The corresponding operator is `|`:

     df[(df &lt; 3) | (df == 5)]

would elementwise check if value is less than 3 or equal to 5.

---

If you need a function to do this, we have [**`np.logical_or`**](https://docs.scipy.org/doc/numpy/reference/generated/numpy.logical_or.html). For two conditions, you can use

    df[np.logical_or(df&lt;3, df==5)]

Or, for multiple conditions use the `logical_or.reduce`,

    df[np.logical_or.reduce([df&lt;3, df==5])]

Since the conditions are specified as individual arguments, **parentheses grouping is not needed.**
 
More information on logical operations with pandas can be found [here](https://stackoverflow.com/a/54358361/4909087).

To take the element-wise logical OR of two Series `a` and `b` just do

    a | b

I have a project written in VB, and I need to convert the whole project to C# project. I don&#39;t want to do it file by file, I found some online converters, but they convert only lines of codes, not the whole project. I found on researches, and here was written that 
I saw [this question][1] , there was written that 

&quot;The best part about Sharpdevelop is it converts the whole project (source files) from VB and C# and the other way&quot; 

And I also researched on Google, that if you Right Click on the project, I found [here][2] , that 
&quot;Right click on the project in the Projects window and select Convert - From VB.NET to C#&quot;. But when I right click on The Project in the Projects window, I can&#39;d find that &quot;Convert&quot;. Is there any tool I have to install for SharpDevelop and then I will be able to convert my project ? 
Thanks

  [1]: https://stackoverflow.com/questions/4559950/how-to-convert-full-c-sharp-project-to-vb-net
  [2]: http://community.sharpdevelop.net/forums/t/12434.aspx

How Convert VB Project to C# Project

I need to retrieve the visible height of a div within a scrollable area. I consider myself pretty decent with jQuery, but this is completely throwing me off.

Let&#39;s say I&#39;ve got a red div within a black wrapper:

&lt;img src=&quot;http://i.imgur.com/oiFk3n4.png&quot; /&gt;

In the graphic above, the jQuery function would return 248, the visible portion of the div.

&lt;img src=&quot;http://i.imgur.com/qzHJLZo.png&quot; /&gt;

Once the user scrolls past the top of the div, as in the above graphic, it would report 296.

&lt;img src=&quot;http://i.imgur.com/m8lkMcj.png&quot; /&gt;

Now, once the user has scrolled past the div, it would again report 248.

Obviously my numbers aren&#39;t going to be as consistent and clear as they are in this demo, or I&#39;d just hard code for those numbers.

I have a bit of a theory:

 - Get the height of the window
 - Get the height of the div
 - Get the initial offset of the div from the top of the window
 - Get the offset as the user scrolls.
   - If the offset is positive, it means the top of the div is still visible.
   - if it&#39;s negative, the top of the div has been eclipsed by the window. At this point, the div could either be taking up the whole height of the window, or the bottom of the div could be showing
   - If the bottom of the div is showing, figure out the gap between it and the bottom of the window.

It seems pretty simple, but I just can&#39;t wrap my head around it. I&#39;ll take another crack tomorrow morning; I just figured some of you geniuses might be able to help.

Thanks!

UPDATE: I figured this out on my own, but looks like one of the answers below is more elegant, so I&#39;ll be using that instead. For the curious, here&#39;s what I came up with:

    $(document).ready(function() {
    	var windowHeight = $(window).height();
    	var overviewHeight = $(&quot;#overview&quot;).height();
    	var overviewStaticTop = $(&quot;#overview&quot;).offset().top;
    	var overviewScrollTop = overviewStaticTop - $(window).scrollTop();
    	var overviewStaticBottom = overviewStaticTop + $(&quot;#overview&quot;).height();
    	var overviewScrollBottom = windowHeight - (overviewStaticBottom - $(window).scrollTop());
    	var visibleArea;
    	if ((overviewHeight + overviewScrollTop) &lt; windowHeight) {
    		// alert(&quot;bottom is showing!&quot;);
    		visibleArea = windowHeight - overviewScrollBottom;
    		// alert(visibleArea);
    	} else {
    		if (overviewScrollTop &lt; 0) {
    			// alert(&quot;is full height&quot;);
    			visibleArea = windowHeight;
    			// alert(visibleArea);
    		} else {
    			// alert(&quot;top is showing&quot;);
    			visibleArea = windowHeight - overviewScrollTop;
    			// alert(visibleArea);
    		}
    	}
    });

