How to check if a string is one of the known values?

Question 1

<?php
$a = 'abc';

if($a among array('are','abc','xyz','lmn'))
    echo 'true';
?>

Suppose I have the code above, how to write the statement "if($a among...)"? Thank you

Question 2

Use the in_array() function.

Manual says:

Searches haystack for needle using loose comparison unless strict is set.

Example:

<?php
$a = 'abc';

if (in_array($a, array('are','abc','xyz','lmn'))) {
    echo "Got abc";
}
?>

Question 3

Here's what worked best for me when trying to script this (in case anyone else comes across this like I did):

$ pecl -d php_suffix=5.6 install <package>
$ pecl uninstall -r <package>

$ pecl -d php_suffix=7.0 install <package>
$ pecl uninstall -r <package>

$ pecl -d php_suffix=7.1 install <package>
$ pecl uninstall -r <package>

The -d php_suffix=<version> piece allows you to set config values at run time vs pre-setting them with pecl config-set. The uninstall -r bit does not actually uninstall it (from the docs):

vagrant@homestead:~$ pecl help uninstall
pecl uninstall [options] [channel/]<package> ...
Uninstalls one or more PEAR packages.  More than one package may be
specified at once.  Prefix with channel name to uninstall from a
channel not in your default channel (pecl.php.net)

Options:
  ...
  -r, --register-only
        do not remove files, only register the packages as not installed
  ...

The uninstall line is necessary otherwise installing it will remove any previously installed version, even if it was for a different PHP version (ex: Installing an extension for PHP 7.0 would remove the 5.6 version if the package was still registered as installed).

Question 4

One option is just to use the regex | character to try to match each of the substrings in the words in your Series s (still using str.contains).

You can construct the regex by joining the words in searchfor with |:

>>> searchfor = ['og', 'at']
>>> s[s.str.contains('|'.join(searchfor))]
0    cat
1    hat
2    dog
3    fog
dtype: object

As @AndyHayden noted in the comments below, take care if your substrings have special characters such as $ and ^ which you want to match literally. These characters have specific meanings in the context of regular expressions and will affect the matching.

You can make your list of substrings safer by escaping non-alphanumeric characters with re.escape:

>>> import re
>>> matches = ['$money', 'x^y']
>>> safe_matches = [re.escape(m) for m in matches]
>>> safe_matches
['\$money', 'x\^y']

The strings with in this new list will match each character literally when used with str.contains.

Question 5

TL;DR For simple substrings contains() is best but for only matching whole words Regular Expression are probably better.

The best way to see which method is more efficient is to test it.

You can use String.contains() instead of String.indexOf() to simplify your non-regexp code.

To search for different words the Regular Expression looks like this:

apple|orange|pear|banana|kiwi

The | works as an OR in Regular Expressions.

My very simple test code looks like this:

public class TestContains {

   private static String containsWord(Set<String> words,String sentence) {
     for (String word : words) {
       if (sentence.contains(word)) {
         return word;
       }
     }

     return null;
   }

   private static String matchesPattern(Pattern p,String sentence) {
     Matcher m = p.matcher(sentence);

     if (m.find()) {
       return m.group();
     }

     return null;
   }

   public static void main(String[] args) {
     Set<String> words = new HashSet<String>();
     words.add("apple");
     words.add("orange");
     words.add("pear");
     words.add("banana");
     words.add("kiwi");

     Pattern p = Pattern.compile("apple|orange|pear|banana|kiwi");

     String noMatch = "The quick brown fox jumps over the lazy dog.";
     String startMatch = "An apple is nice";
     String endMatch = "This is a longer sentence with the match for our fruit at the end: kiwi";

     long start = System.currentTimeMillis();
     int iterations = 10000000;

     for (int i = 0; i < iterations; i++) {
       containsWord(words, noMatch);
       containsWord(words, startMatch);
       containsWord(words, endMatch);
     }

     System.out.println("Contains took " + (System.currentTimeMillis() - start) + "ms");
     start = System.currentTimeMillis();

     for (int i = 0; i < iterations; i++) {
       matchesPattern(p,noMatch);
       matchesPattern(p,startMatch);
       matchesPattern(p,endMatch);
     }

     System.out.println("Regular Expression took " + (System.currentTimeMillis() - start) + "ms");
   }
}

The results I got were as follows:

Contains took 5962ms
Regular Expression took 63475ms

Obviously timings will vary depending on the number of words being searched for and the Strings being searched, but contains() does seem to be ~10 times faster than regular expressions for a simple search like this.

By using Regular Expressions to search for Strings inside another String you're using a sledgehammer to crack a nut so I guess we shouldn't be surprised that it's slower. Save Regular Expressions for when the patterns you want to find are more complex.

One case where you may want to use Regular Expressions is if indexOf() and contains() won't do the job because you only want to match whole words and not just substrings, e.g. you want to match pear but not spears. Regular Expressions handle this case well as they have the concept of word boundaries.

In this case we'd change our pattern to:

b(apple|orange|pear|banana|kiwi)b

The b says to only match the beginning or end of a word and the brackets group the OR expressions together.

Note, when defining this pattern in your code you need to escape the backslashes with another backslash:

 Pattern p = Pattern.compile("\b(apple|orange|pear|banana|kiwi)\b");

Question 6

Never used any of those, but they look interesting..

Take a look at Gearman as well.. more overhead in systems like these but you get other cool stuff :) Guess it depends on your needs ..

How to check if a string is one of the known values?

Answers

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