1. Why `TRUE == &quot;TRUE&quot;` is `TRUE` in R?
2. Is there any equivalent for `===` in R?

**Update:**

These are all returning `FALSE`:

    TRUE == &quot;True&quot;
    TRUE == &quot;true&quot;
    TRUE == &quot;T&quot;

The only `TRUE` value is `TRUE == &quot;TRUE&quot;`.

In case of checking with `identical()` everything works fine.

**Second Update:**

By `===` operator I meant the process of checking the _Value_ and the _Data Type_ of a _variable_. In this case I assumed that the `==` operator will only compare the _Values_ of variables, not their _Data Type_ as well.

Why TRUE == &quot;TRUE&quot; is TRUE in R?

I have a pandas `Series` object containing boolean values. How can I get a series containing the logical `NOT` of each value?

For example, consider a series containing:

    True
    True
    True
    False

The series I&#39;d like to get would contain:

    False
    False
    False
    True

This seems like it should be reasonably simple, but apparently I&#39;ve misplaced my mojo =(


How can I obtain the element-wise logical NOT of a pandas Series?

In C# we can use `&amp;&amp;` (boolean and) like this:
			
	int i = 5;
	int ii = 10;
	if(i == 5 &amp;&amp; ii == 10) {
		Console.WriteLine(&quot;i is 5, and ii is 10&quot;);
	}
	Console.ReadKey(true);
		
But try that with python:

    i = 5
    ii = 10
    if i == 5 &amp;&amp; ii == 10:
        print &quot;i is 5 and ii is 10&quot;;      

I get an error: `SyntaxError: invalid syntax`

If I use a single `&amp;`, at least I get no syntax error.  How do I do a boolean `&amp;&amp;` in Python?

How to use boolean &#39;and&#39; in Python

I am filtering rows in a dataframe by values in two columns.

For some reason the OR operator behaves like I would expect AND operator to behave and vice versa.

My test code:

    import pandas as pd
    
    df = pd.DataFrame({&#39;a&#39;: range(5), &#39;b&#39;: range(5) })
    
    # let&#39;s insert some -1 values
    df[&#39;a&#39;][1] = -1
    df[&#39;b&#39;][1] = -1
    df[&#39;a&#39;][3] = -1
    df[&#39;b&#39;][4] = -1
    
    df1 = df[(df.a != -1) &amp; (df.b != -1)]
    df2 = df[(df.a != -1) | (df.b != -1)]
    
    print pd.concat([df, df1, df2], axis=1,
                    keys = [ &#39;original df&#39;, &#39;using AND (&amp;)&#39;, &#39;using OR (|)&#39;,])


And the result:

          original df      using AND (&amp;)      using OR (|)    
                 a  b              a   b             a   b
    0            0  0              0   0             0   0
    1           -1 -1            NaN NaN           NaN NaN
    2            2  2              2   2             2   2
    3           -1  3            NaN NaN            -1   3
    4            4 -1            NaN NaN             4  -1
    
    [5 rows x 6 columns]

As you can see, the `AND` operator drops every row in which at least one value equals `-1`. On the other hand, the `OR` operator requires both values to be equal to `-1` to drop them. I would expect exactly the opposite result. Could anyone explain this behavior, please?

I am using pandas 0.13.1.

pandas: multiple conditions while indexing data frame - unexpected behavior

I&#39;m trying to make a function that will compare multiple variables to an integer and output a string of three letters. I was wondering if there was a way to translate this into Python. So say:

    x = 0
    y = 1
    z = 3
    mylist = []

    if x or y or z == 0 :
    	mylist.append(&quot;c&quot;)
    if x or y or z == 1 :
        mylist.append(&quot;d&quot;)
    if x or y or z == 2 :
        mylist.append(&quot;e&quot;)
    if x or y or z == 3 : 
        mylist.append(&quot;f&quot;)

which would return a list of:

    [&quot;c&quot;, &quot;d&quot;, &quot;f&quot;]

Is something like this possible?
    
    

How to test multiple variables for equality against a single value?

I want to return `True` if and only if 3 out of 4 boolean values are true.

The closest I&#39;ve gotten is [`(x ^ y) ^ (a ^ b)`][1]:

![][2]

What should I do?