Get the visible height of a div with jQuery

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

I would like the element-wise logical OR operator. I know "or" itself is not what I am looking for.
I am aware that AND corresponds to <code>&#x26;</code> and NOT, <code>~</code>. But what about OR?
[1]: <a href="https://stackoverflow.com/questions/21415661/logic-operator-for-boolean-indexing-in-pandas" target="_blank" rel="noopener noreferrer">https://stackoverflow.com/questions/21415661/logic-operator-for-boolean-indexing-in-pandas</a> "here"
[2]: <a href="https://stackoverflow.com/questions/15998188/how-can-i-obtain-the-element-wise-logical-not-of-a-pandas-series" target="_blank" rel="noopener noreferrer">https://stackoverflow.com/questions/15998188/how-can-i-obtain-the-element-wise-logical-not-of-a-pandas-series</a>

I am getting an interesting error while trying to use `Unpickler.load()`, here is the source code:

    open(target, &#39;a&#39;).close()
    scores = {};
    with open(target, &quot;rb&quot;) as file:
        unpickler = pickle.Unpickler(file);
        scores = unpickler.load();
        if not isinstance(scores, dict):
            scores = {};

Here is the traceback:

    Traceback (most recent call last):
    File &quot;G:\python\pendu\user_test.py&quot;, line 3, in &lt;module&gt;:
        save_user_points(&quot;Magix&quot;, 30);
    File &quot;G:\python\pendu\user.py&quot;, line 22, in save_user_points:
        scores = unpickler.load();
    EOFError: Ran out of input

The file I am trying to read is empty.
How can I avoid getting this error, and get an empty variable instead?

Why do I get &quot;Pickle - EOFError: Ran out of input&quot; reading an empty file?

In Selenium with Python is it possible to get all the children of a WebElement as a list?

Get all child elements

I have a very simple Django app in order to record the lectures given my colleagues.Since it is quite elementary,I am using the Django admin itself. Here is my models.py:

    #models.py
    from django.db import models

    class Lecture(models.Model):
        topic = models.CharField(max_length=100)
        speaker = models.CharField(max_length=100)
        start_date = models.DateField()
        end_date = models.DateField()

I need to ensure that nobody enters the start date after the end date in the admin forms,so I read the django docs for custom validation in the admin and implemented the following in my admin.py:

    #admin.py
    from models import Lecture
    from django.contrib import admin
    from django import forms


    class LectureForm(forms.ModelForm):
        class Meta:
            model = Lecture

            def clean(self):
                start_date = self.cleaned_data.get(&#39;start_date&#39;)
                end_date = self.cleaned_data.get(&#39;end_date&#39;)
                if start_date &gt; end_date:
                    raise forms.ValidationError(&quot;Dates are incorrect&quot;)
            return self.cleaned_data


    class LectureAdmin(admin.ModelAdmin):
        form = LectureForm
        list_display = (&#39;topic&#39;, &#39;speaker&#39;, &#39;start_date&#39;, &#39;end_date&#39;)

    admin.site.register(Lecture, LectureAdmin)

However,this has no effect whatsoever on my admin and I am able to save lectures where start_date is after end_date as seen in the image:![enter image description here][1]


  [1]: http://i.stack.imgur.com/3pvc7.png

What am I doing wrong ??

Custom validation in Django admin

I do this:

    a = &#39;hello&#39;

And now I just want an independent copy of `a`:

    import copy
 
    b = str(a)
    c = a[:]
    d = a + &#39;&#39;
    e = copy.copy(a)

    map( id, [ a,b,c,d,e ] )

Out[3]:

    [4365576160, 4365576160, 4365576160, 4365576160, 4365576160]

Why do they all have the same memory address and how can I get a copy of `a`?



How can I copy a Python string?

