Remove empty string elements from string array in Java

Question

I am trying to get the contents of a text file, remove everything other than alphabets and then convert it into an array of Strings for individual processing of words. I do this for getting the text file :

String temp1= IOUtils.toString(FIS,"UTF-8");
String temp2=temp1.replaceAll("[,.!;:\\r\\n]"," ");

And then to tokenize the string, I do this:

String[] tempStringArray = temp2.split(" ");

The problem is that when the array is created, there are empty String at various indices.These empty String are at the position of linebreak, more than one whitespace, replaced punctuation marks, etc in the text file.
I want these empty Strings to be removed from my String array or in a way which they are unable to enter the String array.
How can this be done?

Young Emil · Accepted Answer · 2017-05-21 10:58:03Z

5

Split by all whitespaces like: String[] tempStringArray = temp2.split("\\s+")

answered May 21, 2017 at 10:58

Young Emil

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3 Comments

nayan dhabarde Over a year ago

But this leaves us with and empty "" in the array!

Young Emil Over a year ago

@nayandhabarde, is this true? solution works well, at least for the above question. Maybe yours is a bit different. can you paste you string?

nayan dhabarde Over a year ago

truecaller.blog/2018/01/22/life-as-an-android-engineer try requesting this page html as a response, at the end it has -->\n

Daniel Arthur · Accepted Answer · 2017-05-21 10:57:57Z

2

In your example, if you have more than one character from your character set [,.!;:\r\n] in a row, it will replace it with more than one empty space. When you call the split() method, it then places empty occurrences in the array that refer to the multiple blank spaces in a row.

You can use a regex in the split() method, which will work a lot better for your example.

Try repacing temp2.split(" ") with temp2.split("\\s+"). This will look for multiple spaces in a row, and just tokenise the text around the large gaps of empty space.

answered May 21, 2017 at 10:57

Daniel Arthur

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Comments

Tobias Walter · Accepted Answer · 2017-05-21 11:12:16Z

2

While the answers of Daniel Arthur and Young Millie are correct, one can replace the two steps by directly splitting at the Characters you want to avoid:

String[] tempStringArray = temp1.split("[,.!;:\\s]+");

answered May 21, 2017 at 11:12

Tobias Walter

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1 Comment

Shankha057 Over a year ago

Works as well.Thanks for the answer

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