• 9
name

A PHP Error was encountered

Severity: Notice

Message: Undefined index: userid

Filename: views/question.php

Line Number: 191

Backtrace:

File: /home/prodcxja/public_html/questions/application/views/question.php
Line: 191
Function: _error_handler

File: /home/prodcxja/public_html/questions/application/controllers/Questions.php
Line: 433
Function: view

File: /home/prodcxja/public_html/questions/index.php
Line: 315
Function: require_once

name Punditsdkoslkdosdkoskdo

`elif` in list comprehension conditionals

Can we use elif in list comprehension?

Example :

l = [1, 2, 3, 4, 5]

for values in l:
    if values==1:
        print 'yes'
    elif values==2:
        print 'no'
    else:
        print 'idle'

Can we include the elif in our list comprehension, in a similar fashion to the code above?

For example, an answer like:

['yes', 'no', 'idle', 'idle', 'idle']

Up until now, I have only used if and else in list comprehension.

Python's conditional expressions were designed exactly for this sort of use-case:

>>> l = [1, 2, 3, 4, 5]
>>> ['yes' if v == 1 else 'no' if v == 2 else 'idle' for v in l]
['yes', 'no', 'idle', 'idle', 'idle']

Hope this helps :-)

  • 246
Reply Report
    • There's some interesting history in the syntax. For many years before their introduction "tertiary expressions" were one of the five most-requested changes in the language. Since Guido van Rossum explicitly designed it as a statement-based language, he firmly resisted for a long time (tertiary expressions, and particularly their abuse, are sources of much obscurity in code). When he finally succumbed, he announced he had deliberately chosen a syntax that discouraged overuse. As usual, he did an elegant design job nevertheless.
      • 2
    • While I up-vote this answer, I want to mention this: for 1 pair of if/else is easy to read, 2 pairs: it's getting harder to understand. Don't even mention 3 pairs. If the expression needs 3 or more pairs, a dictionary or a separate function will make things easier to read and understand.
      • 1
    • I stumbled upon a case where I needed an elif, but only two values. Using this example, I would've needed just ['yes', 'no'] to be made. To do this, you can do: ['yes' if v == 1 else 'no' for v in l if values in [1,2]]. I currently can't think of a cleaner way to do this.
>>> d = {1: 'yes', 2: 'no'}
>>> [d.get(x, 'idle') for x in l]
['yes', 'no', 'idle', 'idle', 'idle']
  • 48
Reply Report
      • 1
    • I think this form is a lot easier to digest than trying to do a really long and complicated if/else logic within the list comp
    • @jdi Though conditional-expressions may not be to your taste, they were specifically designed to handle if-elif-elif-else chains, just as the OP requested. They aren't hard to learn and can gracefully handle situations that aren't as amenable to dictionary lookup logic: 'A' if grade>=90 else 'B' if grade>=80 else 'C' if grade>=70 else 'F'.
    • The reason I like the list comprehension better is that it reads just like English. Even a non-programmer would be able to understand what it does. With this solution you have to understand the dict.get() method.

You can, sort of.

Note that when you use sytax like:

['yes' if v == 1 else 'no' for v in l]

You are using the ternary form of the if/else operator (if you're familiar with languages like C, this is like the ?: construct: (v == 1 ? 'yes' : 'no')).

The ternary form of the if/else operator doesn't have an 'elif' built in, but you can simulate it in the 'else' condition:

['yes' if v == 1 else 'no' if v == 2 else 'idle' for v in l]

This is like saying:

for v in l:
    if v == 1 :
        print 'yes'
    else:
        if v == 2:
            print 'no'
        else:
            print 'idle'

So there's no direct 'elif' construct like you asked about, but it can be simulated with nested if/else statements.

  • 24
Reply Report

You can use list comprehension is you are going to create another list from original.

>>> l = [1, 2, 3, 4, 5]
>>> result_map = {1: 'yes', 2: 'no'}
>>> [result_map[x] if x in result_map else 'idle' for x in l]
['yes', 'no', 'idle', 'idle', 'idle']
  • 2
Reply Report

Another easy way is to use conditional list comprehension like this:

l=[1,2,3,4,5]
print [[["no","yes"][v==1],"idle"][v!=1 and v!=2] for v in l]

gives you the correct anwer:

['yes', 'no', 'idle', 'idle', 'idle']

  • 2
Reply Report

Trending Tags