The oodocx module mentioned in the same page refers the user to an /examples folder that does not seem to be there.  
I have read the documentation of python-docx 0.7.2, plus everything I could find in Stackoverflow on the subject, so please believe that I have done my “homework”.  

Python is the only language I know (beginner+, maybe intermediate), so please do not assume any knowledge of C, Unix, xml, etc.

Task : Open a ms-word 2007+ document with a single line of text in it (to keep things simple) and replace any “key” word in Dictionary that occurs in that line of text with its dictionary value.  Then close the document keeping everything else the same.

Line of text (for example) “We shall linger in the chambers of the sea.”

    from docx import Document

    document = Document(&#39;/Users/umityalcin/Desktop/Test.docx&#39;)

    Dictionary = {‘sea’: “ocean”}

    sections = document.sections
    for section in sections:
        print(section.start_type)

    #Now, I would like to navigate, focus on, get to, whatever to the section that has my
    #single line of text and execute a find/replace using the dictionary above.
    #then save the document in the usual way.

    document.save(&#39;/Users/umityalcin/Desktop/Test.docx&#39;)


I am not seeing anything in the documentation that allows me to do this—maybe it is there but I don’t get it because everything is not spelled-out at my level.

I have followed other suggestions on this site and have tried to use earlier versions of the module (https://github.com/mikemaccana/python-docx) that is supposed to have &quot;methods like replace, advReplace&quot; as follows: I open the source-code in the python interpreter, and add the following at the end (this is to avoid clashes with the already installed version 0.7.2):

    document = opendocx(&#39;/Users/umityalcin/Desktop/Test.docx&#39;)
    words = document.xpath(&#39;//w:r&#39;, namespaces=document.nsmap)
    for word in words:
        if word in Dictionary.keys():
            print &quot;found it&quot;, Dictionary[word]
            document = replace(document, word, Dictionary[word])
    savedocx(document, coreprops, appprops, contenttypes, websettings,
        wordrelationships, output, imagefiledict=None) 

Running this produces the following error message:

NameError: name &#39;coreprops&#39; is not defined 

Maybe I am trying to do something that cannot be done—but I would appreciate your help if I am missing something simple.

If this matters, I am using the 64 bit version of Enthought&#39;s Canopy on OSX 10.9.3

How to use python-docx to replace text in a Word document and save

Is there a way to check if a column exists in a Pandas DataFrame?

Suppose that I have the following DataFrame:

    &gt;&gt;&gt; import pandas as pd
    &gt;&gt;&gt; from random import randint
    &gt;&gt;&gt; df = pd.DataFrame({&#39;A&#39;: [randint(1, 9) for x in xrange(10)],
                           &#39;B&#39;: [randint(1, 9)*10 for x in xrange(10)],
                           &#39;C&#39;: [randint(1, 9)*100 for x in xrange(10)]})
    &gt;&gt;&gt; df
       A   B    C
    0  3  40  100
    1  6  30  200
    2  7  70  800
    3  3  50  200
    4  7  50  400
    5  4  10  400
    6  3  70  500
    7  8  30  200
    8  3  40  800
    9  6  60  200

and I want to calculate `df[&#39;sum&#39;] = df[&#39;A&#39;] + df[&#39;C&#39;]`

But first I want to check if `df[&#39;A&#39;]` exists, and if not, I want to calculate `df[&#39;sum&#39;] = df[&#39;B&#39;] + df[&#39;C&#39;]` instead.


How to check if a column exists in Pandas

I have noticed very poor performance when using iterrows from pandas.

Is this something that is experienced by others? Is it specific to iterrows and should this function be avoided for data of a certain size (I&#39;m working with 2-3 million rows)?

[This discussion][1] on GitHub led me to believe it is caused when mixing dtypes in the dataframe, however the simple example below shows it is there even when using one dtype (float64). This takes 36 seconds on my machine:

    import pandas as pd
    import numpy as np
    import time
    
    s1 = np.random.randn(2000000)
    s2 = np.random.randn(2000000)
    dfa = pd.DataFrame({&#39;s1&#39;: s1, &#39;s2&#39;: s2})
    
    start = time.time()
    i=0
    for rowindex, row in dfa.iterrows():
        i+=1
    end = time.time()
    print end - start

Why are vectorized operations like apply so much quicker? I imagine there must be some row by row iteration going on there too. 