  [1]: https://www.wolframalpha.com/input/?i=(x+xor+y)+xor+(a+xor+b)+truth+table
  [2]: http://i.stack.imgur.com/X6XCi.gif

Logic to test that 3 of 4 are True

My question is:

&lt;!-- language: c --&gt;
 
    if (/* condition A */)
    {
        if(/* condition B */)
          {
             /* do action C */
          }
        else
          /* ... */
    }
    else
    {
       /* do action C */
    }

Is it possible to just write the code of action C one time instead of twice?

How to simplify it?

If condition A is matched, condition B needs to be matched in order to do action C

I just want to flip a boolean based on what it already is. If it&#39;s true - make it false. If it&#39;s false - make it true.

Here is my code excerpt:

    switch(wParam) {
    
    case VK_F11:
      if (flipVal == true) {
         flipVal = false;
      } else {
        flipVal = true;
      }
    break;
    
    case VK_F12:
      if (otherVal == true) {
         otherValVal = false;
      } else {
        otherVal = true;
      }
    break;
    
    default:
    break;
    }

Easiest way to flip a boolean value?

I&#39;m someone who writes code just for fun and haven&#39;t really delved into it in either an academic or professional setting, so stuff like these bitwise operators really escapes me.

I was reading an article about JavaScript, which apparently supports bitwise operations. I keep seeing this operation mentioned in places, and I&#39;ve tried reading about to figure out what exactly it is, but I just don&#39;t seem to get it at all. So what are they? Clear examples would be great! :D

Just a few more questions - what are some practical applications of bitwise operations? When might you use them?


What are bitwise operators?

Are `not` and `!` synonyms, or are they evaluated differently?

Difference between &quot;not&quot; and &quot;!&quot; in ruby

An interviewer recently asked me this question: given three boolean variables, a, b, and c, return true if at least two out of the three are true.

My solution follows:

    boolean atLeastTwo(boolean a, boolean b, boolean c) {
        if ((a &amp;&amp; b) || (b &amp;&amp; c) || (a &amp;&amp; c)) {
            return true;
        }
        else{
            return false;
        }
    }

He said that this can be improved further, but how?


Check if at least two out of three booleans are true

I know the rules for `&amp;&amp;` and `||` but what are `&amp;` and `|`? Please explain these to me with an example.

Differences in boolean operators: &amp; vs &amp;&amp; and | vs ||

I was refactoring old code and encountered several IF conditions that were way too complex and long and I&#39;m certain they can be simplified. My guess is that those conditions grew so much because of later modifications.

Anyway, I was wondering if any of you know of a good online simplifier I can use. I&#39;m not interested in any specific language, just a simplifier that would take in for example:

**((A OR B) AND (!B AND C) OR C)**

And give me a simplified version of the expression, if any.

I&#39;ve looked at the other similar questions but none point me to a good simplifier.

Thanks.

Any good boolean expression simplifiers out there?

I&#39;m trying to evaluate `((x == a and y == b) or (x == b and y == a))` in Python, but it seems a bit verbose. Is there a more elegant way?

Is there a more elegant way to express ((x == a and y == b) or (x == b and y == a))?

I have a method which is connecting to a database via Odbc. The stored procedure which I&#39;m 
calling has a return value which from the database side is a &#39;Char&#39;. Right now I&#39;m grabbing
that return value as a string and using it in a simple if statement. I really don&#39;t like the idea
of comparing a string like this when only two values can come back from the database, 0 and 1.

    OdbcCommand fetchCommand = new OdbcCommand(storedProc, conn);

    fetchCommand.CommandType = CommandType.StoredProcedure;
    fetchCommand.Parameters.AddWithValue(&quot;@column &quot;, myCustomParameter);
    fetchCommand.Parameters.Add(&quot;@myReturnValue&quot;, OdbcType.Char, 1)
                .Direction = ParameterDirection.Output;
    fetchCommand.ExecuteNonQuery();

    string returnValue = fetchCommand.Parameters[&quot;@myReturnValue&quot;].Value.ToString();
    if (returnValue == &quot;1&quot;)
    {
	    return true;
    } 

What would be the proper way to handle this situation. I&#39;ve tried &#39;Convert.ToBoolean()&#39; which seemed
like the obvious answer but I ran into the &#39;String was not recognized as a valid Boolean. &#39; exception being thrown. Am I missing something
here, or is there another way to make &#39;1&#39; and &#39;0&#39; act like true and false?

Thanks!

