Path: csiph.com!news.swapon.de!eternal-september.org!feeder.eternal-september.org!mx02.eternal-september.org!.POSTED!not-for-mail
From: Nobody <nobody@nowhere.invalid>
Newsgroups: comp.lang.python
Subject: Re: What does a list comprehension do (was: Late-binding of function defaults (was Re: What is a function parameter =[] for?))
Date: Thu, 26 Nov 2015 11:13:44 +0000
Organization: A noiseless patient Spider
Lines: 77
Message-ID: <pan.2015.11.26.11.13.43.440000@nowhere.invalid>
References: <CAPTjJmpwjWnF=d6mpgbKS1biVLoR4APutgyH0n9t6CJ=Kh4dCg@mail.gmail.com> <877fldnm9z.fsf@handshake.de> <mailman.73.1448459485.20593.python-list@python.org>
Mime-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
Injection-Info: mx02.eternal-september.org; posting-host="152d4dee6f95183c956ee6fa63d7e69f"; logging-data="12935"; mail-complaints-to="abuse@eternal-september.org";	posting-account="U2FsdGVkX1+vKZlI8eIRSwpD7xqIpdeX"
User-Agent: Pan/0.14.2 (This is not a psychotic episode. It's a cleansing moment of clarity.)
Cancel-Lock: sha1:t/V2x9dO6KmFzDoAoUODZRwM66k=
Xref: csiph.com comp.lang.python:99555

On Wed, 25 Nov 2015 14:51:23 +0100, Antoon Pardon wrote:

> Am I missing something?

The issue is with lambdas rather than with list comprehensions per se.

Python's lambdas capture free variables by reference, not value.

	> x = 3
	> f = lambda y: x + y
	> f(0)
	3
	> x = 7
	> f(0)
	7

The same issue applies to nested functions:

	> def foo():
	=    x = 3
	=    def f(y):
	=       return x + y
	=    print f(0)
	=    x = 7
	=    print f(0)
	= 
	> foo()
	3
	7

And also to non-nested functions (but most people expect that):

	> x = 3
	> def f(y,x=x):
	=   return x + y
	=
	> print f(0)
	3
	> x=7
	> print f(0)
	3

If you want to capture a variable by value, add a parameter with a default
value using that variable:

	> def foo():
	=    x = 3
	=    def f(y, x=x):
	=       return x + y
	=    print f(0)
	=    x = 7
	=    print f(0)
	= 
	> foo()
	3
	3

This also works for lambdas:

	> x = 3
	> f = lambda y,x=x: x + y
	> f(0)
	3
	> x = 7
	> f(0)
	3

Returning to the original expression:

	> q = [lambda x: i * x for i in range(4)]
	> q[0](1), q[3](1)
	(3, 3)
	> q = [lambda x,i=i: i * x for i in range(4)]
	> q[0](1), q[3](1)
	(0, 3)