Find out how many times a regex matches in a string in Python




Is there a way that I can find out how many matches of a regex are in a string in Python? For example, if I have the string `"It actually happened when it acted out of turn."`


I want to know how many times `"t a"` appears in the string. In that string, `"t a"` appears twice. I want my function to tell me it appeared twice. Is this possible?
The existing solutions based on `findall` are fine for non-overlapping matches (and no doubt optimal except maybe for HUGE number of matches), although alternatives such as `sum(1 for m in re.finditer(thepattern, thestring))` (to avoid ever materializing the list when all you care about is the count) are also quite possible. Somewhat idiosyncratic would be using `subn` and ignoring the resulting string...:



```
def countnonoverlappingrematches(pattern, thestring):
  return re.subn(pattern, '', thestring)[1]

```

the only real advantage of this latter idea would come if you only cared to count (say) up to 100 matches; then, `re.subn(pattern, '', thestring, 100)[1]` might be practical (returning 100 whether there are 100 matches, or 1000, or even larger numbers).


Counting **overlapping** matches requires you to write more code, because the built-in functions in question are all focused on NON-overlapping matches. There's also a problem of definition, e.g, with pattern being `'a+'` and thestring being `'aa'`, would you consider this to be just one match, or three (the first `a`, the second one, both of them), or...?


Assuming for example that you want possibly-overlapping matches *starting at distinct spots in the string* (which then would give TWO matches for the example in the previous paragraph):



```
def countoverlappingdistinct(pattern, thestring):
  total = 0
  start = 0
  there = re.compile(pattern)
  while True:
    mo = there.search(thestring, start)
    if mo is None: return total
    total += 1
    start = 1 + mo.start()

```

Note that you do have to compile the pattern into a RE object in this case: function `re.search` does not accept a `start` argument (starting position for the search) the way *method* `search` does, so you'd have to be slicing thestring as you go -- definitely more effort than just having the next search start at the next possible distinct starting point, which is what I'm doing in this function.
if statements always check if an expression evaluates to the value of true. since you’re checking whether the value true  value true then it’ll run every time. as for the “purpose” of the if statement, it’s here to create a “block scope” within the “function scope”.

interestingly, if you remove … you can use the repmat function here
if you have to print the patter n (here 10) times, use the following code:
a = [5 4 3 2 1 0]';
n = 10;
b =repmat(a, n,1) How you do this will depend on a number of factors: since you haven't tried to explain in any detail, I'll assume you know nothing, and just give you the basics.  

  

The easiest way to do this is to read the text file into a single string, and then use the .NET framework to find the match:  


```
Dim text As String = File.ReadAllText("D:\Temp\MyFile.txt")
Dim index As Integer = text.IndexOf("hello")
If index >= 0 Then
   ' String is in file, starting at character "index"
End If
```
  

There are other ways, but that is pretty much the simplest. for this, just use an if function. see https://macmost.com/learn-how-to-use-the-if-function-in-numbers.html so something like this in a2: if(isblank(b2),c2,b2) the condition is to check if b2 is blank. if it is, get the value of c2. if it isn’t blank, then get the value of b2. cheat engine searches through the memory for values that you input, then locates the memory address of them so that you can change that value later on.

for example:

lets say we would cheat in pinball.

in cheat engine you select pinball as the process.

if your score is 10, you search for 10, when cheat engine is done, change the value in the game (get more / less score), then search for that value, after repeating this a few times, it will have found the address to the value of the score. then you can change it to whatever you want, you can do the same for amounts of balls and so on.

i remember i used to cheat in diablo 2 with it, get infinite stat-points and skillpoints, you could even change the experience to instantly get lvl 99, all games uses values for almost everything, (hp, mana, experience, damage) and so on, just gotta find it in ce so you can change it!

there are tutorials on cheat engines webpage.

http://www.cheatengine.org/tutorials.php Solution:


Hello Gavin,

You have got a nice question. No matter how the question is, don’t forget to ask. Thanks.  
There are a couple of ways to find the number in digits in a number. As you told you just started learning and well-known with the for-loop. I am trying to solve the problem using a simple iteration. Please have a look below program. I hope the program will clarify itself to you as you’ve covered the for-loop topic already.


```
public class DigitsInNumber{
    static int countDigit(long x){ 
        int counter = 0; 
        while (x != 0) { 
            x = x / 10; 
            ++counter; 
        } 
        return counter; 
    }

  	public static void main(String[] args){ 
        long x = 1052809467; 
        System.out.print("Number of digits : " + countDigit(x)); 
    } 
}
```
Thanks again. Good Day! i've been meaning to write a blog post on this for a while, but haven't got round to it yet, so i'll give the condensed version here :slight\_smile:

to understand why the if let syntax is the way it is, we should first look at the let syntax. in most cases, you'll see it used for simple variable dec… The problem is that your pair's closure must return two different types of objects: int if you ask for "con", and the next closure on the line if you ask for "crd".


Here's one possible implementation:


Use a speaker so you can come back.



```
public static Func<string, dynamic> Pair(int x, Func<string, dynamic> y)
{
    Func<string, dynamic> pair =
           (a) =>
           {
               if (a == "con") return x;
               if (a == "crd") return y;
               throw new Exception();
           };

    return pair;
}

public static void Print(Func<string, dynamic> pair)
{
    while (true)
    {
        var next = pair("crd");
        Console.WriteLine((next == null ? "-":"+") +"> " + pair("con"));
        if (next != null)
        {
            Console.WriteLine("|");
            pair = (x) => next(x);
            continue;
        }
        break;
    }
}

        

        
        
        
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```


  




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Test:



```
var pair = Pair(1,
            Pair(2,
                  Pair(3,
                        Pair(4, null))));

Print(pair);

        

        
        
        
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```

output:



```
+> 1
|
+> 2
|
+> 3
|
-> 4

        

        
        
        
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``` > 
> how can I detect when these threads are done with their job and dispose them of their resources?
> 
> 
> 


I think the `ThreadState` property will tell you want you need to know. a pattern with twelve data and twelve greedydata is going to be ridiculously expensive if it fails to match. it may well exceed the 30 second timeout. i would suggest you match multiple patterns to pick out parts of the message

grok {

break\_on\_match => false

match => {

…