How to convert &quot;0&quot; and &quot;1&quot; to false and true

I would like the element-wise logical OR operator. I know &quot;or&quot; itself is not what I am looking for.

I am aware that AND corresponds to `&amp;` and NOT, `~`. But what about OR? 

  [1]: https://stackoverflow.com/questions/21415661/logic-operator-for-boolean-indexing-in-pandas &quot;here&quot;
  [2]: https://stackoverflow.com/questions/15998188/how-can-i-obtain-the-element-wise-logical-not-of-a-pandas-series

Element-wise logical OR in Pandas

Does Python support short-circuiting in boolean expressions?

Does Python support short-circuiting?

I can&#39;t understand the logic behind the terms *union types* and *intersection types* in TypeScript.

Pragmatically, if the properties of different types are sets of properties, if I combine them with the `&amp;` operator, the resulting type will be the *union* of the of those sets. Following that logic, I would expect types like this to be called *union types*. If I combine them with `|`, I can only use the common properties of them, the *intersection* of the sets.

[Wikipedia](https://en.wikipedia.org/wiki/Boolean_algebra_(structure)#Examples) seems to back that logic:
&gt; The power set (set of all subsets) of any given nonempty set S forms a Boolean algebra, an algebra of sets, with the two operations ∨ := ∪ (union) and ∧ := ∩ (intersection).

However, according to [typescriptlang.org](https://www.typescriptlang.org/docs/handbook/advanced-types.html#intersection-types), it&#39;s exactly the opposite: `&amp;` is used to produce *intersection types* and `|` is used for *union types*.

I&#39;m sure there is another way of looking at it, but I cannot figure it out.

Naming of TypeScript&#39;s union and intersection types

&lt;!-- language: lang-none --&gt;

    In [20]: print None or False
    -------&gt; print(None or False)
    False

    In [21]: print False or None
    -------&gt; print(False or None)
    None


This behaviour confuses me. Could someone explain to me why is this happening like this? I expected them to both behave the same.

False or None vs. None or False

I was playing around in jsfiddle.net and I&#39;m curious as to why this returns true?

    if(0 &lt; 5 &lt; 3) {
        alert(&quot;True&quot;);
    }

So does this:

    if(0 &lt; 5 &lt; 2) {
        alert(&quot;True&quot;);
    }

But this doesn&#39;t:

    if(0 &lt; 5 &lt; 1) {
        alert(&quot;True&quot;);
    }

Is this quirk ever useful?

Why does (0 &lt; 5 &lt; 3) return true?

    -1 == true;        // false
    -1 == false        // false
    -1 ? true : false; // true

Can anyone explain the above output? I know I could work round this by comparing to 0 but I&#39;m interested. I&#39;d expect at least one of the sloppy equals statements to be true as they do implicit type conversion, and I certainly didn&#39;t expect the ternary to come up with a totally different result.

Why are JavaScript negative numbers not always true or false?

I want to write an if/else statement that tests if the value of a text input does NOT equal either one of two different values. Like this (excuse my pseudo-English code):

&lt;pre&gt;
var test = $(&quot;#test&quot;).val();
if (test does not equal A or B){
    do stuff;
}
else {
    do other stuff;
}
&lt;/pre&gt;

How do I write the condition for the if statement on line 2?

All Boolean Logic Solutions on Gang of Coders

Total of 22 Boolean Logic Solutions

Why TRUE == "TRUE" is TRUE in R?

How can I obtain the element-wise logical NOT of a pandas Series?

How to use boolean 'and' in Python

pandas: multiple conditions while indexing data frame - unexpected behavior

How to test multiple variables for equality against a single value?

Logic to test that 3 of 4 are True

If condition A is matched, condition B needs to be matched in order to do action C

Easiest way to flip a boolean value?

What are bitwise operators?

Difference between "not" and "!" in ruby

Check if at least two out of three booleans are true

Differences in boolean operators: & vs && and | vs ||

Any good boolean expression simplifiers out there?

Is there a more elegant way to express ((x == a and y == b) or (x == b and y == a))?

How to convert "0" and "1" to false and true

Element-wise logical OR in Pandas

Does Python support short-circuiting?

Naming of TypeScript's union and intersection types

False or None vs. None or False

Why does (0 < 5 < 3) return true?

Why are JavaScript negative numbers not always true or false?

How do I test if a variable does not equal either of two values?