I cannot figure out how to not use iterrows in my case (this I&#39;ll save for a future question). Therefore I would appreciate hearing if you have consistently been able to avoid this iteration. I&#39;m making calculations based on data in separate dataframes. Thank you!

---Edit: simplified version of what I want to run has been added below---

    import pandas as pd
    import numpy as np
    
    #%% Create the original tables
    t1 = {&#39;letter&#39;:[&#39;a&#39;,&#39;b&#39;],
          &#39;number1&#39;:[50,-10]}
            
    t2 = {&#39;letter&#39;:[&#39;a&#39;,&#39;a&#39;,&#39;b&#39;,&#39;b&#39;],
          &#39;number2&#39;:[0.2,0.5,0.1,0.4]}
    
    table1 = pd.DataFrame(t1)
    table2 = pd.DataFrame(t2)
    
    #%% Create the body of the new table
    table3 = pd.DataFrame(np.nan, columns=[&#39;letter&#39;,&#39;number2&#39;], index=[0])
    
    #%% Iterate through filtering relevant data, optimizing, returning info
    for row_index, row in table1.iterrows():   
        t2info = table2[table2.letter == row[&#39;letter&#39;]].reset_index()
        table3.ix[row_index,] = optimize(t2info,row[&#39;number1&#39;])
    
    #%% Define optimization
    def optimize(t2info, t1info):
        calculation = []
        for index, r in t2info.iterrows():
            calculation.append(r[&#39;number2&#39;]*t1info)
        maxrow = calculation.index(max(calculation))
        return t2info.ix[maxrow]

  [1]: https://github.com/pydata/pandas/issues/7683

Does pandas iterrows have performance issues?

Say I have a dictionary that looks like this:

    dictionary = {&#39;A&#39; : {&#39;a&#39;: [1,2,3,4,5],
                         &#39;b&#39;: [6,7,8,9,1]},
    
                  &#39;B&#39; : {&#39;a&#39;: [2,3,4,5,6],
                         &#39;b&#39;: [7,8,9,1,2]}}
and I want a dataframe that looks something like this:

 

         A   B
         a b a b
      0  1 6 2 7
      1  2 7 3 8
      2  3 8 4 9
      3  4 9 5 1
      4  5 1 6 2

Is there a convenient way to do this? If I try:

    In [99]:
    
    DataFrame(dictionary)
    
    Out[99]:
         A	             B
    a	[1, 2, 3, 4, 5]	[2, 3, 4, 5, 6]
    b	[6, 7, 8, 9, 1]	[7, 8, 9, 1, 2]

I get a dataframe where each element is a list. What I need is a multiindex where each level corresponds to the keys in the nested dict and the rows corresponding to each element in the list as shown above. I think I can work a very crude solution but I&#39;m hoping there might be something a bit simpler. 

Nested dictionary to multiindex dataframe where dictionary keys are column labels

I&#39;m trying to sort a dataframe by descending.
I put &#39;False&#39; in the ascending argument, but my order is still ascending.

My code is:

    from pandas import DataFrame
    import pandas as pd
    
    d = {&#39;one&#39;:[2,3,1,4,5],
         &#39;two&#39;:[5,4,3,2,1],
         &#39;letter&#39;:[&#39;a&#39;,&#39;a&#39;,&#39;b&#39;,&#39;b&#39;,&#39;c&#39;]}
    
    df = DataFrame(d)
    
    test = df.sort([&#39;one&#39;], ascending=[False])

but the output is

      letter  one  two
    2      b    1    3
    0      a    2    5
    1      a    3    4
    3      b    4    2
    4      c    5    1



python, sort descending dataframe with pandas

I have a dataframe that looks like:

    data = {&#39;index&#39;: [&#39;2014-06-22 10:46:00&#39;, &#39;2014-06-24 19:52:00&#39;, &#39;2014-06-25 17:02:00&#39;, &#39;2014-06-25 17:55:00&#39;, &#39;2014-07-02 11:36:00&#39;, &#39;2014-07-06 12:40:00&#39;, &#39;2014-07-05 12:46:00&#39;, &#39;2014-07-27 15:12:00&#39;],
        &#39;type&#39;: [&#39;A&#39;, &#39;B&#39;, &#39;C&#39;, &#39;A&#39;, &#39;B&#39;, &#39;C&#39;, &#39;A&#39;, &#39;C&#39;],
        &#39;sum_col&#39;: [1, 2, 3, 1, 1, 3, 2, 1]}
    df = pd.DataFrame(data, columns=[&#39;index&#39;, &#39;type&#39;, &#39;sum_col&#39;])
    df[&#39;index&#39;] = pd.to_datetime(df[&#39;index&#39;])
    df = df.set_index(&#39;index&#39;)
    df[&#39;weekofyear&#39;] = df.index.weekofyear
    df[&#39;date&#39;] = df.index.date
    df[&#39;date&#39;] = pd.to_datetime(df[&#39;date&#39;])



    	                 type sum_col weekofyear   date
    index				
    2014-06-22 10:46:00	   A	1	    25	    2014-06-22
    2014-06-24 19:52:00	   B	2    	26   	2014-06-24
    2014-06-25 17:02:00	   C	3    	26	    2014-06-25
    2014-06-25 17:55:00	   A	1    	26    	2014-06-25
    2014-07-02 11:36:00	   B	1    	27    	2014-07-02
    2014-07-06 12:40:00	   C	3    	27    	2014-07-06
    2014-07-05 12:46:00	   A	2    	27	    2014-07-05
    2014-07-27 15:12:00	   C	1    	30	    2014-07-27

I&#39;m looking to groupby the weekofyear, then sum up the sum_col. In addition, I need to find the earliest, and the latest date for the week. The first part is pretty easy:

    gb = df.groupby([&#39;type&#39;, &#39;weekofyear&#39;])
    gb[&#39;sum_col&#39;].agg({&#39;sum_col&#39; : np.sum})

I&#39;ve tried to find the min/max date with this, but haven&#39;t been successful:

    gb = df.groupby([&#39;type&#39;, &#39;weekofyear&#39;])
    gb.agg({&#39;sum_col&#39; : np.sum,
            &#39;date&#39; : np.min,
            &#39;date&#39; : np.max})

How would one find the earliest/latest date that appears?

Max and Min date in pandas groupby

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

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

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

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&#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))?

The short circuiting behaviour of the operators `&amp;&amp;` and `||` is an amazing tool for programmers.

But why do they lose this behaviour when overloaded? I understand that operators are merely syntactic sugar for functions but the operators for `bool` have this behaviour, why should it be restricted to this single type? Is there any technical reasoning behind this?

Is there actually a reason why overloaded &amp;&amp; and || don&#39;t short circuit?

I read the C# Language Specification on the _Conditional logical operators_ `||` and `&amp;&amp;`, also known as the short-circuiting logical operators. To me it seemed unclear if these existed for nullable booleans, i.e. the operand type `Nullable&lt;bool&gt;` (also written `bool?`), so I tried it with non-dynamic typing:

    bool a = true;
    bool? b = null;
    bool? xxxx = b || a;  // compile-time error, || can&#39;t be applied to these types

That seemed to settle the question (I could not understand the specification clearly, but assuming the implementation of the Visual C# compiler was correct, now I knew).

However, I wanted to try with `dynamic` binding as well. So I tried this instead:

    static class Program
    {
      static dynamic A
      {
        get
        {
          Console.WriteLine(&quot;&#39;A&#39; evaluated&quot;);
          return true;
        }
      }
      static dynamic B
      {
        get
        {
          Console.WriteLine(&quot;&#39;B&#39; evaluated&quot;);
          return null;
        }
      }
    
      static void Main()
      {
        dynamic x = A | B;
        Console.WriteLine((object)x);
        dynamic y = A &amp; B;
        Console.WriteLine((object)y);

        dynamic xx = A || B;
        Console.WriteLine((object)xx);
        dynamic yy = A &amp;&amp; B;
        Console.WriteLine((object)yy);
      }
    }

The surprising result is that this runs without exception.

