {
if(str.endsWith("\"")){
if (str.contains(charDash)){
System.out.println(str);
}
}
}
}
Thanks!
I recommend using the version of indexOf that actually takes a char rather than a string, since this method is much faster. (It is a simple loop, without a nested loop.)
I.e.
if (str.indexOf('-')!=-1) {
System.out.println(str);
}
(Note the single quotes, so this is a char, rather than a string.)
But then you have to split the line and parse the individual values. At present, you are testing if the whole line ends with a quote, which is probably not what you want.
The following code works for me (note: I wrote it with no optimization in mind - it's just for testing purposes):
public static void main(String args[]) {
ArrayList<String> numbers = GetNumbers();
}
private static ArrayList<String> GetNumbers() {
String str1 = "833";
String str2 = "79, 869";
String str3 = "56-57, 568";
ArrayList<String> lines = new ArrayList<String>();
lines.add(str1);
lines.add(str2);
lines.add(str3);
ArrayList<String> numbers = new ArrayList<String>();
for (Iterator<String> s = lines.iterator(); s.hasNext();) {
String thisString = s.next();
if (thisString.contains("-")) {
numbers.add(thisString.substring(thisString.indexOf(",") + 2));
} else if (thisString.contains(",")) {
numbers.add(thisString.substring(0, thisString.indexOf(",")));
} else {
numbers.add(thisString);
}
}
return numbers;
}
Output:
833
79
568
Although it gets a lot of hate these days, I still really like the StringTokenizer for this kind of stuff. You can set it up to return the tokens and, at least to me, it makes the processing trivial without interacting with regexes
you'd have to create it using ",- as your tokens, then just kick it off in a loop.
st=new StringTokenizer(line, "\",-", true);
Then you set up a loop:
while(st.hasNextToken()) {
String token=st.nextToken();
Each case becomes it's own little part of the loop:
// Use punctuation to set flags that tell you how to interpret the numbers.
if(token == "\"") {
isQuoted = !isQuoted;
} else if(token == ",") {
...
} else if(...) {
...
} else { // The punctuation has been dealt with, must be a number group
// Apply flags to determine how to parse this number.
}
I realize that StringTokenizer is outdated now, but I'm not really sure why. Parsing regular expressions can't be faster and the syntax is--well split is a pretty sweet syntax I gotta admit.
I guess if you and everyone you work with is really comfortable with Regular Expressions you could replace that with split and just iterate over the resultant array but I'm not sure how to get split to return the punctuation--probably that "+" thing from other answers but I never trust that some character I'm passing to a regular expression won't do something utterly unexpected.
will
if (str.indexOf(charDash.toString()) > -1){
System.out.println(str);
}
do the trick?
which by the way is fastest than contains... because it implements indexOf
Will this work?
if(str.contains("-")) {
System.out.println(str);
}
I wonder if the charDash variable is not what you are expecting it to be.
I think three regexes would be your best bet - because with a match, you also get the bit you're interested in. I suck at regex, but something along the lines of:
.*\-.*, (.+)
.*, (.+)
and
(.+)
ought to do the trick (in order, because the final pattern matches anything including the first two).
I'm curious why the String.indexOf is returning a 0 (instead of -1) when asking for the index of an empty string within a string.
The Javadocs only say this method returns the index in this string of the specified string, -1 if the string isn't found.
To me this behavior seems highly unexpected, I would have expected a -1. Any ideas why this unexpected behavior is going on? I would at the least think this is worth a note in the method's Javadocs...
System.out.println("FOO".indexOf("")); // outputs 0 wtf!!!
System.out.println("FOO".indexOf("bar")); // outputs -1 as expected
System.out.println("FOO".indexOf("F")); // outputs 0 as expected
System.out.println("".indexOf("")); // outputs 0 as expected, I think
The empty string is everywhere, and nowhere. It is within all strings at all times, permeating the essence of their being, yet as you seek it you shall never catch a glimpse.
How many empty strings can you fit at the beginning of a string? Mu
The student said to the teacher,
Teacher, I believe that I have found the nature of the empty string. The empty string is like a particle of dust, and it floats freely through a string as dust floats freely through the room, glistening in a beam of sunlight.
The teacher responded to the student,
Hmm. A fine notion. Now tell me, where is the dust, and where is the sunlight?
The teacher struck the student with a strap and instructed him to continue his meditation.
Well, if it helps, you can think of "FOO" as "" + "FOO".
int number_of_empty_strings_in_string_named_text = text.length() + 1
All characters are separated by an empty String. Additionally empty String is present at the beginning and at the end.
By using the expression "", you are actually referring to a null string. A null string is an ethereal tag placed on something that exists only to show that there is a lack of anything at this location.
So, by saying "".indexOf( "" ), you are really asking the interpreter:
Where does a string value of null exist in my null string?
It returns a zero, since the null is at the beginning of the non-existent null string.
To add anything to the string would now make it a non-null string... null can be thought of as the absence of everything, even nothing.
Using an algebraic approach, "" is the neutral element of string concatenation: x + "" == x and "" + x == x (although + is non commutative here).
Then it must also be:
x.indexOf ( y ) == i and i != -1
<==> x.substring ( 0, i ) + y + x.substring ( i + y.length () ) == x
when y = "", this holds if i == 0 and x.substring ( 0, 0 ) == "".
I didn't design Java, but I guess mathematicians participated in it...
if we look inside of String implementation for a method "foo".indexOf(""), we arrive at this method:
public int indexOf(String str) {
byte coder = coder();
if (coder == str.coder()) {
return isLatin1() ? StringLatin1.indexOf(value, str.value)
: StringUTF16.indexOf(value, str.value);
}
if (coder == LATIN1) { // str.coder == UTF16
return -1;
}
return StringUTF16.indexOfLatin1(value, str.value);
}
If we look inside of any of the called indexOf(value, str.value) methods we find a condition that says:
if the second parameter (string we are searching for) length is 0 return 0:
public static int indexOf(byte[] value, byte[] str) {
if (str.length == 0) {
return 0;
}
...
This is just defensive coding for an edge case, and it is necessary because in the next method that is called to do actual searching by comparing bytes of the string (string is a byte array) it would otherwise have resulted in an ArrayIndexOutOfBounds exception:
public static int indexOf(byte[] value, int valueCount, byte[] str, int strCount, int fromIndex) {
byte first = str[0];
...
This question is actually two questions:
Why should a string contain the empty string?
Why should the empty string be found specifically at index zero?
Answering #1:
A string contains the empty string in order to be in accordance with Set Theory, according to which:
The empty set is a subset of every set including itself.
This also means that even the empty string contains the empty string, and the following statement proves it:
assert "".indexOf( "" ) == 0;
I am not sure why mathematicians have decided that it should be so, but I am pretty sure they have their reasons, and it appears that these reasons can be explained in layman's terms, as various youtube videos seem to do, (for example, https://www.youtube.com/watch?v=1nBKadtFViM) although I have not actually viewed any of those videos, because #AintNoBodyGotNoTimeFoDat.
Answering #2:
The empty string can be found specifically at index zero of any string, because why not? In other words, if not at index zero, then at which index? Index zero is as good as any other index, and index zero is guaranteed to be a valid index for all strings except for the trifling exception of the empty string.