Well, `x` and `y` are not surprising, their declarations lead to both properties being retrieved, and the resulting values are as expected, `x` is `true` and `y` is `null`.

But the evaluation for `xx` of `A || B` lead to no binding-time exception, and only the property `A` was read, not `B`. Why does this happen? As you can tell, we could change the `B` getter to return a crazy object, like `&quot;Hello world&quot;`, and `xx` would still evaluate to `true` without binding-problems...

Evaluating `A &amp;&amp; B` (for `yy`) also leads to no binding-time error. And here both properties are retrieved, of course. Why is this allowed by the run-time binder? If the returned object from `B` is changed to a &quot;bad&quot; object (like a `string`), a binding exception does occur.

___Is this correct behavior?___ (How can you infer that from the spec?)

If you try `B` as first operand, both `B || A` and `B &amp;&amp; A` give runtime binder exception (`B | A` and `B &amp; A` work fine as everything is normal with non-short-circuiting operators `|` and `&amp;`).

(Tried with C# compiler of Visual Studio 2013, and runtime version .NET 4.5.2.)

Do short-circuiting operators || and &amp;&amp; exist for nullable booleans? The RuntimeBinder sometimes thinks so

The logical expression `( a &amp;&amp; b )` *(both `a` and `b` have boolean values)* can be written like `!(!a || !b)`, for example. Doesn&#39;t this mean that `&amp;&amp;` is &quot;unneccesary&quot;? Does this mean that *all* logical expressions can be made only using `||` and `!`? 

Are || and ! operators sufficient to make every possible logical expression?

Is it possible in Handlebars to check if a string is equal to another value without registering a helper? I can&#39;t seem to find anything relevant to this in the Handlebars reference.

For example:

    {{#if sampleString == &quot;This is a string&quot;}}
    ...do something
    {{/if}}

Handlebarsjs check if a string is equal to a value

I&#39;m trying to learn python and came across some code that is nice and short but doesn&#39;t totally make sense

the context was:

    def fn(*args):
        return len(args) and max(args)-min(args)

I get what it&#39;s doing, but why does python do this - ie return the value rather than True/False?

    10 and 7-2

returns 5. Similarly, changing the and to or will result in a change in functionality. So

    10 or 7 - 2

Would return 10.

Is this legit/reliable style, or are there any gotchas on this?


How do &quot;and&quot; and &quot;or&quot; act with non-boolean values?

In pandas, I&#39;d like to create a computed column that&#39;s a boolean operation on two other columns. 

In pandas, it&#39;s easy to add together two numerical columns. I&#39;d like to do something similar with logical operator `AND`. Here&#39;s my first try:

    In [1]: d = pandas.DataFrame([{&#39;foo&#39;:True, &#39;bar&#39;:True}, {&#39;foo&#39;:True, &#39;bar&#39;:False}, {&#39;foo&#39;:False, &#39;bar&#39;:False}])
    
    In [2]: d
    Out[2]: 
         bar    foo
    0   True   True
    1  False   True
    2  False  False

    In [3]: d.bar and d.foo   ## can&#39;t
    ...
    ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

So I guess logical operators don&#39;t work quite the same way as numeric operators in pandas. I tried doing what the error message suggests and using `bool()`:


    In [258]: d.bar.bool() and d.foo.bool()  ## spoiler: this doesn&#39;t work either
    ...
    ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().


I found a way that works by casting the boolean columns to `int`, adding them together and evaluating as a boolean.
    
    In [4]: (d.bar.apply(int) + d.foo.apply(int)) &gt; 0  ## Logical OR
    Out[4]: 
    0     True
    1     True
    2    False
    dtype: bool
    
    In [5]: (d.bar.apply(int) + d.foo.apply(int)) &gt; 1  ## Logical AND
    Out[5]: 
    0     True
    1    False
    2    False
    dtype: bool
    
This is convoluted. Is there a better way?


Content Type	Original Author	Original Content on Stackoverflow
Question	Keith	View Question on Stackoverflow
Solution 1 - Python	deinonychusaur	View Answer on Stackoverflow
Solution 2 - Python	Jonathan Stray	View Answer on Stackoverflow

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